Fundamentals of Probability With Stochastic Processes, Fourth Edition Saeed Ghahramani Complete stud

SOLUTION MANUAL STUDENTS VERSION

COMPANION FOR CHAPTER 1 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 1 Axioms of Probability 1A 1B

Additional Examples 3 Applications of Probability to Genetics 8 Answers to Odd-Numbered Exercises of Section 1B

3

13

Section 1A

Additional Examples

3

Chapter 1

A xioms of Probability 1A

ADDITIONAL EXAMPLES

Example 1a In a mathematics department of 46 voting faculty members, there are two candidates, Ryan and Nolan, running for the chair position. Assuming that each of the 46 faculty members votes for exactly one of the two candidates, and the candidate with a majority wins, (a) define a sample space for the outcome of the election; (b) describe the event that Nolan receives at least three times as many votes as Ryan.

Solution: (a) Let x be the number of faculty members who vote for Ryan. Then 46 − x is the number of those who vote for Nolan. A sample space is S = {(x, 46 − x) : x = 0, 1, 2, . . . 46}. (b)

The event that Nolan receives at least three times as many votes as Ryan is

E = (0, 46), (1, 45), (2, 44), . . .(11, 35) .

Example 1b Three numbers a, b, and c are selected randomly from the intervals (−1, 1), (0, 1), (−1, 0), respectively. Define a sample space for this experiment and describe the event that the quadratic equation ax 2 + bx + c = 0 has two distinct real roots.

Solution: Let R be the set of all real numbers. Then the sample space is

(a, b, c) ∈ R × R × R : − 1 < a < 1, 0 < b < 1, −1 < c < 0 ,

and the desired event is E = (a, b, c) ∈ S : b2 > 4ac , since the quadratic equation ax2 + bx + c = 0 has two distinct real roots if and only if its discriminant, b 2 − 4ac, is strictly positive.

Example 1c Define a sample space for the experiment of flipping a coin 12 times, and describe the following events: (a)

the fifth flip is heads;

(b)

at least one outcome is heads.

Section 1A

Additional Examples

4

Solution: A sample space for this experiment is

S = ω1 ω2 · · · ω12 : ωi = H or T, 1 ≤ i ≤ 12 .

The event that the fifth flip is heads is

A = ω1 ω2 ω3 ω4 Hω6 · · · ω12 : ωi = H or T, 1 ≤ i ≤ 12, i = 5 .

Let

A1 = Hω2 ω3 · · · ω12 : ωi = H or T, 2 ≤ i ≤ 12 ;

for 2 ≤ j ≤ 11, let

Aj = ω1 ω2 · · · ωj−1 Hωj+1 · · · ω12 : ωi = H or T, 1 ≤ i ≤ 12, i = j ; and let

A12 = ω1 ω2 · · · ω11 H : ωi = H or T, 1 ≤ i ≤ 11 .

The event of at least one heads is 12 j=1 Aj .

Example 1d A farmer decides to build a pen in the shape of a triangle for his chickens.

He sends his son out to cut the lumber and the boy, without taking any thought as to the ultimate purpose, makes two cuts at two points selected at random. (a) Describe a sample space for this experiment. (b) Describe the event that the resulting three pieces of lumber can be used to form a triangular pen.

Solution: (a) Let x be the distance from the left end of the lumber to the leftmost break.

Let y be the distance from the left end

of the lumber to the rightmost break. The sample space is S = (x, y) : 0 < x < y < . (b)

The three pieces x, y − x, and − y form a triangle if and only if x < (y − x) + ( − y) (y − x) < x + ( − y) − y < x + (y − x),

or, equivalently, x<

, 2

y <x+ , 2

and y >

.

2

So the event that the resulting three pieces of lumber can be used to form a triangular pen

. is E = (x, y) : 0 < x < < y < x + 2 2 Example 1e Travis picks up darts to shoot toward an 18 -diameter dartboard aiming at the bullseye on the board. Consider a rectangular coordinate system with origin at the center of the dartboard. The face of the dartboard is divided into 20 equal sectors (wedged-shaped slices) with the positive x-axis being a straight side of the first sector (see Figure 1a). Define a sample space for the point at which a dart hits the board. For this sample space, describe the event that the dart hits a sector that has the positive y-axis as a side.

Section 1A

Figure 1a

Additional Examples

5

Dart Board of Example 1e.

Solution: An appropriate way to define a sample space for this experiment is to use polar coordinates:

S = (r, θ) : 0 ≤ r < 9, 0 ≤ θ < 2π .

The angle of each sector of the board is 2π/20 = π/10. The fifth and sixth sectors of the board are the only sectors that have the positive y-axis as one of their straight sides. So the desired event is

6π 2π 3π 4π ≤θ≤ = (r, θ) : 0 ≤ r < 9, ≤θ≤ .

E = (r, θ) : 0 ≤ r < 9, 10 10 5 5 Example 1f When cell phones were not common, in almost all towns, public pay phones were installed on the sidewalks. Once upon a time, in Berkeley, California, one such a phone was not working properly. It returned coin deposits 50% of the time whether or not the calls went through, it connected 30% of the outgoing calls to wrong numbers, and 23% of the time it took the coin but connected the call to an incorrect number. What is the probability that a call made by this phone was connected for free to the right number?

Solution: Let A be the event that the coin deposited was returned, and B be the event that the call made was connected to the right number. We are interested in P (AB). We have

P (AB) = P (A) + P (B) − P A ∪ B = P (A) + P (B) − 1 − P (Ac B c ) = 0.50 + 0.70 − (1 − 0.23) = 0.43.

Example 1g Show that for three events A, B, and C, the probability that exactly one of them occurs is P (A) + P (B) + P (C) − 2P (AB) − 2P (AC) − 2P (BC) + 3P (ABC).

Section 1A

Additional Examples

6

Solution: The event that exactly one of the three events A, B, or C occurs is

A ∪ B ∪ C − AB ∪ AC ∪ BC).

Since AB ∪ AC ∪ BC ⊆ A ∪ B ∪ C, we have that the desired probability is

P A ∪ B ∪ C − P AB ∪ AC ∪ BC = P (A) + P (B) + P (C) − P (AB) − P (AC) − P (BC) + P (ABC)− P (AB) + P (AC) + P (BC) − P (ABC) − P (ABC) − P (ABC) + P (ABC) = P (A) + P (B) + P (C) − 2P (AB) − 2P (AC) − 2P (BC) + 3P (ABC).

Example 1h In Italy, buses that travel between Rome and Palermo usually stop in a certain village for lunch. Let E, F, and H be the events that a randomly selected entrée from the restaurants of this village is salty, spicy hot, and fatty, respectively.

Suppose that P (E) = 0.08, P (F ) = 0.15, P (H) = 0.17, P E ∪ F = 0.19, P F ∪ H = 0.25, and P (EF H) = 0.02. Lucrezia orders an entrée from a restaurant in this village randomly.

Find the probability that (a) it is not salty; (b) it is salty and spicy hot; (c) it is spicy hot and fatty, but not salty.

Solution: (a) P (E c ) = 1 − 0.08 = 0.92; (b) P (EF ) = P (E) + P (F ) − P E ∪ F = 0.08 + 0.15 − 0.19 = 0.04. (c) To find the desired probability, P (E cF H), note that E cF H = F H − EF H, and EF H ⊂ F H. Thus P (E c F H) = P (F H) − P (EF H).

But

P (F H) = P (F ) + P (H) − P F ∪ H = 0.15 + 0.17 − 0.25 = 0.07.

Therefore, the desired probability is P (E c F H) = 0.07 − 0.02 = 0.05.

Example 1i For events E, F, and G show that (a)

P (EF ) ≥ P (E) − P (F c ).

(b)

P (EF G) ≥ 1 − P (E c ) − P (F c ) − P (Gc ).

Solution: (a)

By Theorem 1.7 of the book, we have that P (E) = P (EF ) + P (EF c ). So P (EF ) = P (E) − P (EF c ).

Now EF c ⊆ F c , so P (EF c ) ≤ P (F c ). Hence P (EF ) = P (E) − P (EF c ) ≥ P (E) − P (F c ).

Section 1A

(b)

Additional Examples

7

By part (a) P (EF G) ≥ P (EF ) − P (Gc ) ≥ P (E) − P (F c ) − P (Gc ) = 1 − P (E c) − P (F c ) − P (Gc ).

Example 1j A coin is tossed infinitely many times. Is the sample space of this experiment countably infinite or uncountably infinite?

Solution: In a toss of the coin, let 1 denote an outcome of heads and 0 denote an outcome of tails. The sample space of the experiment of tossing a coin indefinitely is the set of all sequences of 0’s and 1’s. Clearly, there is a one-to-one correspondence between such a sequence and the numbers of the interval [0, 1] expressed in the binary numeral system; i.e., expressed in the base-2 numeral system. Since [0, 1] is an uncountable set of points, the experiment of tossing a coin infinitely many times has a sample space that is uncountably infinite.

Example 1k A point is chosen at random from the interval (−1, 1). Let E 1 be the event that it falls in the interval (−1, 1/3], E 2 be the event that it falls in the interval (−1, 1/9], in general, for 1 ≤ i < ∞, E i be E3 be the event that it falls in the interval (−1, 1/27] and, ∞ the event that the point is in the interval (−1, 1/3 i]. Find ∞ i=1 Ei and i=1 Ei .

Solution: Clearly, E 1 ⊃ E2 ⊃ E3 ⊃ · · · ⊃ Ei ⊃ · · · . Hence ∞ i=1 Ei = E1 = (−1, 1/3]. Now all the points of the interval (−1, 0] belong to all E i’s, and for any other

point, x, x ∈ (0, 1), there is an i for which x ∈ / (−1, 1/3 i). So ∞ i=1 Ei = (−1, 0].

Section 1B

Applications of Probability to Genetics

8

In the following section, no prior knowledge of genetics is assumed. If it is skipped, then all exercises and examples in this and future companions for various chapters marked “(Genetics)” should be skipped as well.

1B

APPLICATIONS OF PROBABILITY TO GENETICS

In 1854, the Austrian monk Greger Johann Mendel (1822–1884) proposed a mechanism for the inheritance of physical traits. Since then probability has played a central role in the understanding of certain aspects of genetics. In this section, we will examine some elementary applications of probability in the foundations of modern genetics.

Biologists estimate that the human body contains approximately 60 trillion cells. The width of a cell is approximately one-hundredth of a millimeter. Within each cell, there is a large oval structure called the nucleus. In the nucleus, the control center of the cell, are threadlike bodies called chromosomes. Chromosomes contain genes. In the human, there are about 100,000 genes. A gene is the fundamental physical unit of heredity. Genes code for characteristics of the individual and are passed on from one generation to another. For example, a person with green eyes has genes that code for green eyes, and a person with a bent little finger has genes that code for that.

Human beings have 46 chromosomes (23 pairs) in each cell. This number is constant for a particular species. For example, chickens have 36 and peas have 14 chromosomes in each of their cells, respectively. In humans, the reproductive cells are the egg and sperm cells. Except for the reproductive cells, in each human cell, 23 chromosomes come from the father and 23 from the mother. Each member of a pair, is the same length and has the same appearance. It is important to know that the genes that control a certain characteristic occupy the same location on the paired chromosomes. Of all the genetic material on the chromosomes, there are between 20,000 and 50,000 genes that code for all characteristics of the individual. For nonreproductive cells, every cell carries every single gene. Therefore, all of the genetic material of individuals can be found in each and every cell. In each nonreproductive cell, there is one pair of chromosomes called the sex chromosome, which determines a person’s gender. Unlike all other pairs of chromosomes, in males, the sex chromosomes do not look alike. One is called the X chromosome and one is called the Y .

They differ in shape and size. Females have two X chromosomes. Furthermore, in males, the X chromosome has no corresponding genetic information on the Y chromosome. The reproductive cells contain only 23 unpaired chromosomes. This is so that during fertilization each parent can contribute 23 chromosomes to the zygote or fertilized egg. We will discuss sex chromosomes and reproductive cells in Companion for Chapter 3. For now, we concentrate on nonreproductive cells and nonsex chromosomes.

Genes appear in alternate forms. For example, for seed color of the garden pea, the gene appears in one form for yellow and another form for green. Similarly, the gene for the stem height of the garden pea has one form for “tall” and another form for “short.” Alternate forms of genes are called alleles. Therefore, the gene for the seed color of the garden pea has two alleles, the allele for yellow and the allele for green. The gene for the stem height of a garden pea has an allele for tall and an allele for short.

During fertilization, for each trait, each parent contributes only one of a pair of alleles.

That is how the paired conditions of chromosomes and genes, discussed previously, are

Section 1B

Applications of Probability to Genetics

9

transferred from one generation to another. The genes contributed by the parents may or may not be the same allele. For the seed color of the garden pea, during reproduction, both parents might contribute their yellow producing alleles, both might contribute their green producing alleles, or one might contribute its green and the other its yellow producing allele. In the former case, after reproduction, it does not matter which parent contributed the yellow producing allele and which the green producing allele. The result is the same: The offspring has one yellow and one green producing allele. It often happens that when the genes contributed by the parents are different alleles, then, in the offspring, only one allele is fully expressed and the other one is masked. That is, one allele is dominant over the other. In such cases, one can observe the full effect of the dominant allele, but no effect of other alleles can be detected. The allele that is masked is called recessive.

As Mendel did, suppose that a biologist takes a number of pea plants that have inherited one dominant allele for purple flower color from father and one dominant allele for purple flower color from mother. Suppose that the biologist crosses each of these plants with a pea plant that has inherited two recessive alleles for white flower color, one from each parent. Then she will observe that the resulting first generation offspring all will have purple flowers and, as a result of fertilization, the colors will not change to lighter purple. This is an observation that proves that inheritance is not necessarily a blending phenomenon. At this stage, one might think that the absence of white flowers among the first generation offspring indicates that its heritable factor is annihilated. However, if the biologist continues her experiment and crosses the new generation of the plants between each other, she will observe that approximately 1/4th of their offspring will have white flowers. This phenomenon demonstrates the presence of the recessive white flower allele in the first generation of the plants.

The organism’s genetic constitution (makeup); that is, the set of genes that it possesses, is its genotype. Thus, if in an organism, an inheritable characteristic is controlled by a single pair of genes, one from each parent, then the pair is the genotype of that characteristic. To discuss applications of probability to genetics, we use uppercase letters to represent dominant alleles and lowercase to represent recessive alleles. For the pea plants, let A represent the dominant allele for purple flowers and a represent the recessive allele for white flowers. Then each of the pea plants in the experiment above belongs to one of the three genotypes AA, Aa, aa. Recall that Aa and aA are identical. They both mean that the pea plant has inherited a dominant purple flower allele from one parent and a recessive white flower allele from another parent. AA and aa are defined similarly. For Mendel’s experiments with pea plants, to determine all the possible genotypes of offsprings, the possible results for the first set of experiments is conveniently represented in the following table: a a

A aA aA

A aA aA

This table shows that the only outcome is aA, and since A is dominant, the probability that all offsprings have only purple flower is 1. In the next stage, these offsprings are crossed with each other. That is, pea plants of genotype aA are crossed with pea plants of genotype aA. The possible results are given by the following table.

Section 1B

a A

a aa Aa

Applications of Probability to Genetics

10

A aA AA

Since all four genotypes in the center of this table are equally likely and aA and Aa are the same, the probability of genotypes aa, aA, and AA are 1/4, 1/2, and 1/4, respectively. This shows that the probability is 1/4 that a second generation offspring has white flowers.

Example 1 Suppose that lima bean plants with green pods were crossed with lima bean plants with yellow pods. Furthermore, suppose that the parents of all the plants crossed came from pure lines. Therefore, each plant crossed had identical alleles. If all these crossings yielded 420 offsprings of which 100 had yellow pods and 320 had green pods, which allele is dominant, the allele for green pods or the allele for yellow pods?

Solution: Intuitively, it should be clear that the allele for a green pod is dominant. To show this mathematically, let A represent the dominant allele and a represent the recessive allele. Since the parents of all the plants crossed came from pure lines, as the following table shows, each lima bean plant crossed belonged to the genotype aA. a a

A aA aA

A aA aA

Therefore, by the following table, each of the 420 offsprings belongs to one of the genotypes aa, aA, and AA. a A

a aa Aa

A aA AA

Since a is the recessive allele, by this table, with probability 1/4, an offspring is a lima bean that has inherited two recessive alleles. Therefore, approximately 1/4th of the 420 offsprings; that is, approximately 105 of the plants should be lima bean plants that inherited two recessive alleles for their pod color. Since, in the experience, of the 420 offsprings, 100 had yellow pods and 320 had green pods, and 100 is statistically close enough to 105, the recessive allele is the allele for yellow pod, and hence the dominant one is the allele for green pod.

Remark In this exercise, the fact that, approximately 1/4th of the 420 offsprings should be lima bean plants that inherited two recessive alleles for their pod color should be intuitively clear. However, it is mathematically justified by the formula for the expected value of binomial random variables, which is discussed in Section 5.1 of the book.

Example 1m In human beings, eye colors and existence or nonexistence of dimples are physical attributes that have an influence on our perception of facial beauty. As for eye colors, brown eyes (B) are dominant over both green eyes (G) and blue eyes (b). However, whenever there is no allele for brown eyes, but there are alleles for green and blue eyes, green is dominant over blue. When it comes to dimples, people with dimples have dominant

Section 1B

Applications of Probability to Genetics

11

alleles (D) over those without dimples who have recessive alleles (d). In this example, we are interested in two characteristics transferred from one generation to another. Namely, eye color and presence or absence of dimples. Suppose that a man with genotype Bbdd marries a woman with genotype bbDd. Find the probability that a child of theirs (a) has dimples and brown eyes; (b) has no dimples, but brown eyes; (c) has blue eyes.

Solution: The child inherits Bd or bd from the father with equal probabilities, and bD or bd from the mother also with equal probabilities. So the child belongs to one of the four genotypes in the following table with equal probabilities. bD bd

Bd BbDd Bbdd

bd bbDd bbdd

This table shows that the child has (a) dimples and brown eyes with probability 1/4; (b) no dimples and brown eyes with probability 1/4; and (c) blue eyes with probability 1/2.

EXERCISES FOR SECTION 1B

1.

For the shape of a pea seed, the allele for a round shape, denoted by R, is dominant to the allele for a wrinkled shape, denoted by r. A pea plant with a round seed shape is crossed with a pea plant with a wrinkled seed shape. If half of the offspring have round seed shapes and half have wrinkled seed shapes, what were the genotypes of the parents?

2.

In a study, 200 children of the parents who all have an allele for a cleft chin and an allele for a smooth chin are considered. If only 48 of them have a smooth chin, show that the gene for a cleft chin must be dominant over the gene for a smooth chin.

3.

In human beings, the gene for naturally curly hair (H) is dominant over the gene for straight hair (h), and the gene for right-handedness (R) is dominant over the gene for left-handedness (r). Suppose that a man with genotype Hhrr marries a woman with genotype hhRr. Find the probability that a child of theirs (a) is right-handed with curly hair; (b) is right-handed; (c) has straight hair.

4.

In human beings, unattached ear lobes (E) are dominant over ear lobes that are attached to the sides of the head (e). The ability to roll the lateral edge of the tongue together (T ) is a dominant trait over the inability to do so (t). As for dimples, people with dimples have dominant alleles D over those without dimples having recessive alleles d. Suppose that a man with genotype EettDd marries a woman with genotype Eettdd. Find the probability that a child of theirs has dimples and unattached ear lobes.

5.

A hitchhiker’s thumb is a thumb that can bend backwards when extended. In human beings, no-hitchhiker’s thumb (H) is dominant over hitchhiker’s thumb (h), face freckles (F ) are dominant over no face freckles (f ), and brown eye color (B)

Section 1B

Applications of Probability to Genetics

12

is dominant over blue eye color (b). Suppose that a man with genotype Hhf f Bb marries a woman with genotype HhF f bb. Find the probability that a child of theirs has (a) blue eyes and hitchhiker’s thumb; (b) blue eyes and no-hitchhiker’s thumb; (c) blue eyes, hitchhiker’s thumb, and face freckles.

Answers to Odd-Numbered Exercises of Section 1B

Answers to Odd-Numbered Exercises of Section 1B 1. The genotypes of the parents were rr and Rr. 3. The answers to parts (a), (b), and (c) are 1/4, 1/2, and 1/2, respectively.

5. The answers to parts (a), (b), and (c) are 1/8, 3/8, and 3/16, respectively.

13

COMPANION FOR CHAPTER 2 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 2 Combinatorial Methods 2A

Additional Examples

3

3

Section 2A

Additional Examples

3

Chapter 2

C ombinatorial Methods 2A

ADDITIONAL EXAMPLES

Example 2a One of Olivia’s iPod playlists has only 12 songs, 6 by Lionel Richie and 6

by Adele. She sets the ipod to play the songs of that playlist. However, before doing that, Olivia unchecks and then rechecks the shuffle button of her iPod, and the iPod plays the playlist’s songs in a random order, without repeating a song. What is the probability that no two songs of the same singer will be played back to back?

Solution: There are 12! ways for the songs to be played. In order that no two songs of the singers be played back to back, the 1st, 3rd, 5th, 7th, 9th, and 11th songs the iPod will play should be Lionel Richie songs, and the 2nd, 4th, 6th, 8th, 10th, and 12th songs should be Adele songs, or vice versa. So the desired probability is 1 2 × 6! × 6! = ≈ 0.00216.

12!

462 Example 2b An 8-letter code is made by choosing distinct letters from the set of the English alphabet {a, b, c, . . . , z}. What is the probability that the letters a and b do not show up consecutively in either order?

Solution: The total number of possible 8-letter codes with distinct letters is 26 P8 . Now we will count the number of codes in which the letters a and b show up consecutively in either order. When forming 8-letter codes, there are 7 ways in which b can immediately follow a, and 7 ways in which a can immediately follow b. For each of these 14 cases, there are 24 P6 possibilities for the remaining 6 letters of the 8-letter random code. So a and b can appear consecutively in either order in 14 × 24 P6 ways. Hence the desired probability is 1−

318 14 × 24 P6 ≈ 0.98. = 325 26 P8

Example 2c An old lady claims that she can predict the sex of a baby by just looking at the shape of the face of the pregnant mother. At a Lamaze class, the lady meets 7 pregnant women and predicts the sexes of their children. Assuming that she does not have the ability that she claims she has, what is the probability that she predicts only 3 of the 7 sexes of the babies correctly?

Solution: Suppose that by CCCIIII we mean that the lady predicts the sexes of the first 3 pregnant women she met correctly and the sexes of the remaining 4 pregnant ladies

Section 2A

Additional Examples

4

incorrectly. We represent other cases similarly. The number of possibilities in which she is correct in exactly 3 cases is equal to the number of distinguishable sequences with three C’s and four I’s. So the probability that she predicts only 3 of the 7 sexes of the babies 7! 7 2 ≈ 0.273. correctly is 3! 4!

Example 2d In an international college, jewelry design is a very popular course, and Chaowei Yuan and 6 other Chinese students, 8 American, and 3 Indians are trying to take the course. However, there is room only for 5 more students to enroll. If the professor teaching the course chooses 5 of these students randomly and enrolls them, what is the probability that (a) Chaowei Yuan and exactly 2 of the 3 Indians are included? (b) Only one of the 8 American students is included?

3 14 13 2 2 = ≈ 0.0319.

Solution: (a) 18 408 5

8 10 10 1 4 = ≈ 0.196. (a) 18 51 5 Example 2e In the Congress of the United States of America, regardless of population or size, each of the 50 states is represented by two senators. For an interview concerning an important pending legislation, a TV station chooses 8 senators at random for interview.

What is the probability that no two of the selected senators are from the same state.

100 Solution: The TV station has choices for 8 senators to interview. If we want no 8 two senators to be from 8 states from the 50, for the same state, first we need to choose

50 50 which we have choices. Then for each of these choices, we have 28 ways to 8 8 choose one senator from each of the chosen 8 states. So the desired probability is

50 8 2 8

≈ 0.739.

100 8 Example 2f In a medical study, 25 patients with a duodenal ulcer have volunteered to be treated by an experimental drug. The researcher has decided to treat 18 of the patients, randomly selected, by the new drug and to give the remaining 7 a placebo as control. If the duodenal ulcers of 17 of these patients are caused by the bacterium H. pylori, and the rest by nonsteroidal anti-inflammatory drugs such as aspirin, what is the probability that the duodenal ulcers of at least 13 of the experiment group (the patients treated by the drug) are caused by the bacterium H. pylori?

Section 2A

Solution: The answer is

17

i=13

Additional Examples

5

17 8 859 i 18 − i

≈ 0.393. = 25 2185 18

Example 2g In the spring semester of 2016, Zelma asked her probability professor why she had never been included in a presidential poll conducted by media. The estimated population of the United States, at that time, was 323,000,000, of which approximately 219,000,000 were eligible to vote (those who were at least 18 years old). The professor calculated the probability that Zelma be included in a random sample of 1000, which was the average size of a poll:

1 219, 000, 000 − 1 1 1 999

≈ 0.0000046. = 219, 000, 000 219, 000 1000 Based on this, he concluded that the chances were extremely small for Zelma to be included in a presidential random sample of size 1000.

Example 2h By mistake, a student who is running a computer program enters with negative signs two of the six positive numbers and with positive signs two of the four negative numbers. If at some stage the program chooses three distinct numbers from these 10 at random and multiplies them, what is the probability that at that stage no mistake will occur?

6 6 6 6 2 2 + + + 3 1 1 1 1 1

= 0.467.

Solution: 10 3

6 6 4 + 3 1 2

= 0.467.

Another Solution: 10 3 Example 2i According to the 1998 edition of Encyclopedia Britannica, “there are at least 15,000 to as many as 35,000 species of orchids.” These species have been found naturally and are distinct from each other. Suppose that hybrids can be created by crossing any two existing species. Furthermore, suppose that hybrids themselves can be continued to be hybridized with each other or with an original species. Orchid lovers develop thousands and thousands of hybrids for use as garden or greenhouse ornamental and for the commercial flower trade. Suppose that all species are crossed, two at a time, to create the first generation of hybrids. Then the first generation of hybrids are crossed with each other and with the original species, two at a time, to develop the second generation of hybrids. The second generation of hybrids are crossed with each other, with the first generation hybrids, and with the original species, two at a time, to generate the third generation of hybrids, and so on. Let n be the total number of original species of orchids. Let n k be the number of hybrids in the kth generation.

Section 2A

Additional Examples

6

(a)

Find nk in terms of n, n1 , . . . , nk−1 .

(b)

For n = 25, 000, find the largest possible total number of all hybrids in the first four generations.

Solution: (a)

(b)

It must be clear that

n n1 = 2

n1 + nn1 n2 = 2

n2 + n2 (n + n1 ) n3 = 2

n3 + n3 (n + n1 + n2 ) n4 = 2 .. .

nk−1 + nk−1 (n + n1 + · · · + nk−1 ). nk = 2

For n = 25, 000, successive calculations of n k ’s yield, n1 = 312, 487, 500, n2 = 48, 832, 030, 859, 381, 250, n3 = 1, 192, 283, 634, 186, 401, 370, 231, 933, 886, 715, 625, n4 = 710, 770, 132, 174, 366, 339, 321, 713, 883, 042, 336, 781, 236, 550, 151, 462, 446, 793, 456, 831, 056, 250.

For n = 25, 000, the total number of all possible hybrids in the first four generations, n1 +n2 +n3 +n4 , is 710,770,132,174,366,339,321,713,883,042,337,973,520,184,337, 863,865,857,421,889,665,625. This number is approximately 710 × 10 63 .

Example 2j In Maryland’s lottery, players pick six different integers between 1 and 49, the order of selection being irrelevant. The lottery commission then randomly selects six of these as the winning numbers. What is the probability that at least two consecutive integers are selected among the winning numbers?

Hint: Let A be the set of all 6-element combinations of 1, 2, . . . , 49 with no consecutive integers. Let B be the set of all 6-element combinations of 1, 2, . . ., 44 . Begin by showing that there is a one-to-one correspondence between A and B. Solution: Let a 6-element combination of a set of integers be denoted by {a 1 , a2 , . . . , a6 }, where a1 < a2 < · · · < a6 . It can be easily verified that the function h : B → A defined by

h {a1 , a2 , . . . , a6 } = {a1 , a2 + 1, . . ., a6 + 5}

Section 2A

Additional Examples

7

is one-to-one and onto. Therefore, there is a one-to-one correspondence between B and

44 A . This shows that the number of elements in A is . Thus the probability that no 6

44 49 consecutive integers are selected among the winning numbers is ≈ 0.505.

6 6 This implies that the probability of at least two consecutive integers among the winning numbers is approximately 1−0.505 = 0.495. Given that there are 47 integers between 1 and 49, this high probability might be counter-intuitive. Even without knowledge of expected value, a keen student might observe that, on the average, there should be (49 − 1)/7 = 6.86 numbers between each ai and ai+1 , 1 ≤ i ≤ 5. Thus he or she might erroneously think that it is unlikely to obtain consecutive integers frequently.

Example 2k In how many ways can 10 different photographs be placed in six different envelopes, no envelope remaining empty?

Hint: An easy way to do this problem is to use the following version of the inclusionexclusion principle: Let A 1 , A2 , . . ., An be n subsets of a finite set Ω with N elements. Let N (Ai) be the number of elements of Ai , 1 ≤ i ≤ n, and N (Aci ) = N − N (Ai ). Let Sk be the sum of the elements of all those intersections of A 1 , A2 , . . ., An that are formed of exactly k sets. That is, S1 = N (A1 ) + N (A2 ) + · · · + N (An ), S2 = N (A1 A2 ) + N (A1 A3 ) + · · · + N (An−1 An ), and so on. Then N (Ac1 Ac2 · · · Acn ) = N − S1 + S2 − S3 + · · · + (−1)n Sn .

Solution: Let N be the number of ways that 10 different photographs can be placed in six different envelopes, allowing for the possibility of empty envelopes. Let A i be the set of c c all situations in which envelope i is empty. Then the desired quantity is N (A c1 A 2 · · · A6 ).

6 equal Clearly, N = 610 , N (Ai) = 510 , N (AiAj ) = 410 , i = j, and so on. So S1 has 1

6 equal terms, and so on. Therefore, the solution is terms, S2 has 2

6 10 6 10 6 10 6 10 6 10 6 10 10 5 + 4 − 3 + 2 − 1 + 0 = 16, 435, 440.

6 − 1 2 3 4 5 6 Example 2 We are given n (n > 5) points in space, no three of which lie on the same straight line. Let Ω be the family of planes defined by any three of these points. Suppose that the points are situated in a way that no four of them are coplanar, and no two planes of Ω are parallel. From the set of the lines of the intersections of the planes of Ω, a line is selected at random. What is the probability that it passes through none of the n points?

Hint: For i = 0, 1, 2, let A i be the set of all lines of the intersections that are determined

Section 2A

Additional Examples

8

by planes having i of the given n points in common. If |A i | denotes the number of elements of Ai , the answer is |A0 |/ |A0 | + |A1 | + |A2 | .

Solution: For i = 0, 1, 2, let A i be the set of all lines of the intersections that are deterthe number mined by planes having i of the given n points in common. If |A i | denotes

1 n n−3 , of elements of Ai , the answer is |A0 |/ |A0 | + |A1 | + |A2 | . |A0 | = 2 3 3

1 n 3 n−3 1 n 3 n−3 , |A2 | = . |A1 | = 2 3 1 2 2 3 2 1 The answer is

(n − 4)(n − 5) |A0 | = . |A0 | + |A1 | + |A2 | n2 + 2

COMPANION FOR CHAPTER 3 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 3 Conditional Probability and Independence 3A 3B

Additional Examples 3 More on Applications of Probability to Genetics 10 Hardy-Weinberg Law 12 Sex-Linked Genes 14 Answers to Odd-Numbered Exercises of Section 3B 18

3

Section 3A

Additional Examples

3

Chapter 3

C onditional Probability and I ndependence 3A

ADDITIONAL EXAMPLES

Example 3a What is the probability that among n unrelated people, selected at random, no two have the same birthday? Assume that the birth rates are constant throughout the year and that each year has 365 days.

Solution: The probability that n unrelated randomly selected people have n different birthdays is

2 3 n − 1 1 1− 1− ··· 1− .

1− 365 365 365 365 Example 3b At a party, there are three coolers. The first one has 10 regular colas, 12 diet colas, 5 orange drinks, and 7 ginger ales. The second one has 13 diet colas and 12 ginger ales. The third cooler has 15 regular colas, 5 ginger ales, and 8 orange drinks. Mali chooses a cooler at random and draws a drink from the cooler without looking. What is the probability that she draws a diet cola?

Solution: For 1 ≤ i ≤ 3, let A i be the event that Mali chooses the ith cooler. Let D be the event that she draws a diet cola. The desired probability is P (D) = P (D | A1 )P (A1 ) + P (D | A2 )P (A2 ) + P (D | A3 )P (A3 ) =

1 371 12 1 13 1 · + · +0· = ≈ 0.29.

34 3 25 3 3 1275

Example 3c Suppose that for infectious diseases, when a patient is tested for a specific disease, the result is not 100% reliable, and 7% of the time it is false negative. That is, a patient that has the disease is not actually diagnosed with it. However, if the patient does not have the disease, the test is 100% reliable. Fontana is experiencing common symptoms of three infectious diseases A, B, and C. Her doctored tested her for disease A, and the result was negative. If the chances are 60%, 27% and 13% that patients with those symptoms have disease A, B, and C, respectively, what is the probability that Fontana has disease A? Assume that no one gets infected with more than one of these infectious diseases.

Section 3A

Additional Examples

4

Solution: Let E, F and G be the events that Fontana has disease A, B, and C, respectively.

Let H be the event that she has tested negative for disease A. The desired probability is P (E | H) = =

P (H | E)P (E) P (H | E)P (E) + P (H | F )P (F ) + P (H | G)P (G) (0.07)(0.60) = 0.095.

(0.07)(0.60) + 1 · (0.27) + 1 · (0.13)

Example 3d A number is selected at random from the set {1, 2, . . . , 20}. Then a second number is selected randomly between 1 and the first number. If the second number is 5, what is the probability the first number is 15?

Solution: For 1 ≤ i ≤ 20, let A i be the event that the first number selected is i. Let F be the event that the second number is 5. By Bayes’ Theorem, 1 1 · P (F | A15 )P (A15 ) = 2015 20 ≈ 0.044.

P (A15 | F ) = 20

1 1 · P (F | Ai )P (Ai ) i 20 i=1

i=5

Example 3e For its cars, an automobile assembly manufacturing plant purchases engines from two different auto engine manufactures: 35% from Wilbraham Motors and 65% from Antonio Engines. Even though the warranties offered by both companies are identical, experience shows that only 87% of the engines manufactured by Wilbraham Motors last at least for 150,000 miles whereas 96% of those manufactured by Antonio Engines last for that many miles. (a) Find the probability that the engine of a randomly selected car assembled by this manufacturing plant lasts at least 150,000 miles. (b) If the engine of a random car manufactured by this plant failed before functioning for 150,000 miles, what is the probability that it was manufactured by Wilbraham Motors?

Solution: (a)

Let A be the event that the engine of the random car manufactured by this plant lasts at least 150,000 miles. Let B be the event that the engine of the randomly selected car was manufactured by Wilbraham Motors. We have P (A) = P (A | B)P (B) + P (A | B c )P (B c ) = (0.87)(0.35) + (0.96)(0.65) ≈ 0.93.

(b) P (B | Ac ) = =

P (Ac | B)P (B) P (Ac | B)P (B) + P (Ac | B c )P (B c ) (1 − 0.87)(0.35) ≈ 0.64.

(1 − 0.87)(0.35) + (1 − 0.96)(0.65)

Section 3A

Additional Examples

5

Example 3f Solve the following problem, asked of Marilyn Vos Savant in the “Ask Marilyn” column of Parade Magazine, January 3, 1999.

You’re at a party with 199 other guests when robbers break in and announce that they are going to rob one of you. They put 199 blank pieces of paper in a hat, plus one marked “you lose.” Each guest must draw, and the person who draws “you lose” will get robbed. The robbers offer you the option of drawing first, last, or at any time in between. When would you take your turn?

Assume that those who draw earlier will not announce whether they drew the “you lose” paper.

Solution: Let A i be the event that the ith person draws the “you lose” paper. Clearly, P (A1 ) =

1 , 200

1 199 1 · = , 200 199 200 1 199 198 1 · · = , P (A3 ) = P (Ac1 Ac2 A3 ) = P (Ac1 )P (Ac2 | Ac1 )P (A3 | Ac1 Ac2 ) = 200 199 198 200 P (A2 ) = P (Ac1 A2 ) = P (Ac1 )P (A2 | Ac1 ) =

and so on. Therefore, P (Ai ) = 1/200 for 1 ≤ i ≤ 200. This means that it makes no difference if you draw first, last or anywhere in the middle. Here is Marilyn Vos Savant’s intuitive solution to this problem: It makes no difference if you draw first, last, or anywhere in the middle. Look at it this way: Say the robbers make everyone draw at once. You’d agree that everyone has the same change of losing (one in 200), right? Taking turns just makes that same event happen in a slow and orderly fashion. Envision a raffle at a church with 200 people in attendance, each person buys a ticket. Some buy a ticket when they arrive, some during the event, and some just before the winner is drawn. It doesn’t matter. At the party the end result is this: all 200 guests draw a slip of paper, and, regardless of when they look at the slips, the result will be identical: one will lose. You can’t alter your chances by looking at your slip before anyone else does, or waiting until everyone else has looked at theirs.

Example 3g At a university, 30% of the students are freshmen, 25% are sophomores, 23% are juniors, and the remaining 22% are seniors. Suppose that 7% of the freshmen, 6% of the sophomores, 5% of the juniors, and 8% of the seniors are A-students. If a randomly selected student from this university is an A-student, what is the probability that he or she is a senior?

Solution: Let E, F, G, and H be the events that the randomly selected student is a freshman, sophomore, junior, and senior, respectively. Let L be the event that the student is an

Section 3A

Additional Examples

6

A-student. We are interested in P (H | L) . By Bayes theorem, P (H | L) =

P (L | H)P (H) P (L | E)P (E) + P (L | F )P (F ) + P (L | G)P (G) + P (L | H)P (H)

=

(0.08)(0.22) = 0.27.

(0.07)(0.30) + (0.06)(0.25) + (0.05)(0.23) + (0.08)(0.22)

Example 3h Suppose that 16% of an insurance company’s automobile policyholders are male and under the age of 25, while 12% are female and under the age of 25. The following table lists the percentages of various groups of policyholders who were involved in a car accident last a year.

Group

Male Under 25

Female Under 25

Between 25 and 65

Over 65

Percentage of Accidents

20%

8%

5%

10%

Find the range of the percentages of this company’s policyholders who got involved in a car accident during the previous year.

Solution: Let M, F, B, and O be, respectively, the events that a randomly selected automobile policyholder of this company is a male and under the age of 25, a female and under the age of 25, between the ages 25 and 65, and over the age of 65. Clearly, {M, F, B, O} is a partition of the sample space. Thus, P (M ) + P (F ) + P (B) + P (O) = 1, implying that 0.16 + 0.12 + P (B) + P (O) = 1.

This gives P (B)+P (O) = 0.72. Let A be the event that the randomly selected policyholder got involved in an accident last year. We want to find the greatest lower bound and the least upper bound for P (A). By the law of total probability, P (A) = P (A | M )P (M ) + P (A | F )P (F ) + P (A | B)P (B) + P (A | O)P (O) = (0.20)(0.16) + (0.08)(0.12) + (0.05)P (B) + (0.10)P (O) = 0.0416 + (0.05)P (B) + (0.10)P (O).

The greatest lower bound for P (A) is obtained, from this relation, by substituting 0.72 for P (B) and 0 for P (O): P (A) > 0.0416 + (0.05)(0.72) = 0.0776.

Similarly, the least upper bound for P (A) is obtained, from the relation, by substituting 0 for P (B) and 0.72 for P (O): P (A) < 0.0416 + (0.10)(0.72) = 0.1136.

Section 3A

Additional Examples

7

Therefore, the range of the percentages we are interested in is between 7.76% and 11.36%.

Note that the data in the table implies that neither P (O) nor P (B) can be 0. So the inequalities above are both strict.

Example 3i Suppose that n ≥ 2 missiles are fired at a target and hit it independently. If the probability that the ith missile hits it is p i , i = 1, 2, . . ., n, find the probability that at least two missiles will hit the target.

Solution: For convenience, let p 0 = 0; the desired probability is 1−P (none of the missiles hits the target)−P (exactly one of the missiles hits the target) n n

(1 − p1 )(1 − p2 ) · · · (1 − pi−1 )pi(1 − pi+1 ) · · · (1 − pn ). = 1 − (1 − pi ) − i=1

i=1

Example 3j Avril has certain standards for selecting her future husband. She has n suitors and knows how to compare any two and rank them. She decides to date one suitor at a time randomly. When she knows a suitor well enough, she can marry or reject him. If she marries the suitor, she can never know the ones not dated. If she rejects the suitor, she will not be able to reconsider him. In this process, Avril will not choose a suitor she ranks lower than at least one of the previous ones dated. Avril’s goal is to maximize the probability of selecting the best suitor. To achieve this, she adopts the following strategy: For some m, 0 ≤ m < n, she dumps the first m suitors she dates after knowing each of them well no matter how good they are. Then she marries the first suitor she dates who is better than all those preceding him. In terms of n, find the value of m which maximizes the probability of selecting the best suitor.

Remark: This example is a model for many real-world problems in which we have to reject “choices” or “offers” until the one we think is the “best.” Note that it is quite possible that we reject the best “offer” or “choice” in the process. If this happens, we will not be able to make a selection.

Solution: Let E m be the event that Avril selects the best suitor given her strategy. Let B i be the event that the best suitor is the ith of Avril’s dates. By the law of total probability, P (Em ) =

n

n

P (Em | Bi )P (Bi ) =

i=1

1 P (Em | Bi ). n i=1

Clearly, P (Em | Bi ) = 0 for 1 ≤ i ≤ m. For i > m, if the ith suitor is the best, then Avril chooses him if and only if among the first i − 1 suitors Avril dates, the best is one of the first m. So m .

P (Em | Bi ) = i−1 Therefore, n n m 1 m 1 = .

P (Em ) = n i−1 n i−1 i=m+1

i=m+1

Section 3A

Now

n

i=m+1

1 ≈ i−1

Additional Examples

8

n

n 1 dx = ln . m m x

m n ln . n m To find the maximum of P (E m ), consider the differentiable function x n h(x) = ln . n x

Thus

P (Em ) ≈

Since

1 n 1 ln − =0 n x n implies that x = n/e, the maximum of P (E m ) is at m = [n/e], where [n/e] is the greatest integer less than or equal to n/e. Hence Avril should dump the first [n/e] suitors she dates and marry the first suitor she dates afterward who is better than all those preceding him.

The probability that with such a strategy she selects the best suitor of all n is approximately n 1 1 = ln e = ≈ 0.368. h e e e h (x) =

Example 3k (Shrewd Prisoner’s Dilemma) Because of a prisoner’s constant supplication, the king grants him this favor: He is given 2N balls, which differ from each other only in that half of them are green and half are red. The king instructs the prisoner to divide the balls between two identical urns. One of the urns will then be selected at random, and the prisoner will be asked to choose a ball at random from the urn chosen. If the ball turns out to be green, the prisoner will be freed. How should he distribute the balls in the urn to maximize his chances of freedom?

Hint: Solution: Let g be the number of green balls and r be the number of red balls in the first urn. The corresponding numbers in the second urn are N − g and N − r. The probability that a green ball is drawn is f (g, r): N −g 1 g + . f (g, r) = 2 g + r 2N − g − r We will find the maximum of this function of two variables (r and g). Note that the maximum need not occur at an interior point of the domain. Let N be the set of nonnegative integers. The domain of f is

(g, r) ∈ N × N : 0 ≤ g ≤ N, 0 ≤ r ≤ N, 0 < g + r < 2N . ∂f ∂f = = 0 gives ∂g ∂r g = r = N/2 and f (N/2, N/2) = 1/2. However, this is not the maximum value because on the boundary of the domain of f along r = 0, we find that N −g 1 1+ f (g, 0) = 2 2N − g

Extending the domain of f to all points (g, r) ∈ R × R, we find that

Section 3A

Additional Examples

9

is maximum at g = 1 and

1 3N − 2 1 ≥ .

2 2N − 1 2 We also find that on the boundary along r = N, f (1, 0) =

f (g, N ) =

1 g +1 2 g+N

is maximum at g = N − 1 and f (N − 1, N ) =

1 3N − 2 1 ≥ .

2 2N − 1 2

1 3N − 2 . There2 2N − 1 fore, there are exactly two maximums and they occur at (1, 0) and (N − 1, N ). That is, the maximum of f occurs if one urn contains one green and 0 red balls and the other one contains N − 1 green and N red balls. For large N, the probability that the prisoner is freed 1 3N − 2 3 ≈ . is 2 2N − 1 4 The maximums of f along other sides of the boundary are all less than

Section 3B

3B

More on Applications of Probability to Genetics

10

MORE ON APPLICATIONS OF PROBABILITY TO GENETICS

As we discussed in Companion for Chapter 1, inheritance patterns usually involve exactly two alleles per gene. However, there are genes that have multiple alleles. It is important to know that even when, in a population, multiple alleles of a gene exist, still, for that gene, each individual carries at most two different alleles. As an example, for blood groups in humans, there are three alleles: A, B, and O. Since each person inherits one of these alleles from each parent, there are six genotypes. Blood groups A and B are dominant to the O. Therefore, people who inherit type A from one parent and type O from another will have type A blood. Similarly, people who inherit type B blood from one parent and type O from another will have type B blood. Obviously, inheriting type A blood from both parents results in type A blood, inheriting type B blood from both parents results in type B blood, and inheriting type O blood from both parents results in type O blood. Interestingly enough, neither blood group A is dominant to blood group B nor vice versa. Individuals who inherit type A blood from one parent and type B blood from another have a blood type in which both A and B alleles are expressed. That blood type is called type AB blood. A and B are said to be codominant to each other.

In general, during fertilization, for each characteristic, each parent contributes randomly one member of its alleles independently of other characteristics and independently of the reproduction of other offspring. Therefore, in the pea plant example discussed previously, where A represents the dominant allele for purple flowers and a represents the recessive allele for white flowers, when the biologist crosses a pea plant of genotype AA with a pea plant of genotype aa, the resulting plant is Aa with probability 1. However, when crossing a plant of genotype Aa with another plant of genotype Aa, the resulting plant is AA with probability 1/4, Aa with probability 1/2, and aa with probability 1/4. That is why, in the second generation, approximately 1/4th of the pea plants will have white flowers.

Example 3 In guinea pigs, having long hair is a recessive trait. Suppose that an offspring of two short-haired guinea pigs has long hair. What is the probability that at least one of the next three offspring of these two guinea pigs also has long hair?

Solution: Let A represent the dominant allele for short hair and a represent the recessive allele for long hair. Since the offspring with long hair is aa, both parents must be Aa. Hence any other offspring is Aa with probability 1/2, is AA with probability 1/4, and is aa with probability 1/4. Therefore, the probability is 3/4 that any offspring is shorthaired. The probability that the next three offspring are all short-haired is (3/4) 3. The probability that at least one of them is long-haired is 1 − (3/4) 3 = 37/64 = 0.578.

Example 3m Consider the three alleles A, B, and O for blood types. Suppose that a man of genotype AB marries a woman of genotype AO. If they have six children, what is the probability that none will have type AB blood?

Solution: The genotype of a child of these parents will be one of the following with equal probabilities: AA, AB, AO, BO. Since the alleles A and B are codominant to each other and dominant to O, a child of these parents will have blood type A with probability 1/2,

Section 3B

More on Applications of Probability to Genetics

11

blood type AB with probability 1/4, and blood type B with probability 1/4. Therefore, by independence, the probability that none of the six children will have blood type AB is (3/4)6 ≈ 0.178.

Example 3n Consider the three alleles A, B, and O for the human blood types. Suppose that, in a certain population, the frequencies of these alleles are 0.45, 0.20, and 0.35, respectively. Kim and John, two members of the population, are married and have a son named Dan. Kim and Dan both have blood types AB. John’s blood type is B. What is the probability that John’s genotype is BB? Solution: Clearly, John’s genotype is either BB or BO. Let E be the event that it is BB. Then E c is the event that John’s genotype is BO. Let F be the event that Dan’s genotype is AB. By Bayes’ formula, the probability we are interested in is P (E | F ) = =

P (F | E)P (E) P (F | E)P (E) + P (F | E c)P (E c ) (1/2)(0.20)2 = 0.53.

(1/2)(0.20)2 + (1/4)(0.20)(0.35)

Note that P (E) = (0.20) 2 and P (E c) = (0.20)(0.35) since for the two B alleles in John’s blood type, each is coming from one parent independently of the other parent.

Certain characteristics such as human height or human skin color result from polygenic inheritance. Unlike the color of the pea plant flowers, such characteristics are not controlled by a single gene. For example, it is the additive effect of several genes that determines the level of darkness of human skin. Apart from the environmental factors such as nutrition and light, in humans, variation in darkness level of the skin is continuous mainly because, as a heritable character, skin color is controlled by several genes. In humans, the same is true for the height and most other quantitative inheritance. Suppose that human skin color is controlled by four genes independently inherited. For those genes, let A, B, C, and D represent the dominant alleles for dark skin. For the respective genes, let a, b, c, and d represent the recessive alleles for light skin. Then individuals who are AABBCCDD are the darkest members of the population, and individuals who are aabbccdd are the lightest.

Moreover, all the dominant alleles contribute to the skin color. For example, an individual who is AAbbccdd is darker than an individual who is Aabbccdd, and those who are Aabbccdd are darker than those with genotypes aabbccdd. Depending on the degree to which each of the four genes contributes to the darkness of the skin, there is a very wide range of skin colors among the members of the population. As an example, assuming that each gene contributes the same amount of darkness to the skin, people with the following genotypes are ordered from lightest to darkest: Aabbccdd,

aaBbCcdd,

AABbccDd,

AaBbCcDD,

whereas people with the following genotypes are indistinguishable in skin color phenotype: AAbbCcDd,

AaBbCcDd,

aaBbCcDD,

AABBccdd.

Section 3B

More on Applications of Probability to Genetics

12

As another example, suppose that on an island, the zeroth generation is the entire population, which consists of a male population all members having the lightest skin and a female population all members having the darkest skin. Suppose that a member belongs to the second generation if all of his or her grandparents belong to the zeroth generation. To find the probability that a member of the second generation has the lightest possible skin, of the four genes that control skin color, consider the gene whose dominant allele for dark skin is denoted by A. In the zeroth generation, all females are AA and all males are aa. Therefore, in the first generation, all members, regardless of gender, are Aa. In the second generation, a member is AA with probability 1/4, Aa with probability 1/2, and aa with probability 1/4.

Similarly, the probability is 1/4 that in the second generation a member is bb, the probability is 1/4 that he or she is cc, and the probability is 1/4 that he or she is dd. Since genes are independently inherited, the probability that, in the second generation, a person has the lightest color (i.e., has genotype aabbccdd) is (1/4) 4 = 1/256.

Hardy-Weinberg Law In organisms having two sets of chromosomes, called diploid organisms, which we all are, each hereditary character of each individual is carried by a pair of genes. A gene has alternate alleles that usually are dominant A or recessive a, so that the possible pairs of genes are AA, Aa (same as aA), and aa. Under certain conditions, by the Hardy-Weinberg law, to be discussed later, the probabilities of AA, Aa, and Aa in all future generations remain the same as those in the second generation. To show this important fact, we will first discuss what is meant by natural selection, mutation, random mating, and migration.

Roughly speaking, natural selection is the process by which forms of life best adapted to the environment are the most fit. They live longer and are most likely to reproduce and hence perpetuate the survival of their type. Lack of natural selection implies that, in a form of life, all types are equally adapted to the environment. There is no type that survives longer and produces in greater number.

A heritable change to the genetic material is called mutation. It can happen by radiation, chemicals, or any carcinogen. A mutagen is a substance or preparation that can cause mutation.

In certain populations, individuals choose mates on the basis of their own genotypes.

For example, among humans, short and tall individuals may marry more often within their own groups. In other situations, individuals might often choose to mate with individuals of different genotypes. In such cases, mating is not random. We say that, in a population, random mating occurs if mating between genotypes occur totally at random. Under random mating, between genotypes, mating will occur in proportion to the frequencies of the genotypes. Therefore, the probability of a mating between any two genotypes is the product of the frequencies of those genotypes in the population.

To prevent gene flow between populations of the same species, individuals should be prevented from migration. Clearly, immigrants and emigrants can change the balance of the frequencies of the genotypes. Existing alleles may be removed from the population by emigration. New alleles might be added by immigration. In both cases, the frequencies of the remaining alleles might increase or decrease.

Let r be the probability that a randomly selected person in the population is AA. Let s be the probability that he or she is Aa, and let t be the probability that he or she is aa. Then

Section 3B

More on Applications of Probability to Genetics

13

r + s + t = 1. Let E be the event that a randomly selected father transmits A to a child.

Then P (E) = P (E | the father is AA)P (the father is AA) + P (E | the father is Aa)P (the father is Aa) + P (E | the father is aa)P (the father is aa) s 1 = 1·r+ ·s+0·t = r+ .

2 2 Similarly, the probability that the father transmits a is t+s/2, the probability that the mother transmits A is r + s/2, and the probability that the mother transmits a is t + s/2. Now suppose that we have a very large population in which natural selection does not occur and is free from mutation and migration. Then, under random mating, the contributions of the parents to the genetic heritage of their offspring are independent of each other. Therefore, the probability that an offspring selected at random is AA is r 1 = (r+s/2)2 , the probability that he or she is Aa is s 1 = 2(r + s/2)(t + s/2), and the probability that he or she is aa is t1 = (t + s/2)2 . These formulas imply that, r 2 , the probability that a randomly selected offspring from the third generation is AA is given by (r 1 + s1 /2)2 , the probability that he or she is Aa is s 2 = 2(r1 + s1 /2)(t1 + s1 /2), and the probability that he or she is aa is t2 = (t1 + s1 /2)2 . Some algebra shows that r 2 = r1 , t2 = t1 and s2 = s1 . For example,

s 2 s1 2 s 2 s t+ = r+ + r+ r2 = r1 + 2 2 2 2

s s 2 s r+ +t+ = r+ 2 2 2

2 s s 2 (r + t + s) = r + = r+ = r1 .

2 2 Therefore, the proportions of AA, Aa, and aa in the third generation are the same as in the s s second generation. Let p = r + and q = t + . Then p is the frequency of allele A and 2 2 q is the frequency of allele a in the population. We have shown that, under the conditions explained previously, when the population is in equilibrium, the genotype frequencies of AA, Aa, and aa are p2 , 2pq, and q 2 , respectively. Note that p + q = 1 implies that p2 + 2pq + q 2 = (p + q)2 = 1, as expected. We have established the following important law: The Hardy-Weinberg Law: For a diploid organism with alternate dominant allele A and recessive allele a, the possible pairs of genes are AA, Aa, and aa. Under random mating conditions, in a very large population of a diploid organism in which natural selection does not occur and is free from mutation and migration, the proportions of AA, Aa, and aa in all future generations remain the same as those in the second generation. Moreover, if the gene frequency of A is p and the gene frequency of a is q, then, in all future generations, the genotype frequencies of AA, Aa, and aa remain in the proportions p 2 , 2pq, and q 2 .

Section 3B

More on Applications of Probability to Genetics

14

Example 3o Suppose that 64% of the sheep in a certain region have straight horn. If, for the sheep in that region, the allele for curly horn is dominant to the allele for straight horn, under the conditions of Hardy-Weinberg law, find the percentage of sheep who have two different alleles of the gene for horn shape.

Solution: Let the dominant allele for the curly horn be denoted by H and the recessive allele for straight horn be h. Since the genotype of a sheep with straight horn is hh, applying the Hardy-Weinberg’s law, we have that q 2 = 0.64, or q = 0.8. Therefore, p = 0.2, and the percentage of sheep who have Hh is 2pq = 2(0.2)(0.8) = 0.32.

Sex-Linked Genes In this subsection, we will discuss sex chromosomes and reproductive cells. Recall that the reproductive cells are the egg and sperm cells, collectively called gametes. We know that except for gametes, in each of the human cells, 23 chromosomes come from the father and 23 from the mother. Among the 23 pairs of chromosomes in each nongamete cell, there is one pair of chromosomes, called the sex chromosomes, which determines a person’s gender. The pair of sex chromosomes in human females are called X chromosomes. Similar to other pairs of chromosomes, the pair of female sex chromosomes come one from each parent and look alike. The genes also appear in pairs, one from each parent, and those controlling the same characteristic occupy the same position on the paired X chromosomes inherited from father and mother. Concerning dominance and recessivity, the rules for the paired X chromosomes are the same as those followed by the 22 pairs of nonsex chromosomes of nongametes. The pair of sex chromosomes in human males do not look alike.

They differ in shape and size. One of the chromosomes, inherited from the mother, is an X chromosome. The other one, inherited from the father, is said to be a Y chromosome.

For most genes lying on the X chromosome, there is no corresponding gene on the Y chromosome. The only genes that lie on the Y chromosomes are those that are important in determining the male structures and organs. All genes on the X chromosome are called Xlinked genes. Similarly, all genes on the Y chromosome are called Y -linked genes. For X-linked genes in males, we do not have recessivity and dominance. This is because males have exactly one X chromosome. Therefore, males carry only one allele of each X-linked gene. As a result, for example, if a male is color-blind, since the gene for color blindness is on the X chromosome, he must have inherited it from the mother. Some people bleed excessively when injured. They might have a genetic disease called hemophilia. If a male is a hemophiliac, he must have inherited it from his mother, since the genes for hemophilia lie on the X chromosome. There are many other characteristics such as X-linked baldness where a male inherits the characteristic from his mother.

Like humans, in all mammals, the sex chromosomes determine a mammal’s gender.

Mammalian males have one X chromosome and one Y chromosome. Mammalian females have two X chromosomes. In such cases, males are represented as XY, females as XX. It is important to know that the XY and XX patterns are not true for all males and females of all species. For example, female birds carry two different chromosomes, whereas male birds carry two identical chromosomes.

We will now discuss gametes. Unlike nonproductive cells, which have 23 pairs of chromosomes, gametes carry 23 single chromosomes. Therefore, no gamete contains any paired

Section 3B

More on Applications of Probability to Genetics

15

chromosomes. A gamete carries every single gene of the individual, but in only one copy.

When gametes are formed, with equal probabilities and independent of other genes, only one of the alleles of each gene will end up in the gamete. Clearly, a gamete has only one sex chromosome. In human females, the unique chromosome is necessarily an X chromosome.

However, in males, it is either an X chromosome or a Y chromosome with equal probabilities. This follows because, in females, both of the sex chromosomes in nonreproductive cells are X and, in males, one of them is X and the other one is Y . Therefore, an egg contains only one X chromosome, while a sperm cell contains either an X chromosome or a Y chromosome with equal probability. If the sperm cell that fertilizes the egg carries an X chromosome, then the offspring is female. If it carries a Y chromosome, then the offspring is male. That explains why in humans it is equally likely that an offspring is a boy or a girl.

Example 3p What is the probability that 12 of the chromosomes in a randomly selected human gamete are paternal and 11 are maternal?

Solution: There is a one-to-one correspondence between the set of all possible combinations of 23 chromosomes of which 12 are paternal (P ) and 11 are maternal (M ) and the set of all distinguishable sequences of 12 P ’s and 11 M ’s. By the generalized counting principle, the number of sequences of P and M of length 23 is 2 23 . By Theorem 2.4 of the book, the number of distinguishable permutations of 23 letters of which 12 are P and 11 23!

23! . Therefore, the desired probability is 223 = 0.16. are M is 12! 11!

12! 11!

Example 3q Hemophilia is a sex-linked disease with normal allele H dominant to the mutant allele h. Kim and John are married and their son, Dan, has hemophilia. If the frequencies of H and h are 0.98 and 0.02, respectively, what is the probability that Kim has hemophilia?

Solution: Dan has inherited all of his sex-linked genes from his mother. Since Dan is hemophiliac, Kim is either Hh or hh. Let E be the event that she is hh. Then E c is the event that she is Hh. Let F be the event that Dan has hemophilia. By Bayes’ formula, the desired probability is P (E | F ) = =

P (F | E)P (E) P (F | E)P (E) + P (F | E c)P (E c ) 1 · (0.02)(0.02) = 0.02.

1 · (0.02)(0.02) + (1/2) · 2(0.98)(0.02)

Hence the conditional probability that Kim has hemophilia given that her son, Dan, has hemophilia is 0.02. Note that the unconditional probability that Kim has hemophilia is (0.02)(0.02) = 0.0004.

Section 3B

More on Applications of Probability to Genetics

16

EXERCISES FOR SECTION 3B 1.

Kim has blood type O and John has blood type A. The blood type of their son, Dan, is O. What is the probability that John’s genotype is AO?

2.

Suppose that a gene has k (k > 2) alleles. For that gene, how many genotypes can an individual have?

3.

For the shape of pea seed, the allele for round shape, denoted by R, is dominant to the allele for wrinkled shape, denoted by r. A pea plant with round seed shape is crossed with a pea plant with wrinkled seed shape. If half of the offspring have round seed shape and half have wrinkled seed shape, what were the genotypes of the parents?

4.

In humans, the presence of freckles and having free earlobes are independently inherited dominant traits. Kim and John both have free earlobes and both have freckles, but their son, Dan, has attached earlobes and no freckles. What is the probability that Kim and John’s next child has free earlobes and no freckles?

5.

For Drosophila (a kind of fruit fly), B, the gray body, is dominant over b, the black body, and V, the wild-type wing is dominant over v, vestigial (a very small wing). A geneticist, T. H. Morgan, when mating Drosophila of genotype BbV v with Drosophila of genotype bbvv, observed that 42% of the offspring were BbV v, 41% were bbvv, 9% were Bbvv, and 8% were bbV v. Based on these results, should we expect that the body color and wild-type genes of Drosophila are on different chromosomes (that is, are unlinked and hence independent)? Why or why not?

6.

In a population, 1% of the people suffer from hereditary genetic deafness. The hearing allele is dominant and is denoted by D. The recessive allele for deafness is denoted by d. Kim and John, from the population, are married and have a son named Dan. If Kim and Dan are both deaf, what is the probability that John is also deaf?

7.

In the United States, among the Caucasian population, cystic fibrosis is a serious genetic disease. For cystic fibrosis, the normal allele is dominant to the mutant allele.

Suppose that, in a certain region of the country, 5.29% of the people have cystic fibrosis. Under the Hardy-Weinberg conditions, determine the percentage of people in that region who carry at least one mutant allele of the disease.

8.

Hemophilia is a sex-linked disease with normal allele H dominant to the mutant allele h. Kim and John are both phenotypically normal. If the frequencies of H and h are 0.98 and 0.02, respectively, what is the probability that Dan, their son, has hemophilia?

9.

Color blindness is a sex-linked hereditary disease with normal allele C dominant to the mutant allele c. Kim and John are married and their son, Dan, is color-blind. If, in the population, the frequencies of C and c are 0.83 and 0.17, respectively, what is the probability that Kim is color-blind?

10.

Hemophilia is a sex-linked hereditary disease with normal allele H dominant to the mutant allele h. Kim and John are married and their daughter, Ann, has hemophilia.

Section 3B

More on Applications of Probability to Genetics

17

If John has hemophilia, and the frequencies of H and h, in the population, are 0.98 and 0.02, respectively, what is the probability that Kim does not have hemophilia but she is a carrier for that disease?

11.

In a certain country, all children with cystic fibrosis die before reaching adulthood.

For this lethal disease, the normal allele, denoted by C, is dominant to the mutant allele, denoted by c. In that country, the probability is p that a person is only a carrier; that is, he or she is Cc. If Mr. J, a resident of the country had a brother who died of cystic fibrosis, what is the probability that Mr. J’s next child will have cystic fibrosis?

12.

Let p and q be positive numbers with p + q = 1. For a gene with dominant allele A and recessive allele a, let p 2 , 2pq, and q 2 be the probabilities that a randomly selected person has genotype AA, Aa, and aa, respectively. If a man is of genotype AA, what is the probability that his brother is also of genotype AA?

13.

In human beings, the gene for naturally curly hair (H) is dominant over the gene for straight hair (h), and the gene for right-handedness (R) is dominant over the gene for left-handedness (r). Suppose that a left-handed man with curly hair marries a righthanded woman with straight hair. Find the probability that their child is (a) righthanded with curly hair; (b) right-handed with straight hair; (c) left-handed with curly hair; (d) left-handed with straight hair.

14.

In human beings, brown eyes (B) are dominant over blue eyes (b). The probability that Stanley has genotype BB is 5/8, and the probability that he has genotype bB is 3/8. If Stanley marries Stella, a blue-eyed woman, what is the probability that their first child has brown eyes? Note that since all eye colors are dominant over blue, Stella’s genotype is necessarily bb.

15.

In genetics, a person or other organism is said to be a carrier of a certain trait if its genotype includes exactly one copy of a recessive allele of that trait. So a carrier of a trait does not display that trait. It is estimated that 3.2% of the US population is a carrier of the recessive hereditary gene of cystic fibrosis (c). Natasha is tested and is a carrier. However, we do not know whether her husband, Austin, is a carrier or not.

We only know that he does not have cystic fibrosis. Find the probability that Austin is also a carrier if their first child does not have cystic fibrosis.

16.

In human beings, the gene for naturally curly hair (H) is dominant over the gene for straight hair (h), and the gene for right-handedness (R) is dominant over the gene for left-handedness (r). Julie is a right-handed woman with curly hair. However, her genotype is not determined. The probability that her genotype is HhRr is 2/3, and the probability that it is HhRR is 1/3. If Julie marries a man with genotype Hhrr, what is the probability that their first child is right-handed with straight hair?

Answers to Odd-Numbered Exercises of Section 3B

Answers to Odd-Numbered Exercises of Section 3B 1. 1.

3. The parents are rr and Rr. 5. the observed data shows that the genes are linked.

7. 40.71%.

9. 0.17.

11. p/6.

13. The answers to parts (a), (b), (c), and (d) are 4/9, 2/9, 2/9, and 1/9, respectively.

15. 0.024.

18

COMPANION FOR CHAPTER 4 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 4 Distribution Functions and Discrete Random Variables 4A 4B 4C

Additional Examples 3 An Investment Application Quantiles and the Median

11 13

3

Section 4A

Additional Examples

3

Chapter 4

D istribution F unctions and Discrete R andom Variables 4A

ADDITIONAL EXAMPLES

Example 4a Flip a fair coin, if it lands heads let X = 2 and if it lands tails let X be a random point from the interval (1, 3). Find the distribution function of X and use it to calculate P (3/2 < X < 5/2).

Solution: The random variable X can only assume values of the interval (1, 3), hence F (t) = P (X ≤ t) = 0 if t ≤ 1. To calculate F (t) for other values of t, we use Theorem 3.3 of the book, the law of total probabilities. Let H and T stand for the events of heads and tails, respectively, when flipping the coin. Then F (t) = P (X ≤ t) = P (X ≤ t | H)P (H) + P (X ≤ t | T )P (T ).

Now from this relation, for 1 ≤ t < 2 we get, F (t) = 0 ×

t−1 1 t−1 1 + × = , 2 3−1 2 4

F (t) = 1 ×

t−1 1 t+1 1 + × = .

2 3−1 2 4

and for 2 ≤ t < 3 we get,

For t ≥ 3, F (t) = P (X ≤ t) = 1. Hence ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨(t − 1)/4 F (t) = ⎪ ⎪ (t + 1)/4 ⎪ ⎪ ⎪ ⎪ ⎩ 1

t<1 1≤t<2 2≤t<3 t ≥ 3.

Observe that the magnitude of the jump at t = 2 is 1/2. Hence P (X = 2) = 1/2 as we expect. To calculate P (3/2 < X < 5/2), note that P (3/2 < X < 5/2) = P (X < 5/2) − P (X ≤ 3/2).

Section 4A

4

Additional Examples

Now P (X < 5/2) = P (X ≤ 5/2) since P (X = 5/2) = 0. Therefore, P (3/2 < X < 5/2) = P (X ≤ 5/2) − P (X ≤ 3/2) 3 (5/2) + 1 (3/2) − 1 − = = 0.75. = F (5/2) − F (3/2) = 4 4 4 Example 4b In a certain country, where the resources are very limited, to test individuals for a specific disease, they divide them into groups of 15, take blood samples of all, mix them, and analyze the mixture. If the test is negative, none of the individuals in the group has the disease. If it is positive, then they must test each of the members of the group separately. If the probability that a person from this country has the disease, independently from others, is 0.07, and if for each group, it is chemically possible to mix their blood samples for the test, what is the expected number of tests for each group of 15 people?

Solution: In a random group, let X be the number of tests required to identify the individuals with the disease. Since the probability that a random person from this country does not have the disease is 1 − 0.07 = 0.93, ⎧ with probability (0.93) 15 ⎨1 X= ⎩ 16 with probability 1 − (0.93) 15 Therefore,

Example 4c

E(X) = 1 · (0.93)15 + 16 · 1 − (0.93)15 ≈ 10.95.

Let X be a discrete random variable with distribution function ⎧ ⎪ t<1 ⎨0 F (t) = n ⎪ ⎩1 − 4 n ≤ t < n + 1, n = 1, 2, 3, . . ..

5

Find p, the probability mass function of X and E(X).

Solution: The set of possible values of X is {1, 2, 3, . . .}, since it is at the points of this set that F has jumps. P (X = i) is the magnitude of jump at i. We have

1 4 −0= , P (X = 1) = F (1) − F (1−) = 1 − 5 5

4 2 4 4 2 4 4 1 4 4 = − 1− = . = − 1− P (X = 2) = F (2) − F (2−) = 1 − 5 5 5 5 5 5 5 5 In general, for n ≥ 1, P (X = n) = F (n) − F (n−) = 1 − 4 n−1 4 = = 1− 5 5 5 4

4 5

n

− 1−

n−1 1

5

.

4 5

n−1

=

4 5

n−1

−

4 5

n

Section 4A

So the probability mass function of X is 4 n−1 1 , P (n) = 5 5

Additional Examples

5

n = 1, 2, 3, . . . .

By definition of expected value, ∞

E(X) = n=1

Since for |r| < 1,

4 n 5

n−1 1

5

=

1/5 4/5

∞ n=1

4 n 5

n

.

∞ x 2 n=1 xr = r/(1 − r) , we have

1 E(X) = · 4

4 5 1−

4 5

2 = 5.

Example 4d A box contains 5 red and 3 blue balls. The balls are identical except for their colors. Balls are drawn at random, one by one, and without replacement until a blue ball is drawn. Let X be the number of balls drawn. Find E(X) and Var(X).

Solution: The set of possible values for X is {1, 2, 3, 4, 5, 6}. The probabilities associated with these values are 3 P (X = 1) = ; 8 15 5 3 ; P (X = 2) = · = 8 7 56 5 5 4 3 P (X = 3) = · · = ; 8 7 6 28 3 5 4 3 3 ; P (X = 4) = · · · = 8 7 6 5 28 3 5 4 3 2 3 P (X = 5) = · · · · = ; 8 7 6 5 4 56 1 5 4 3 2 1 3 .

P (X = 6) = · · · · · = 8 7 6 5 4 3 56

Therefore, 15 5 3 3 1 9 3 +2· +3· +4· +5· +6· = = 2.25.

8 56 28 28 56 56 4 15 5 3 3 1 27 3 +9· + 16 · + 25 · + 36 · = ; E(X 2) = 1 · + 4 · 8 56 28 28 56 56 4

2 27 9 2 27 − ≈ 1.69. = Var(X) = E(X 2) − E(X) = 4 4 16 E(X) = 1 ·

Section 4A

Additional Examples

6

Alternative Solution: To calculate the probability mass function of X, p(i) = P (X = i), i = 1, 2, . . ., 6, let Ai be the event of i − 1 red balls in i draws and B i be the event that in i draws, the last draw is a blue ball. Then

5 3 1 i−1 1

· . pi = P (X = i) = P (Ai Bi ) = P (Ai )P (Bi | Ai ) = 8 i i Therefore,

5 3 6 9 1 i−1 1

E(X) = i· · = = 2.25.

8 i 4 i=1 i

5 3 6 27 1 i−1 1

; i2 · · = E(X 2) = 8 i 4 i=1 i

2 27 9 2 27 − ≈ 1.69. = Var(X) = E(X 2) − E(X) = 4 4 16

Example 4e Let X be a discrete random variable with the set of possible values {1, 2, 3 . . .} and finite expected value. Show that ∞

P (X ≥ n).

E(X) = n=1

Solution: By the definition of expected value, ∞

∞

i

iP (X = i) =

E(X) = i=1 ∞

P (X = i) i=1 n=1

∞

∞

P (X = i) =

= n=1 i=n

P (X ≥ n).

n=1

Example 4f Friday the 13th, How Often Do They Occur? Not only superstitious people, but also non-superstitious mathematicians have shown interest in the number of Friday the 13th’s that appear in a given year. Let X be the number of Friday the 13th’s in a randomly chosen year. (a)

Find the probability mass function of the random variable X.

(b)

What is the probability that the 13th of a randomly selected month is a Friday?

Section 4A

(c)

7

Additional Examples

What is the probability that a randomly selected year has two consecutive months with their 13th’s falling on Friday?

Solution: (a) The following table determines the months of any given year that have their 13th on a Friday.

Day Number 0 1 2 3 4 5 6

M ONTHS WITH A F RIDAY THE 13 TH Day Non-Leap Year Leap Year Sunday January, October January, April, July Monday April, July September, December Tuesday September, December June Wednesday June March, November Thursday February, March, November February, August Friday August May Saturday May October

This table is given by Ritter et al. in their paper “An Aid to the Superstitious,” which was published in vol. 70, no. 5 issue of the Mathematics Teacher in 1977. To use the table above, all we need to know is the first day of the year. For example, if the first day of a non-leap year is a Thursday, then that year has three months with their 13th’s falling on Friday, namely, February, March, and November. To find the first day of a given year, Y, let f (Y ) = Y + [(Y − 1)/4] − [(Y − 1)/100] + [(Y − 1)/400],

(1)

where the symbol [x] denotes the greatest integer less than or equal to x. Then find the D, the number that satisfies f (Y ) ≡ D (mod 7), (2) where by a ≡ b (mod n) we mean that when a is divided by n, the remainder is b. If D is 0, the year begins on a Sunday. If it is 1, the year begins on a Monday, and so on. As an example, consider the year Y = 1992. From (1) we have, f (1992) = 1992 + [1991/4] − [1991/100] + [1991/400] = 1992 + 497 − 19 + 4 = 2474.

Now dividing 2474 by 7, the remainder obtained is 3. Thus f (1992) ≡ 3 (mod 7), and hence the first day of 1992 is a Wednesday. But 1992 is a leap year, so from the table above we have that this year has two months with their 13th’s falling on Friday, namely, March and November. The underlying reason for (1) is that the Gregorian Calendar has a leap year every four years except century years that are not divisible by 400. Thus all non-century years divisible by 4 and all century years divisible by 400 are leap years. For example, 1988, 2000, 2060, 2296, and 2800 are leap years while 1300, 1977, 1998, 2067, 2100, and 2700 are not.

Now the set of possible values of X, the number of the Friday the 13th’s in a randomly selected year, is {1, 2, 3}. To find P (X = i), i ∈ {1, 2, 3}, we will use the knowledge that the Gregorian Calendar is periodic with period of 400. By this knowledge, for any year Y, let Z be one of the years Y through Y + 399. Then whatever day of the week on which

Section 4A

Additional Examples

8

January 1st of Z falls, the January 1st of Z + 400 will fall on the same day of week. Thus the respective numbers of Friday the 13th’s in the years Y through Y + 399 are the same as the numbers of Friday the 13th’s in the years Y + 400 through Y + 799. Let us restrict ourselves to the period from the year 2000 to the year 2399. For 0 ≤ i ≤ 6, let |L i | and |N Li| be the respective numbers of leap and non-leap years in this period with their January 1st a number i day. Then after using (1) and (2) to determine the days of the week that the January 1st’s of the years 2000 to 2399 fall on, we obtain |N L0 | = |N L1 | = |N L3 | = |N L5 | = |N L6 | = 43, |N L2 | = |N L4 | = 44, |L0 | = |L5 | = 15, |L1 | = |L4 | = |L6 | = 13, |L2 | = |L3 | = 14.

Now, it must be clear that, for i = 1, 2, 3, the fraction of all years with exactly i Friday the 13th’s is equal to the fraction of the years in one period with exactly i Friday the 13th’s. So |N L3 | + |N L5 | + |N L6 | + |L2 | + |L5 | + |L6 | 400 171 43 + 43 + 43 + 14 + 15 + 13 = . = 400 400

P (X = 1) =

Similar arguments show that, P (X = 2) = 170/400 and P (X = 3) = 59/400. (b) Using the same method, we have that the probability that the 13th of a randomly selected month will be Friday is 688/(400 × 12) = 43/300, where 688 is obtained from the following formula: 2|N L0| + 2|N L1| + 2|N L2 | + |N L3 | + 3|N L4| + |N L5| + |N L6|+ 3|L0 | + 2|L1 | + |L2 | + 2|L3 | + 2|L4 | + |L5 | + |L6 | = 688. (c) The probability that a randomly selected year has two consecutive months with their 13th’s falling on Friday is |N L 4|/400 = 44/400 = 11/100.

The following, a truly challenging problem, is an example of a game in which despite a low probability of winning, the expected length of the play is high.

Example 4g (The Clock Solitaire) An ordinary deck of 52 cards is well shuffled and dealt face down into 13 equal piles. The first 12 piles are arranged in a circle like the numbers on the face of a clock. The 13th pile is placed at the center of the circle. Play begins by turning over the bottom card in the center pile. If this card is a king, it is placed face up on the top of the center pile, and a new card is drawn from the bottom of this pile. If the card drawn is not a king, then (counting the jack as 11 and the queen as 12) it is placed face up on the pile located in the hour position corresponding to the number of the card.

Whichever pile the card drawn is placed on, a new card is drawn from the bottom of that

Section 4A

Additional Examples

9

pile. This card is placed face up on the pile indicated (either the hour position or the center depending on whether the card is or is not a king) and the play is repeated. The game ends when the 4th king is placed on the center pile. If that occurs on the last remaining card, the player wins. The number of cards turned over until the 4th king appears determines the length of the game. Therefore, the player wins if the length of the game is 52. (a)

Find p(j), the probability that the length of the game is j. That is, the 4th king will appear on the jth card.

(b)

Find the probability that the player wins.

(c)

Find the expected length of the game.

Solution: (a)

The sample space has 52! elements because when the cards are dealt face down, any ordering of the cards is a possibility. To find p(j), the probability that the 4th king 4 will appear on the jth card, we claim that in · P · 48! ways the 4th king 1 (j − 1) 3 will appear on the jth card, and the remaining 3 kings earlier. To see this, note that

4 we have combinations for the king that appears on the jth card, and (j − 1) P 3 1 different permutations for the remaining 3 kings that appear earlier. The last term 48!, is for the remaining 48 cards that can appear in any order in the remaining 48 positions. Therefore,

j−1 j−1 4 · (j − 1) P · 48!

3 3 3 1 = . = p(j) = 52!

52 52!

4! 48!

4

(b)

The probability that the player wins is p(52) =

(c)

To find

51 52 = 1/13.

3 4

52 1 j−1 jp(j) = j , E= 52 3 j=4 j=4 4 52

the expected length of the game, we use a technique introduced by Jenkyns and Muller in Mathematics Magazine, 54, (1981), page 203. We have the following relation which can be readily checked.

j j −1 j−1 4 (j + 1) −j , j ≥ 5. j = 5 4 4 3

Section 4A

Additional Examples

10

This gives 52

52 j −1 4 j j −1 j = (j + 1) − j 5 3 4 4 j=5 j=5 j=5

52 4 4 53 −5 = 11, 478, 736, = 5 4 4 52

where the next-to-the-last equality follows because terms cancel out in pairs. Thus

52 52 j−1 1 j−1 1 j = 4+ j E= 52 52 3 3 j=4 j=5 4 4 1 = (4 + 11, 478, 736) = 42.4.

52 4 As Jenkyns and Muller have noted, “This relatively high expectation value is what makes the game interesting. However, the low probability of winning makes it frustrating!”

Section 4B

4B

An Investment Application

11

AN INVESTMENT APPLICATION

In business, financial assets are instruments of trade or commerce valued only in terms of the national medium of exchange (i.e., money) having no other intrinsic value. These instruments can be freely bought and sold. They are, in general, divided into three categories. Those guaranteed with a fixed income or income based on a specific formula are fixed-income securities. An example would be a bond. Those that depend on a firm’s success through the purchase of shares are called equity. An example would be common stock. If the financial value of an asset is derived from other assets or depends on the values of other assets, then it is called a derivative security. For example, the interest rate of an adjustable-rate mortgage depends on a combination of factors concerning various other interest rates that determine an interest rate index, which in turn determines the periodical adjustments to the mortgage rate.

For the purposes of simplicity we are going to assume that the fixed-income securities are in the form of zero-coupon bonds, meaning they pay no annual interest but provide a return to the investor based on the capital gain over the purchase price. Similarly, we will assume that the instruments based on the firm’s success represent businesses that pay no annual dividends but instead promise a return to the investor based on the growth of the company and thus the value of the stock.

Let X be the amount paid to purchase an asset, and let Y be the amount received from the sale of the same asset. Putting fixed-income securities aside, the ratio Y /X is a random variable called the total return and is denoted by R. Obviously, Y = RX. The ratio r = (Y − X)/X is a random variable called the rate of return. If we purchase an asset for $100 and sell it for $120, the total return is 1.2, whereas the rate of return is 0.2. The latter quantity shows that the value of the asset was increased 20%, whereas the former shows that its value reached 120% of its original price. Clearly, r = (Y /X) − 1 = R − 1, or R = 1 + r.

Every investor has a collection of financial assets, which can be kept, added to, or sold. This collection of financial assets forms the investor’s portfolio. Let X be the total investment. Suppose that the portfolio of the investor consists of a total of n financial assets.

Furthermore, let wi be the fraction of investment in the ith financial asset. Then X i = wi X is the amount invested in the ith financial asset, and w i is called the weight of asset i.

Clearly, n

n

Xi = i=1

wi X = X i=1

implies that ni wi = 1. If Ri is the total return of financial asset i, then asset i sells for Ri wiX, and R, the total return is obtained from n

Ri wi X R=

i=1

X

n

wi Ri .

= i=1

(3)

Section 4B

An Investment Application

12

Similarly, r, the rate of return of the portfolio is n

Ri wi X − X r=

i=1

X

n i=1 n

n

wi(Ri − 1) =

= i=1

n

Ri wi X − 1 =

=

n

Ri wi − i=1

wi i=1

wi ri .

(4)

i=1

We have made the following important observations: The total return of the portfolio is the weighted sum of the total returns from each financial asst of the portfolio.

The rate of return of the portfolio is the weighted sum of the rates from the assets of the portfolio.

Putting fixed-income securities aside, for each i, R i and ri are random variables, and we have the following formulas for the expected values of R and r: n

E(R) =

wi E(Ri),

(5)

wi E(ri).

(6)

i=1 n

E(r) = i=1

These formulas are simple results of Theorem 10.1 proved in Chapter 10 of the book. That theorem is a generalization of the corollary following Theorem 4.2 of the book.

Section 4C

4C

Quantiles and the Median

13

QUANTILES AND THE MEDIAN

Let X be a random variable with the distribution function F, then Q p is said to be a quantile of order p of X if P (X ≤ Qp ) ≥ p and P (X ≥ Qp ) ≥ 1 − p.

This is equivalent to P (X < Qp) ≤ p ≤ P (X ≤ Qp ) or F (Qp −) ≤ p ≤ F (Qp ).

Therefore Qp is a value at which either F is p or F jumps from under p to over p. If X is a continuous random variable, and f is its probability density function, then the area under the graph of f to the left of Q p is p and the area to the right of Q p is 1 − p.

Whether X is continuous or discrete, the quantity Q .5 is called the median, Q .25 , the first quartile and Q .75 the third quartile of X. The difference Q .75 − Q.25 is called the interquartile range. If X is continuous, then since half of the area under the graph of f lies to the left of the median and the other half to its right, the median is often used as a center of the distribution.

Quantiles are important measures of location and are very useful both in the analysis of data and random variables. To demonstrate how they are calculated, we will give two examples.

Example 4h. From the Webster’s New World Dictionary we selected 30 words at random and found that 6 of them had 1 vowel, 6 of them 2, 10 of them 3, 6 of them 4, and 2 of them 5 vowels. Let X be the number of vowels in a random word from these 30. Then P (X = 1) = P (X = 2) = P (X = 4) = 6/30, P (X = 3) = 10/30, P (X = 5) = 2/30.

Therefore F, the distribution function of X is given by ⎧ 0 x<1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 6/30 1 ≤ x < 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨12/30 2 ≤ x < 3 F (x) = ⎪ ⎪ 22/30 3 ≤ x < 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 28/30 4 ≤ x < 5 ⎪ ⎪ ⎪ ⎪ ⎩ 1 x ≥ 5.

Now Q.5 , the median of X is a value at which either F is .5 or F jumps from under .5 to over .5. Therefore Q.5 = 3. Note that P (X < 3) ≤ .5 ≤ P (X ≤ 3) is valid only at 3, so the median in this case is unique. However, Q .2 , for instance, is not unique. Q .2 is a value

Section 4C

Quantiles and the Median

14

at which either F is .2 or F jumps from under .2 to over .2. All of the points of the interval [1, 2) have this property. Therefore each of them can be taken as Q .2 .

Example 4i. Let X be the amount of time (in minutes) that a randomly selected customer waits to be served in a restaurant. Suppose that the distribution of X is given by F (t) = 1−e−t/15 . To determine Q.5 , the median of F, note that since F (Q .5 −) ≤ .5 ≤ F (Q.5 ) and F is continuous, we have that F (Q .5 ) = .5. This gives 1 − e −Q.5 /15 = 1/2. or e−Q.5 /15 = 1/2. The solution to this equation is Q .5 = 15 ln 2 ≈ 10.40. Therefore the probability that a customer has to wait more than 10.40 minutes is .5 and less than 10.40 is also .5. In other words, roughly, half of the customers wait more than 10.40 minutes and half of them wait less.

COMPANION FOR CHAPTER 5 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 5 Special Discrete Distributions 5A

Additional Examples

3

3

Section 5A

Additional Examples

3

Chapter 5

Special Discrete Distributions 5A

ADDITIONAL EXAMPLES

Example 5a A realtor claims that only 30% of the houses in a certain neighborhood are appraised at less than $200,000. A random sample of 20 houses from that neighborhood is selected and appraised. The results in (thousands of dollars) are as follows: 285 156 202 306 276 562 415 245 185 143 186 377 225 192 510 222 264 198 168 363 Based on these data, is the realtor’s claim acceptable?

Solution: Suppose that the realtor’s claim is acceptable, and p = 0.30 is the probability that a randomly selected house is appraised at less than $200,000. In a randomly selected sample of 20 houses, let X be the number that are less than $200,000. Since in the data given there are seven such houses, we will calculate P (X ≥ 7). If this probability is too small, we will reject the claim of the realtor; otherwise, we will accept it. To find P (X ≥ 7), note that X is binomial with parameters 20 and 0.30. Therefore, P (X ≥ 7) = 1 − P (X ≤ 6) = 1 −

6

20 i=0

i

(0.30)i(0.70)20−i ≈ 0.392,

which shows that the event that seven or more houses are appraised at less than $200,000 is highly probable. So the claim of the realtor should be accepted.

Example 5b For a manufacturer of Go-Onto-Location-in-Space (GOLIS) missiles, there are two companies that build missile trackers for guided missiles that are designed for moving targets. The manufacturers purchase 40% of their missile trackers from company I and the remaining 60% from company II. Suppose that 8.7% of the missile trackers designed by company I and 7.3% of the missile trackers designed by company II are defective. In a GOLIS missile test, a batch of 100 such missiles, all equipped with missile trackers designed

Section 5A

4

Additional Examples

by the same company, were launched attacking moving targets. If 9 of the missiles failed to reach their intended targets, all because of the defects in their missile trackers, what is the probability that the batch of 100 missiles were all manufactured by company I? Hint: Use Bayes’ formula.

Solution: Let X be the number of missiles that failed to reach their intended targets. Let A be the event that the batch of 100 missiles all had missile trackers designed by company I. We are interested in P (A | X = 9). By Bayes’ formula, P (X = 9 | A)P (A) P (X = 9 | A)P (A) + P (X = 9 | Ac )P (Ac )

100 (0.087)9(1 − 0.087)91 · 0.4 9

≈ 0.81. = 100 100 9 91 (0.073)9(1 − 0.073)91 · 0.6 (0.087) (1 − 0.087) · 0.4 + 9 9

P (A | X = 9) =

Example 5c In a certain population, one percent of the people are colorblind. In order to calculate the probability of exactly i colorblind people among a random sample of n, we use the binomial probability function with parameters (n, 0.01) and let

n (0.01)i(0.99)n−i . bi = i Then we use Poisson probability function with parameter λ = n(1/100) = n/100 and let e−(n/100)(n/100)i . i!

Both of these can be used only as approximations. They are not exact, since in binomial (and hence in Poisson approximations) the “trials” are independent. Here the “trials” are dependent, because the number of colorblind people in the population is fixed. Therefore, for, say, 0 < i, j < n, the probability that the jth person in the sample is colorblind depends on whether the ith person is colorblind or is not. The probability that the jth person is colorblind is less if the ith person is colorblind than if he or she is not colorblind. However; if n is not too small, then this dependence is very weak, and both binomial and Poisson are good approximations.

For different values of n and i, the following tables show how b i’s and pi ’s are in relative agreement. pi =

(i) n = 20 i 0 1 2 3 4

bi .8180 .1652 .0159 .0010 .00004

pi .8187 .1638 .0164 .0011 .00005

Section 5A

Additional Examples

5

(ii) n = 30 i 0 1 2 3 4 5

bi .7397 .2242 .0328 .0031 .0002 .00001

pi .7408 .2222 .0333 .0033 .0003 .00002

i 0 1 2 3 4 5 6

bi .6050 .3056 .0757 .0122 .0015 .00013 .00001

pi .6065 .3033 .0759 .0126 .0016 .00016 .00001

(iii) n = 50

Example 5d The probability of winning a certain game is p and the probability of losing it is 1 − p. Players A and B keep playing this game until one of them wins n games. The games are independent and the person who wins n games first is the winner. Suppose that after k + r (k < n, r < n) games A has won k games and B has won r games. What is the probability that A wins?

Solution: In order that A wins, he must win n − k games before losing n − r. This happens if and only if the (n − k)th win of A occurs before or at the (n − r) + (n − k) − 1 = 2n − (r + k + 1)th game. This follows since if the (n − k)th win for A occurs after the 2n − (r + k + 1)th game, then B wins more than n − r games that he needs. Let X be the number of games required that A wins (n − k) games. Then X is a negative binomial with parameter (n − k, p). The desired probability is 2n−(r+k+1)

2n−(r+k+1)

i=n−k

P (X = i) =

i=n−k

i−1 pn−k (1 − p)i−n+k . n−k−1

Example 5e Peggy and Norma are two basketball players. The probability that Peggy makes a foul shot, independently of her other shots, is p. The probability that Norma makes a foul shot is also p, independently of her other shots. Peggy shoots free throws until a successful one. Then she gives the ball to Norma, who shoots free throws until she makes a foul shot too. After that they both quit shooting basketball shots. If Peggy and Norma together attempted 8 shots before quitting, for 1 ≤ j ≤ 7, what is the probability that Peggy made her foul shot on her jth attempt, j = 1, 2, . . ., 7?

Section 5A

Additional Examples

6

Solution: Let X be the number of times Peggy attempts until she makes a foul shot. Let Y be the number of times Norma attempts until she makes a foul shot. Since X and Y are independent and identically distributed, by symmetry, we must have 1 P (X = j | X + Y = 8) = , 7

j = 1, 2, . . .7.

Here is a formal proof: P (X = j | X + Y = 8) =

P (X = j, Y = 8 − j) P (X = j, X + Y = 8) = 7 P (X + Y = 8) i=1 P (X = i, Y = 8 − i)

P (X = j)P (Y = 8 − j) = 7 i=1 P (X = i)P (Y = 8 − i) (1 − p)j−1 p · (1 − p)8−j−1 p = 7 i−1 p · (1 − p)8−i−1 p i=1 (1 − p) 1 (1 − p)6 p2 (1 − p)6 p2 = . = = 7 7 6 2 6 2 7 (1 − p) p i=1 (1 − p) p i=1 1 Example 5f Let X be a Poisson random variable with parameter λ > 0. Show that (a)

for n > λ − 1, P (X ≥ n) <

(b)

n > 2λ − 1,

Solution: (a) P (X ≥ n) = =

n+1 P (X = n); n+1−λ

P (X > n) < P (X = n).

We have that ∞ −λ

e · λi

i!

i=n e−λ · λn

n!

1+

λ λ2 λ3 + + +··· . n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)

Now λ2 λ2 < (n + 1)(n + 2) (n + 1)2 λ3 λ3 < (n + 1)(n + 2)(n + 3) (n + 1)3 .. .

Section 5A

Additional Examples

7

Thus ∞

P (X ≥ n) <

λi e−λ · λn n! (n + 1)i i=0 ∞

= P (X = n)

i=0

λ n+1

i

∞

i λ λ < 1, and is a geometric series that n+1 n+1 i=0 n+1 = . Therefore, for n > λ − 1, n+1−λ

Now, for n > λ − 1, we have 1

converges to 1−

λ n+1

P (X ≥ n) < (b)

n+1 P (X = n). n+1−λ

Note that, by the result of part (a), n+1 P (X = n) − P (X = n) P (X > n) = P (X ≥ n) − P (X = n) < n+1−λ n+1 λ − 1 P (X = n) = P (X = n). = n+1−λ n+1−λ

This shows that P (X > n) < P (X = n) if

λ < 1 or if n > 2λ − 1. n+1−λ

Example 5g Suppose that, from a box containing D defective and N − D non-defective items, n n ≤ min(D, N − D) are drawn at random and without replacement. Let X be the number of defective items drawn. Then X is a hypergeometric random variable. (a)

Prove the following recurrence relation for X. P (X = x + 1) =

(n − x)(D − x) P (X = x), (x + 1)(N − D − n + x + 1)

where max 0, n − (N − D) ≤ x ≤ min(n, D). (b)

Using part (a), calculate the probability mass function of the number of aces in a bridge hand. A bridge hand consists of 13 randomly selected cards from an ordinary deck of 52 cards.

Solution: (a)

Note that

P (X = x + 1) = P (X = x)

D N −D N x+1 n−x−1 n

D N −D N x n−x n

Section 5A

=

Additional Examples

8

(N − D)!

D! · (x + 1)! (D − x − 1)! (n − x − 1)! (N − D − n + x + 1)!

(N − D)!

D! · x! (D − x)! (n − x)! (N − D − n + x)!

=

x! (D − x)! (n − x)! (N − D − n + x)! · (x + 1)! (D − x − 1)! (n − x − 1)! (N − D − n + x + 1)!

=

n−x D−x · . x+1 N −D−n+x+1

This establishes the relation. (b) Let X be the number of aces in a bridge hand. Clearly, X is a hypergeometric random variable with parameters N = 52, D = 4, and n = 13. Let p be the probability mass function of X. Clearly,

4 48 0 13 p(0) = P (X = 0) = ≈ 0.3038.

52 13 Now by part (a), p(1) =

52 13 × 4 p(0) ≈ × 0.3038 ≈ 0.4388, 1 × 36 36

p(2) =

36 12 × 3 p(1) ≈ × 0.4388 ≈ 0.2135, 2 × 37 74

p(3) =

22 11 × 2 p(2) ≈ × 0.2135 ≈ 0.0412, 3 × 38 114

p(4) =

10 10 × 1 p(3) ≈ × 0.0412 ≈ 0.0026.

4 × 39 156

Example 5h Suppose that independent Bernoulli trials with parameter p, 0 < p < 1, are performed and X denotes the number of trials until the rth success. In this example, we explain the reason why X is called negative binomial. The advantage of this explanation is that it also enables us to calculate the expectation of X and prove analytically that, for 0 < p < 1,

x−1 r p(x) = P (X = x) = p (1 − p)x−r , , x = r, r + 1, r + 2, . . . , r−1

n is a probability mass function. In Chapter 2 of the book, the notion of is defined by r

n n!/[r! (n − r)!] for positive integers n and non-negative integers r. The same notion, , r

Section 5A

Additional Examples

9

can also be defined for real numbers n and non-negative integers r by

n n(n − 1)(n − 2) · · · (n − r + 1) , = r! r which obviously agrees with n!/[r! (n − r)!] in the case of positive integer n. Now let Y be the number of failures before the rth success. Then X = Y + r, and, for y = 0, 1, 2, . . .,

y+r−1 r (1) P (Y = y) = P (X = y + r) = p (1 − p)y . r−1 But

y+r−1 (y + r − 1)! r(r + 1)(r + 2) · · · (y + r − 1) = = r−1 (r − 1)! y! y!

(−1)y − r(−r − 1)(−r − 2) · · · (−r − y + 1) = y!

−r . = (−1)y y

Therefore,

−r r −r r y p (p − 1)y , p (1 − p) = P (Y = y) = (−1) y y y

y = 0, 1, 2, . . .

It is because of this relation that X is called “negative binomial”. Using this relation, we will now show that p(x) is a probability function: ∞

p(x) =

x=r

∞

P (X = x) =

x=r

=

P (Y = x − r) =

x=r

∞

−r y=0

∞

y

∞

P (Y = y)

y=0

∞

−r p (p − 1) = p (p − 1)y y r

y

r

y=0

= pr (1 + p − 1)−r = 1, where the next to the last equality comes from calculus: If r is any real number, then for ∞

r y r t . |t| < 1, (1 + t) = y y=0

Another application of relation (1) is that it enables us to find E(X) easily. Note that E(X) =

∞

xp(x) =

x=r

∞

xP (Y = x − r)

x=r

∞ ∞ ∞

(y + r)P (Y = y) = yP (Y = y) + r P (Y = y) = y=0

= E(Y ) + r =

y=0

r r(1 − p) +r= , p p

y=0

Section 5A

where E(Y ) =

Additional Examples

10

∞

−r r y p (p − 1)y = r(1 − p) /p follows from a straightforward y y=0

procedure similar to the one we used to calculate the expected value of a binomial random variable. Note that X is the number of trials until the rth success. Since the average number of trials until the first success is 1/p (the expected value of a geometric random variable with parameter p is 1/p), we should expect that the average number of trials until the rth success be r/p. This is the intuition behind the relation E(X) = r/p obtained above. In Chapter 10 of the book, we have calculated E(X) using a simpler method.

Example 5i (Genetics) In a population of n diploid organisms with alternate dominant allele A and recessive allele a, those inheriting aa will not survive. Suppose that, in the population, the number of AA individuals is α, and the number of Aa individuals is n − α. Suppose that, as a result of random mating, m (m > 2) offspring are produced. (a)

What is the probability that at most two of the offspring are aa?

(b)

For 0 ≤ i ≤ m, what is the probability that exactly i of the offspring are AA and the remaining are all Aa?

Solution: Let p, q, and r be the probabilities that a randomly selected offspring is AA, Aa, and aa, respectively. Note that both parents of the offspring are AA with proba 2 bility (α/n) 2 , they are both Aa with probability 1 − (α/n) , and the probability is 2(α/n) 1 − (α/n) that one parent is AA and the other is Aa. Therefore, by the law of total probability, p=1· q =0· r =0·

α n α n α n

α 2 1 α α 1 α 2 1 α 1 1 · 1− 1− = + , + ·2 + 4 n 2 n n 4 n 2 n 4 2 α 2 1 α α 1 1 α 2 1 1− 1− = − + + ·2 , 2 n 2 n n 2 2 n α 2 α 2 α 1 α 2 1 1− 1− = 1− + +0·2 .

4 n n n 4 n 2

+

The probability that at most two of the offspring are aa is 2

m i=0

i

r i (1 − r)m−i .

The probability that exactly i of the offspring are AA and the remaining are all Aa is

m i m−i .

pq i

COMPANION FOR CHAPTER 6 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 6 Continuous Random Variables 6A

Additional Examples

3

3

Section 6A

Additional Examples

3

Chapter 6

C ontinuous R andom Variables 6A

ADDITIONAL EXAMPLES

Example 6a Let X be a random variable (discrete or continuous) that satisfies 0 ≤ X ≤ 1 with probability 1. Show that Var(X) ≤ 1/4.

Solution: Let μ = E(X). Since 0 ≤ X ≤ 1, we have that X 2 ≤ X, which yields E(X 2) ≤ E(X). Thus

2

2 Var(X) = E(X 2) − E(X) ≤ E(X) − E(X) = μ − μ2 .

Now the function g(μ) = μ − μ 2 is maximum for a value of μ for which g (μ) = 0, or 1 − 2μ = 0. Thus g is maximum at μ = 1/2. The value of g at μ = 1/2 is g(1/2) = 1/2 − 1/4 = 1/4. So μ − μ2 ≤ 1/4 for all values of μ. Therefore, Var(X) ≤ μ − μ2 ≤

1 .

4

Example 6b Let X be a continuous random variable with probability density function ⎧ if −4 < x < 8 ⎨c|x − 2| f (x) = ⎩ 0 elsewhere. (a)

Calculate the value of c.

(b)

Find the distribution function of X.

(c)

Find E(X).

Solution: (a)

We must have 8 2 (2 − x) dx + c (x − 2) dx = 1. c −4

2

Section 6A

Additional Examples

This gives 36c = 1, or c = 1/36. (b)

Note that

⎧ 1 ⎪ ⎪ (2 − x) ⎪ ⎪ 36 ⎪ ⎪ ⎨ 1 f (x) = (x − 2) ⎪ ⎪ 36 ⎪ ⎪ ⎪ ⎪ ⎩ 0

−4 < x < 2 2≤x<8 otherwise.

Let F be the distribution function of X. For −4 ≤ t < 2, t 1 4 1 1 F (t) = (2 − x) dx = − t2 + t + , 72 18 9 −4 36 and, for 2 ≤ t < 8, t 2 1 5 1 1 1 (2 − x) dx + (x − 2) dx = t2 − t + .

F (t) = 72 18 9 −4 36 2 36 ⎧ ⎪ 0 t < −4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 4 1 1 2 ⎪ ⎪ ⎪ −4 ≤ t < 2 ⎨− 72 t + 18 t + 9 F (t) = ⎪ 5 1 1 2 ⎪ ⎪ t − t+ 2≤t<8 ⎪ ⎪ ⎪ 72 18 9 ⎪ ⎪ ⎪ ⎪ ⎩ 1 t ≥ 8.

8 2 1 1 (2 − x) dx + (x − 2) dx = 2. x· x· E(X) = 36 36 −4 2

Hence

(c)

Example 6c

Let X be a continuous random variable with probability density function ⎧ 2 ⎪ ⎪ (x − 2) 2<x≤3 ⎪ ⎪3 ⎪ ⎪ ⎨ f (x) = 1 (5 − x) 3<x<5 ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎪ ⎩ 0 otherwise.

(i)

Find F, the distribution function of X.

(ii)

Calculate E(X), Var(X), and E |X − 4| .

Solution: (i) For 2 ≤ t < 3, F (t) =

t

1 2 (x − 2) dx = (t2 − 4t + 4).

3 2 3

4

Section 6A

Additional Examples

For 3 ≤ t < 5, F (t) = = Therefore,

3 2

2 (x − 2) dx + 3

1 + 3

t

1 (5 − x) dx 3 3

1 7 1 5 − t2 + t − = (−t2 + 10t − 19).

6 3 2 6

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 2 ⎪ ⎪ ⎪ ⎨ 3 (t − 4t + 4) F (t) = ⎪ ⎪1 ⎪ (−t2 + 10t − 19) ⎪ ⎪ ⎪ 6 ⎪ ⎪ ⎪ ⎪ ⎩ 1

t<2 2≤t<3 3≤t<5 t ≥ 5.

(ii) We have 3

2 x · (x − 2) dx + E(X) = 3 2

5

1 x · (5 − x) dx 3 3

10 8 22 + = ≈ 3.33.

9 9 3 5 3 2 1 x2 · (x − 2) dx + x2 · (5 − x) dx E(X 2) = 3 3 2 3 =

23 43 82 + = .

18 9 2

2 23 100 7 − = ≈ 0.389.

Var(X) = E(X 2) − E(X) = 2 9 18 ∞ |x − 4|f (x) dx E |X − 4| = =

−∞

=

3

2 (4 − x) · (x − 2) dx + 3 2

5

4

1 (4 − x) · (5 − x) dx+ 3 3

4 5 1 7 1 + = ≈ 0.778. (x − 4) · (5 − x) dx = + 3 9 18 18 9 4

Example 6d For α > 0, β > 0, let ⎧ ⎨(e−αx − e−βx )/x f (x) = ⎩ 0 Under what condition(s) is f a density function?

if x > 0 if x ≤ 0.

5

Section 6A

Additional Examples

6

Solution: f (x) ≥ 0 implies that we must have e −αx > e−βx , which requires α < β. We must also have that ∞ −αx e − e−βx dx = 1.

(1) x 0 To see under what conditions this is true, let ∞ −tx −αx e (e − e−βx ) dx, H(t) = x 0+

t ≥ 0.

It is easily seen that H is differentiable and

H (t) = =

∞ 0

−βx

(e

−αx

−e

−tx

)e

e−(α+t)x e−(β+t)x − dx = α+t β +t

∞ 0

1 1 − . β+t α+t

Thus, H(t) = ln

β+t + c. α+t

But since limt→∞ H(t) = 0, this relation gives 0 = ln 1 + c or c = 0, so H(t) = ln Now

β +t . α+t

∞

β e−αx − e−βx dx = H(0) = ln . x α 0+

Therefore, for the validity of (1), we must have ln(β/α) = 1, or β = eα. If β = eα, then α < β is also true and f would be a density function.

Example 6e is given by

A random variable X is said to be logistic if its probability density function f (x) =

ex , (1 + ex )2

−∞ < x < ∞.

(a)

Find the distribution function of a logistic random variable X.

(b)

Let Y be a random number between 0 and 1. Show that ln

Solution: (a)

Y 1−Y

is logistic.

Let F be the distribution function of X. For −∞ < t < ∞, t ex dx.

F (t) = x 2 −∞ (1 + e )

Section 6A

Additional Examples

7

To calculate this integral, we make the substitution u = 1 + e x . We have du = ex dx.

Therefore, 1+et 1 1+et 1 et 1 du = − = 1 − = , −∞ < t < ∞.

F (t) = u2 u 1 1 + et 1 + et 1 (b)

We will show that the distribution function of ln

shows that ln

P

Y 1−Y

Y ln 1−Y

Y 1−Y

is

et . By part (a), this 1 + et

is logistic. We have

≤t =P

Y ≤ et = P Y ≤ (1 − Y )et 1−Y

=P Y ≤

et 1 + et

et −0 et 1 + et = . = 1 + et 1−0

Example 6f Let X be a continuous random variable with distribution function F and density function f . Find the distribution and the density functions of Y = |X|.

Solution: Since Y ≥ 0, for t < 0, P (Y ≤ t) = 0. For t ≥ 0, P (Y ≤ t) = P |X| ≤ t = P (−t ≤ X ≤ t) = P (X ≤ t) − P (X < −t) = P (X ≤ t) − P (X ≤ −t) = F (t) − F (−t).

Hence G, the distribution function of |X| is given by

0 G(t) = F (t) − F (−t)

if t < 0 if t ≥ 0

and so g, the density function of |X| is obtained by differentiation of G:

f (t) + f (−t) if t ≥ 0

g(t) = G (t) = 0 if t < 0.

Example 6g

Let X be a continuous random variable with the following density function.

1/2 if x ∈ (−1, 1) fX (x) = 0 otherwise.

Using the method of transformations, find the density function of Y = e X .

Solution: The set of possible values of X is A = (−1, 1). Let h : (−1, 1) → R be defined by h(x) = ex . We want to find the density function of Y = h(X) = e X . The set of possible values of h(X) is B = {h(a) : a ∈ A} = (1/e, e). The function h is invertible

Section 6A

Additional Examples

8

with the inverse x = g(y) = ln y which is differentiable and its derivative is g (y) = 1/y.

Therefore, by Theorem 6.1 of the book,

1 1 1

, y ∈ (1/e, e), fY (y) = fX (g(y))|g (y)| = = 2 y 2y is the density function of Y = e X .

COMPANION FOR CHAPTER 7 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 7 Special Continuous Distributions 7A 7B 7C 7D 7E

Additional Examples 3 Black-Scholes Call Option Formula The Cauchy Distribution 10 The Lognormal Distribution 12 The Hyperexponential Distribution

8

15

3

Section 7A

Additional Examples

3

Chapter 7

Special C ontinuous Distributions 7A

ADDITIONAL EXAMPLES

Example 7a In a measurement, a number is rounded off to the nearest k decimal places.

Let X be the rounding error. Determine the distribution function of X and its parameters.

Solution: Round off error to the nearest integer is uniform over (−0.5, 0.5); round off error to the nearest 1st decimal place is uniform over (−0.05, 0.05); round off error to the nearest 2nd decimal place is uniform over (−0.005, 0.005), and so on. In general, round off error to the nearest k decimal places is uniform over (−5/10 k+1 , 5/10k+1 ). Therefore, the distribution function of X is ⎧ ⎪ 0 t < −5/10k+1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 5 ⎪ ⎪ ⎨ t + k+1 1 10 −5/10k+1 ≤ t < 5/10k+1 = 10k t + F (t) = 5 5 ⎪ 2 ⎪ + ⎪ ⎪ ⎪ 10k+1 10k+1 ⎪ ⎪ ⎪ ⎪ ⎩ 1 t ≥ 5/10k+1 .

Example 7b As discussed in Section 4.5 of the book, for a random variable X, the true

average magnitude of the fluctuations of X from E(X) is E |X − E(X)| . However,

since, mathematically, this quantity is difficult to handle, the quantity E (X − E(X))2 is used and is called the standard deviation of X. Let X be a normal random variable with mean μ and standard deviation σ. Find E |X − μ| , the true average magnitude of the fluctuations of X from μ.

Solution: Clearly,

E |X − μ| =

∞

−(x − μ)2 1 |x − μ| √ exp dx.

2σ 2 σ 2π −∞

Section 7A

Additional Examples

4

Let x − μ = t; then dx = dt, and −t2 1 |t| √ exp dt.

2σ 2 σ 2π −∞ −t2 1 Since the function g(t) = √ |t| exp is an even function; that is g(−t) = g(t), 2σ 2 σ 2π we have −t2 −t2 ∞ ∞

1 2 √ |t| √ exp t exp dt = dt.

E |X − μ| = 2 2σ 2 2σ 2 σ 2π σ 2π 0 0

E |X − μ| =

∞

1 t2 , we have du = 2 t dt. So 2 2σ σ

∞ ∞

1 2 2 2 −u 2 −u −e . e σ du = σ =σ E |X − μ| = σ π 0 π π 0

By making the substitution u =

Example 7c Code 1 for “yes,” and code 0 for “no” are transmitted through a noisy channel regularly. However, the quality of signals and data are degraded by the noise that the channel adds. Suppose that when a code X is sent, independently of other transmissions, another code X + is received, where ∼ N (0, 1/16). If all the codes that are received and are less than or equal to 1/2 are interpreted as 0, and the ones that are received and are greater than 1/2 are interpreted as 1, what is the probability that the code 11001001 sent is decoded correctly when transmitted?

Solution: If the transmitted code sent is 0, when received, it will be interpreted as 0 with probability P ( ≤ 1/2). If the transmitted code is 1, when received, it will be interpreted as 1 with probability P ( > −1/2). Since ∼ N (0, 1/16), we have

1/2 − 0 1 =P Z≤ = P (Z ≤ 2) = Φ(2) = 0.9772; P ≤ 2 1/4

−1/2 − 0 1 =P Z> = P (Z > −2) = 1 − Φ(−2) = Φ(2) = 0.9772.

P >− 2 1/4 Therefore, when transmitted, each of the codes 0 and 1 is received and interpreted correctly with probability 0.9772, independently of other codes. Hence the probability that 11001001 is transmitted and interpreted as 11001001 is (0.9772) 8 ≈ 0.83.

Example 7d At a prime location, due to the shortage of homes for sale, some homebuyers make offers above the asking prices. Morgan has put her house on the market and has decided to sell it to the first person who offers her at least $350,000. If the offers are exponential random variables with mean $320,000, independently of one another, on the average, how many offers will Morgan receive until she sells her house?

Solution: Call an offer a “success” if it is at least $350,000. Let X be a random offer, in tens of thousands dollars. Then the probability of “success” is p = P (X ≥ 3.5) =

∞

1 −x/3.2 e dx ≈ 0.335.

3.5 3.2

Section 7A

Additional Examples

5

We are interested in the expected value of the number of offers until the first “success.” That 1 1 ≈ 2.99. is a geometric random variable with mean ≈ p 0.335 Example 7e

The hazard rate of device A is given by ⎧ 0 < t < 10 years ⎨t/10 λ(t) = ⎩ 10 t ≥ 10 years.

(i)

What is the probability that device A having lasted for 7 years lasts at least another 2 years?

(ii)

The failure rate of device B is twice the rate at which A fails. What is the probability that device B having lasted 7 years lasts at least another 2 years?

Solution: Let X and Y be the lifetimes of devices A and B, respectively. Then using (7.6), the answer to (i) is

9 t exp − dt P (X ≥ 9) 0 10 = P (X ≥ 9 | X ≥ 7) = ≈ 0.202.

7 P (X ≥ 7) t dt exp − 0 10 Similarly, the answer to (ii) is obtained as follows.

exp − P (Y ≥ 9) = P (Y ≥ 9 | Y ≥ 7) =

P (Y ≥ 7) exp −

t dt 0 5 ≈ 0.041.

7 t dt 0 5 9

Therefore, device B has a much lower chance of surviving 2 or more years than device A. This is because device B has a failure rate that is much larger than the failure rate of A. Example 7f Let X be an exponential random variable with parameter θ > 0. Let ⎧ with probability 1/2 ⎨X Y = ⎩ −X with probability 1/2 Find the probability density function of Y .

Solution: Let f and F be the density and distribution functions of X, respectively. Let g and G be the density and distribution functions of Y, respectively. By the law of total probability, we have

Section 7A

Additional Examples

6

G(t) = P (Y ≤ t) = P (Y ≤ t | Y = X)P (Y = X) + P (Y ≤ t | Y = −X)P (Y = −X) 1 1 + P (−X ≤ t) · 2 2 ⎧ 1 1 ⎪ ⎪ t≥0 ⎨ 2 P (X ≤ t) + 2 = ⎪ ⎪ ⎩ 1 P (X ≥ −t) t≤0 2 ⎧ 1 1 ⎪ ⎪ t≥0 ⎨ 2 P (X ≤ t) + 2 = ⎪ ⎪ ⎩ 1 − 1 P (X ≤ −t) t≤0 2 2 ⎧ 1 1 ⎪ ⎪ t≥0 ⎨ 2 F (t) + 2 = ⎪ ⎪ ⎩ 1 − 1 F (−t) t≤0 2 2

= P (X ≤ t) ·

Differentiating both sides of this relation, we obtain ⎧ 1 ⎪ ⎪ t≥0 ⎨ 2 f (t) g(t) = ⎪ ⎪ ⎩ 1 f (−t) t≤0 2 Therefore,

1 1 f (|t|) = θe−θ|t| − ∞ < t < ∞.

2 2 So Y has Laplace distribution ( See Exercise 16 of Section 7.3 of the book). g(t) =

Example 7g Show that ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎨ √ 2 F (t) = arcsin( t) ⎪ π ⎪ ⎪ ⎪ ⎪ ⎩ 1

t<0 0≤t<1 t≥1

is the distribution function of a beta random variable with parameters (1/2, 1/2).

Solution: Let f be the probability density function of a random variable X with distribution function F . Then ⎧ 2 1 1 1 ⎪ ⎪ 0<t<1 · √ · = ⎪ √ ⎨π 2 t π t(1 − t) 2 1 − ( t)

f (t) = F (t) = ⎪ ⎪ ⎪ ⎩0 otherwise.

Section 7A

Since, for 0 < t < 1,

if we show that Γ(1/2) = f (t) =

√

Additional Examples

7

1 f (t) = √ √ t−1/2 (1 − t)−1/2 , π π π, then we have

1 Γ(1/2 + 1/2) −1/2 t t−1/2 (1 − t)−1/2 , (1 − t)−1/2 = Γ(1/2)Γ(1/2) B(1/2, 1/2)

which is the probability density function of a beta random variable with parameters (1/2, 1/2). √ What is left is to show that Γ(1/2) = π. Note that, Γ(1/2) =

∞

t−1/2 e−t dt.

0

Making the substitution t = y 2 /2, we get 1 √ = 2 Γ 2 =

√

∞

−y2 /2

e 0

1 π·√ 2π

√ dy =

∞ −∞

2

2 2

∞

2

e−y /2 dy

−∞

e−y /2 dy =

√

π.

Section 7B

7B

Black-Scholes Call Option Formula

8

BLACK-SCHOLES CALL OPTION FORMULA

To show the wide range of applications of normal random variables, in the next example, we will present, without proof, a celebrated formula in finance which is constructed based on normal distribution function.

In natural processes, the changes that take place are often the result of a sum of many insignificant random factors. Although the effect of each of these factors, separately, may be ignored, the effect of their sum in general may not. Therefore, the study of distributions of sums of a large number of independent random variables is essential and is the subject of Chapter 11 of the book. In Section 11.5, we have shown that for an immense number of random phenomena, the distribution functions of such sums are approximately normal.

In finance, random fluctuations of prices of shares of stocks, or other assets, are a result of an enormity of political, social, economical, and financial factors independently assessed by a large number of traders. For that reason financial scholars and consultants use normal distributions as models for prices of assets in general, and shares of stocks in particular.

That is why the Black-Scholes call option formula, presented next, is based on normal distribution function.

In finance, a contract that entitles a person to purchase a commodity is called call option. To explain the concept of call option, we will present an example. Suppose that Henry purchases, for a price of $7 per share, the option of buying a block of 100 shares in a company at $80 per share during, say, the next six months. That is, Henry pays $700 to a seller and buys the right of purchasing 100 shares in the company at any time before or including the agreed expiration day at $80 per share. Henry is not obliged to purchase any number of the 100 shares he simply has the option to purchase. However, the premium paid to buy call options is not recovered whether or not he buys any number of the shares. In the language of finance, Henry is the option holder, the seller is the writer, and the predetermined price per share of $80, which is valid until the expiration day is called strike price. If Henry actually does buy a block of at least one share in the company, then we say that he exercises the option.

Suppose that sometime before the expiration day, the price of the shares in the company rise to, say, $95 per share. If at that time Henry purchases 100 shares for $80 per share, and sells them immediately for $95, he makes a profit of $9500 − $8000 − $700 = $800.

On the other hand, if the price does not rise by the expiration day, Henry will lose $700 for having purchased the call option to buy the block of 100 shares. As for the writer, in the latter case, she makes a profit of $700, whereas in the former case, she will end up selling the block of 100 shares, which is worth $9500, for only $8000, losing an opportunity to make $1500 − $700 = $800 more.

From this example, it should be evident that determining a fair price for an option is a major problem in option theory. The question is whether or not the right to buy a share for a price of $7 is fair to both Henry and the writer. Until 1973, it was widely believed that the call option should be determined based on the principle assumption that the average return from the stock should be lower than the average return from the option itself. However, in 1973, Fisher Black, an applied mathematician and a financial consultant, and MIT professor Myron Scholes presented a paper with a new approach. They argued, among other points, that all risk will be eliminated if option holders or writers chose two lines of investments

Section 7B

Black-Scholes Call Option Formula

9

with identical returns, buy one and sell the other simultaneously on the expiration day.

Based on such assumptions, in terms of normal distribution function, they derived formulas for call options that are extremely popular. To state a version of Black-Scholes’ widely used call option formula, note that, since periods remaining to expiration are usually short, those stocks with higher standard deviation will rise above strick price with higher probability.

Therefore, they should have higher option values. Interest rate also has an effect on the value of an option. Usually, as interest rate increases, on average, the growth rate of stock price also increases. This should in turn increase the prices of options. Since dividends essentially reduce the stock price, they have negative effect on the values of the call option.

For this reason, in the next formula we assume that no dividends are paid before or on the expiration day.

We are now ready to state the Black-Scholes call option formula for European options, in which, unlike American options, the option holder can only exercise on the expiration day. The appearance of the normal distribution function in the following formula has something to do with the fact that the number of upward movements in price of a share of a stock is binomial, and the limiting behavior of binomial is normal.

Black-Scholes Call Option Formula: In a European call option with standard deviation σ, let K be the strike price, S be the current price, and τ denote the time period over which the option is valid. Suppose that interest is compounded continuously at a rate r. Then the Black-Scholes formula for the value of call option is given by √ C = SΦ(α) − Ke−rτ Φ(α − σ τ ), where α=

ln(S/K) + (r + σ 2 /2)τ √ , σ τ

and, as usual, Φ is the distribution function of a standard normal random variable.

To find the Black-Scholes call option value, per share, for the block of 100 shares of the stock Henry is interested in, suppose that the current price of the stock, per share, is $85 with standard deviation of 18% per year, and the option is valid for 6 months. The strike price is $80 and the interest rate is 12%. Therefore, S = 85, K = 80, σ = 0.18, r = 0.12, and τ = 6/12 = 1/2. Hence α=

ln(85/80) + (0.12 + 0.182/2)(1/2)

= 1.01, 0.18 1/2

and thus the value of the call option is

C = 85Φ(1.01) − 80e−(0.12)(1/2)Φ 1.01 − 0.18 1/2 = 85Φ(1.01) − 75.34Φ(0.88) = 85(0.8438) − 75.34(0.8106) = 10.65.

Section 7C

7C

The Cauchy Distribution

10

THE CAUCHY DISTRIBUTION

A random variable X is called Cauchy ∗ with parameter α, if its probability density function is given by 1 , −∞ < x < ∞.

(1) f (x) = π [1 + (x − α)2 ] As is shown in Example 6.9 of the book, an interesting property of the Cauchy distribution with parameter 0 is that, although it is symmetric with respect to the origin, E(X) does not exist for a Cauchy random variable X; it has no expected value.

Consider the Cauchy density function f given by (1). The graph of f is symmetric about α, and the parameter α is both the unique median and the unique mode of f . Moreover, the graph of f resembles that of a normal density. For example, the graph of f with α = 0 is very similar to that of the probability density function of the standard normal random variable. While it is flatter than the graph of the density of standard normal, it is not as flat as the density of a normal random variable with the mean 0 and variance 2 (see Figure 7a).

Figure 7a

Cauchy density function with α = 0.

Example 7h. A vertical mirror freely rotates about 0, the origin of the coordinate system.

It projects a horizontal light ray on a wall that has equation x = 1. The direction of the reflected ray is a random number. That is, Θ, the angle between the reflected ray and the x-axis, is a uniform random variable over (−π/2, π/2). Let A be the point (1, 0), B be the intersection of the reflected light ray with the wall, and X be the distance from A to B; that is, X = AB (see Figure 7b). Then X is the random variable tan Θ. To find F, the distribution function of X, note that since Θ is uniform over (−π/2, π/2), P (Θ ≤ t) = ∗

1 1 t − (−π/2) = + t, π/2 − (−π/2) 2 π

−

π π <t< .

2 2

Augustin Louis Cauchy (1789-1857), born in Paris, was one of the leaders in infusing rigor into Analysis.

He is considered to be one of the greatest mathematicians of the 19th century.

Section 7C

The Cauchy Distribution

Since y = tan x is an increasing function over the interval (−π/2, π/2), F (t) = P (X ≤ t) = P (tan Θ ≤ t) = P (Θ ≤ arctan t) =

1 1 + arctan t, 2 π

−∞ < t < ∞.

The density function of X, f, is obtained by differentiation of F : f (t) = F (t) =

1 , π(1 + t2 )

This shows that X is a Cauchy random variable.

Figure 7b

−∞ < t < ∞.

Reflected light ray of Example 7h.

11

Section 7D

7D

The Lognormal Distribution

12

THE LOGNORMAL DISTRIBUTION

As we know from Section 7.2 of the book, it is reasonable to assume that such random variables as the height of a man, the weight of a woman, the error made in a measurement, the I. Q. of a person, the growth distribution of plants, animals or their organs, etc. are normally distributed. However, in examples such as the height of a man and the I. Q. of a person, the random variable is strictly positive while the corresponding normal random variable X attains negative values as well. Although in these and similar examples, usually the mean is large and the standard deviation is small making P (X < 0) negligible, still mathematicians and statisticians prefer to use positive random variables as models for positive random variables. For this reason, they have introduced the lognormal probability density function, which has been proven to be very useful in such studies as the income distributions, stock market prices, pollutant concentrations, the size of oil fields in America, production data, critical doses of drugs in human beings and animals, etc.

Definition 7a A random variable Y is called lognormal if the natural log of Y has a normal distribution. That is, if for some normal random variable X, we have that Y = e X .

Let F be the distribution function of a normal random variable X with parameters μ and σ, and let f be its probability density function. Let Y = e X . Then ln Y ∼ N (μ, σ 2), and, G, the distribution function of Y, is calculated as follows.

G(y) = P (Y ≤ y) = P (eX ≤ y) = P (X ≤ ln y) = F (ln y).

Therefore, g, the density function of Y, which is obtained by differentiating G, is given by ⎧

1 −(ln y − μ)2 ⎪ ⎪ y>0 ⎨ √ exp 1 2σ 2 g(y) = G (y) = F (ln y) = f (ln y) = σy 2π ⎪ y ⎪ ⎩ 0 y ≤ 0.

We will now find the expected value and the standard deviation of a lognormal random 1 variable Y with parameters μ and σ. Let z = ln y − μ; then y = e z+μ , dz = dy, and y z ∈ (−∞, ∞). Thus, for n ≥ 0,

∞ −(ln y − μ)2 1 n n √ exp y · dy E(Y ) = 2σ 2 σy 2π 0

∞ −(ln y − μ)2 1 1 n dy y · √ exp = 2σ 2 y σ 2π 0 2 ∞ −z 1 n(z+μ) √ exp e dz = 2σ 2 σ 2π −∞

∞ z2 1 √ exp n(z + μ) − 2 dz = 2σ −∞ σ 2π

∞ z2 1 2 2 1 2 2 1 √ exp nz + nμ − 2 + σ n − σ n dz = 2σ 2 2 −∞ σ 2π

Section 7D

The Lognormal Distribution

13

z2 1 2 2 1 √ exp nz − 2 − σ n dz 2σ 2 −∞ σ 2π

∞ −(z 2 − 2σ 2 nz + σ 4 n2 ) 1 1 2 2 √ exp dz = exp(nμ + σ n ) 2 2σ 2 −∞ σ 2π

∞ −(z − σ 2 n)2 1 2 2 1 √ exp dz = exp(nμ + σ n ) 2 2σ 2 −∞ σ 2π ∞

1 = exp(nμ + σ 2 n2 ) 2

1 = exp(nμ + σ 2 n2 ), 2

−(z − σ 2 n)2 1 is the density function of where the last equality follows since √ exp 2σ 2 σ 2π N (σ 2n, σ 2 ), and hence its integral over (−∞, ∞) is 1. Therefore, for n ≥ 1,

1 E(Y n ) = exp nμ + σ 2 n2 .

2 This implies that

1

2

2

E(Y ) = eμ+ 2 σ ,

E(Y 2 ) = e2μ+2σ ,

and Var(Y ) = E(Y 2 ) − E(Y )

2

2

2

= e2μ+2σ − e2μ+σ = e2μ+σ

2

2 eσ − 1 .

Example 7i For a certain spice, one attribute with potential health hazards to the consumers is insect fragments. Although insect fragments are very hard to monitor, studies have shown that the insect fragment count in a bag containing one pound of the spice is lognormal with mean 230 and variance 180. From bags each containing one pound of this spice, one bag is selected at random. What is the probability that its insect fragment count is at most 215?

Solution Let Y be the number of insect fragments in the bag. Then X = ln Y ∼ N (μ, σ 2), where 1 2 (2) eμ+ 2 σ = 230, and

e2μ+σ

2

2 eσ − 1 = 180.

(3)

Squaring both sides of (2), we obtain 2

e2μ+σ = 52, 900,

(4)

substituting this in (3), we have 2

eσ − 1 = 2

180 ≈ 0.0034.

52, 900

Therefore, eσ ≈ 1.0034 and σ 2 ≈ ln 1.0034 ≈ 0.0034, which gives σ ≈ 0.0583. Now 2 substituting e σ ≈ 1.0034 in (4), we have e2μ ≈ 52, 900/1.0034 ≈ 52, 720.75, which gives

Section 7D

The Lognormal Distribution

14

2μ ≈ ln 52, 720.75 ≈ 10.873 or μ ≈ 5.44. Therefore, X = ln Y ∼ N (5.44, 0.0034), and thus the desired probability is obtained as follows.

P (Y ≤ 215) = P (ln Y ≤ ln 215) = P (X ≤ ln 215) ≈ P (X ≤ 5.37)

5.37 − 5.44 X − 5.44 ≤ ≈P 0.0583 0.0583 ≈ P (Z ≤ −1.20) = 1 − Φ(1.20) ≈ 1 − 0.8849 = 0.1151, where, as usual, Z ∼ N (0, 1).

Section 7E

7E

Hyperexponential Distribution

15

HYPEREXPONENTIAL DISTRIBUTION

Let X1 and X2 be exponential random variables with means 1/λ 1 and 1/λ2, respectively.

The random variable Y, defined by ⎧ with probability p ⎨ X1 Y = ⎩ X2 with probability 1 − p, is called hyperexponential of degree 2 if for all t ≥ 0 and s ≥ 0, the event that X 1 ≤ t is independent of the event that X 2 ≤ s. The following example is a general model for hyperexponential random variables.

Example 7j Suppose that a company has two plants. Plant A produces items which have exponential lifetimes with mean 1/λ 1 , and, independently of plant A, plant B produces the same type of items but with exponential lifetime distribution with mean 1/λ 2 . If 100p percent of the total items are produced by plant A and the rest by plant B, and the products are mixed in the market, then Y, the lifetime of a random product of this company in the market, is hyperexponential of order 2. Let F be the distribution function of Y, E be the event that the random product was produced by plant A, and F be the event that it was produced by plant B. By the law of total probability, for t ≥ 0, F (t) = P (Y ≤ t) = P (Y ≤ t | E)P (E) + P (Y ≤ t | F )P (F ) = P (X1 ≤ t)P (E) + P (X2 ≤ t)P (F ) = (1 − e−λ1 t )p + (1 − e−λ2 t )(1 − p), and F (t) = 0 for t < 0. Let f be the probability density function of Y ; we have ⎧ t≥0 ⎨pλ1 e−λ1 t + (1 − p)λ2 e−λ2 t

f (t) = F (t) = ⎩ 0 t < 0.

Therefore, as shown in Example 7j, in general, the probability density function, f, of a hyperexponential random variable of order 2 is given by ⎧ t≥0 ⎨pλ1 e−λ1 t + (1 − p)λ2 e−λ2 t f (t) = ⎩ 0 t < 0.

Hyperexponential random variables are generalized in the following obvious way: For i = 1, 2, . . ., n, let pi > 0, ni=1 pi = 1 and λi ≥ 0. The probability density function, ⎧ n t≥0 ⎨ i=1 pi λie−λi t g(t) = ⎩ 0 t < 0, is called hyperexponential of order n.

COMPANION FOR CHAPTER 8 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 8 Bivariate Distributions 8A

Additional Examples

3

3

Chapter 8

Additional Examples

3

Chapter 8

Bivariate Distributions

8A

ADDITIONAL EXAMPLES

Example 8a The joint probability density function of the continuous random variables X and Y is given by ⎧ ⎨12x(1 − x) 0≤x≤y≤1 f (x, y) = ⎩ 0 otherwise.

Find E(X | Y = y).

Solution: Note that E(X | Y = y) =

y 0

where, fX|Y (x|y) = Now fY (y) =

y 0

fX|Y (x|y) = E(X | Y = y) =

f (x, y) . fY (y)

12x(1 − x) dx = −4y 3 + 6y 2 ,

Thus

So

xfX|Y (x|y) dx,

y 0

12x(1 − x) , −4y 3 + 6y 2

0 ≤ y ≤ 1.

0 ≤ y ≤ 1, 0 ≤ x ≤ y.

−3y 2 + 4y 12x2 (1 − x) , dx = −4y 3 + 6y 2 −4y + 6

0 ≤ y ≤ 1.

Example 8b Let X be the fraction of lead and Y be the fraction of zinc in a random piece of a certain sedimentary rock. Suppose that the joint probability density function of

Chapter 8

Additional Examples

4

X and Y is given by ⎧ 16 ⎪ ⎨ (7x + 2y) 3 f (x, y) = ⎪ ⎩ 0

x > 0, y > 0, 2x + 2y < 1 otherwise.

Find E(X) and E(Y ).

Solution: We have that

16 1/2 (1−2x)/2 x(7x + 2y) dy dx 3 0 0 1/2 2 4 x(1 − 2x)(12x + 1) dx = ≈ 0.22. = 3 0 9

E(X) =

To find E(Y ), similarly to the above calculation, we can use the formula E(X) =

16 3

1/2 (1−2x)/2 0

0

y(7x + 2y) dy dx.

However, for calculational simplicity, it is better to proceed as follows

16 1/2 (1−2y)/2 y(7x + 2y) dx dy E(Y ) = 3 0 0

1 4 1/2 y(1 − 2y)(7 − 10y) dy = ≈ 0.25. = 3 0 4 Example 8c

A point is selected at random from the unit disk D = (x, y) : x2 + y 2 ≤ 1 .

Find the expected value of the distance of the point selected from the origin.

Solution: Let X and Y be the coordinates of the point selected. Let f (x, y) be the joint probability density function of X and Y . By the discussion right before Definition 8.5 of the book, ⎧ 1 ⎪ ⎪ if (x, y) ∈ D ⎨ area(D) f (x, y) = ⎪ ⎪ ⎩0 otherwise.

Therefore,

⎧ 1 ⎪ ⎨ f (x, y) = π ⎪ ⎩ 0

if x2 + y 2 ≤ 1 otherwise.

Chapter 8

5

Additional Examples

The distance of the point selected from the origin is x2 + y 2 . So we are interested in x2 + y 2 , which is calculated as follows.

E

1 2 2 x +y = x2 + y 2 · dx dy.

E π x2 +y2 ≤1

To calculate this double integral, we will convert it to polar coordinates. We have x = r cos θ, y = r sin θ, x2 + y 2 = r, dx dy = r dr dθ, 0 ≤ r ≤ 1, 0 ≤ θ < 2π. So

2π 1 1 2 2 x + y = E(r) = r · r dr dθ E π 0 0

1

2 1 2π 1 1 2π dθ = . r 2 dr dθ = = π 0 π 3 3 0 0 Example 8d An engineer, monitoring a system, identifies when a fault occurs and determines whether the fault is Type I or another type. Suppose that Y, the total number of faults occurring in an hour is a Poisson random variable with mean 3. Suppose that, independently of other faults, the probability that a fault is Type I is 1/3. Let X be the number of Type I faults in an hour. Find p(i, j), the joint probability mass function of X and Y, and pX (i), the marginal probability mass function of X. Solution: By the law of multiplication, p(i, j) = P (X = i, Y = j) = P (X = i | Y = j)P (Y = j)

j 1 i 2 j−i e−3 · 3j · = 3 3 j! i =

e−3 · 2j−i , i! (j − i)!

j = 0, 1, 2, . . .; 0 ≤ i ≤ j.

The marginal probability mass function of X is pX (i) =

∞

p(i, j) =

j=i

=

∞ −3

e · 2j−i j=i

i! (j − i)!

∞

∞

j=i

k=0

e−3 2k e−3 2 e−1 e−3 2j−i = = ·e = , i! (j − i)! i! k! i! i!

i = 0, 1, 2, . . ..

Therefore, the marginal probability mass function of X is Poisson with parameter 1.

Example 8e Suppose that in a sequence of random digits the total number of 0’s among the first 100 is 12. What is the expected number of 0’s among the first 10?

Chapter 8

Additional Examples

6

Solution: Let X and Y denote the number of 0’s among the first 10 and the first 100 digits, respectively. To find the desired quantity, E(X | Y = 12), note that E(X | Y = 12) =

10

x pX|Y (x|12) =

x=0

10

x=0

x

p(x, 12) , pY (12)

where p(x, 12) = P (X = x, Y = 12) = P (X = x, Y − X = 12 − x) = P (X = x)P (Y − X = 12 − x)

90 1 12−x 9 90−(12−x) 10 1 x 9 10−x · = 10 10 10 12 − x 10 x

10 90 1 12 9 88 = .

10 x 12 − x 10 This is because X and Y − X, being the number of 0’s among the first 10 and the number of 0’s among the next 90 digits, are independent binomial random variables with the parameters (10, 1/10) and (90, 1/10), respectively. Therefore,

10 90 1 12 9 88 pX|Y (x|12) =

p(x, 12) = pY (12)

12 − x 10 10

12 88 9 100 1 10 10 12

x

10 90 x 12 − x

= .

100 12

This shows that p X|Y (x|12) is a hypergeometric probability mass function with parameters D = 10, N = 100, and n = 12 (see Definition 5.6 of the book). Hence its expected value is nD/N = (12 × 10)/100 = 1.2; that is, E(X | Y = 12) = 1.2, an answer consistent with our intuitive expectation.

Example 8f Let X and Y be two continuous random variables with the joint probability density function f (x, y). Prove that for any subset A of real numbers,

∞ 1 xf (x, y)dy dx. (i) E(X | Y ∈ A) = P (Y ∈ A) −∞ A

1 E(X | Y = y)fY (y)dy. (ii) E(X | Y ∈ A) = P (Y ∈ A) A Proof:

Let H(x) = P (X ≤ x | Y ∈ A) and h(x) = H (x). Then

x 1 P (X ≤ x, Y ∈ A) = f (x, y) dy dx.

H(x) = P (Y ∈ A) P (Y ∈ A) −∞ A

(i)

Chapter 8

Additional Examples

7

Now using fundamental theorem of calculus, we differentiate H to obtain h. We get

1 f (x, y) dy. h(x) = P (Y ∈ A) A Therefore, ∞

1 xh(x) dx = E(X | Y ∈ A) = P (Y ∈ A) −∞

∞ −∞

A

xf (x, y) dy

dx.

(ii) By changing the order of integration in (i) we obtain

∞ xf (x, y) 1 dx fY (y) dy E(X | Y ∈ A) = P (Y ∈ A) A −∞ fY (y)

1 E(X | Y = y)fY (y) dy. = P (Y ∈ A) A Remark 8a Let a and b be two real numbers and F (x, y) be the joint probability distribution function of X and Y . This theorem implies relations such as

∞ b 1 xf (x, y)dy dx.

E(X | a < Y < b) = FY (b) − FY (a) −∞ a and

1 E(X | a < Y < b) = FY (b) − FY (a)

b a

E(X | Y = y)fY (y) dy.

Example 8g Let the joint probability density function of the continuous random variables X and Y be given by ⎧ 20 √ ⎪ 0≤x≤ y≤1 ⎨ (x + y) f (x, y) = 13 ⎪ ⎩ 0 elsewhere.

(i)

Calculate P (0 < X < 3/4, 1/4 < Y < 3/4).

(ii)

Find the marginal probability density functions of X and Y, the conditional probability density function of X given that Y = y, and the conditional probability density function of Y given that X = x.

(iii)

Calculate E(X | Y = y) and E(Y | X = x).

Chapter 8

Additional Examples

3/4 1/4 1/2 3/4

Figure 8g

The shaded area is the region of integration of Example 8g.

Solution: (i) The region of integration for calculation of P (0 < X < 3/4, 1/4 < Y < 3/4) is shown in Figure 8g. Thus P (0 < X < 3/4, 1/4 < Y < 3/4) = 1/2 3/4 0

5 13

1/4

1/2 0

3/4 3/4 20 20 (x + y) dy dx + (x + y) dy dx = 13 13 1/2 x2

(2x + 1) dx +

5 104

3/4 1/2

(−16x4 − 32x3 + 24x + 9) dx =

2979 15 1059 + = ≈ 0.45.

52 6656 6656 (ii) We have fX (x) = fY (y) =

1

10 20 (x + y) dy = (−x4 − 2x3 + 2x + 1), 13 x2 13

√y 0

10 √ 20 (x + y) dx = y(2 y + 1), 13 13

0 ≤ x ≤ 1;

0 ≤ y ≤ 1;

8

Chapter 8

fX|Y (x|y) =

2(x + y) f (x, y) = , √ fY (y) y(2 y + 1)

fY |X (y|x) =

2(x + y) f (x, y) = , 4 fX (x) −x − 2x3 + 2x + 1

0≤x≤

√

Additional Examples

9

y, 0 ≤ y ≤ 1;

x2 ≤ y ≤ 1, 0 ≤ x ≤ 1.

(iii) By relation (8.21) of the book, E(X | Y = y) = = E(Y | X = x) = =

√y 0

√y 0

1 x2

xfX|Y (x|y) dx √ 3y + 2 y) 2x(x + 1) dx = , √ √ y(2 y + 1) 3(2 y + 1)

yfY |X (y|x) dy =

1 x

0 ≤ y ≤ 1;

2y(x + y)

4 3 2 −x − 2x + 2x + 1

−2x6 − 3x5 + 3x + 2 , 3(−x4 − 2x3 + 2x + 1)

0 ≤ x ≤ 1.

dy

In Section 8.4 of the book, we used Theorem 8.8 to find the joint probability density function of functions of continuous random variables. Calculation of joint probability mass functions of functions of discrete random variables is done similarly. An example follows.

Example 8h Let X and Y be independent Poisson random variables with parameters λ1 and λ2 , respectively. Find the joint probability mass function of X + Y and Y, and the probability mass function of X + Y .

Solution: Let p(x, y) be the joint probability mass function of X and Y . Then p(x, y) =

λx λy λx1 e−λ1 λy2 e−λ2 · = 1 2 e−(λ1 +λ2 ) , x! y! x! y!

x = 0, 1, 2, . . ., y = 0, 1, 2, . . .. To find g(u, v), the joint probability mass function of X + Y and Y, let U = X + Y and V = Y, then X = U − V and Y = V . So g(u, v) =

λv2 −(λ1 +λ2 ) λu−v 1 e , (u − v)! v!

u = 0, 1, 2, . . ., v = 0, 1, 2, . . ..

Note that g U (u), the marginal probability mass function of g(u, v) gives the probability mass function of X + Y, the sum of two independent Poisson random variables. Since

Chapter 8

Additional Examples

10

V < U, gU (u) =

u

g(u, v) = e−(λ1 +λ2 )

v=0

=

u

λu−v λv2 1 (u − v)! v! v=0

u −(λ1 +λ2 )

e

u!

v=0

u! λu−v λv2 (u − v)! v! 1

e−(λ1 +λ2 ) (λ1 + λ2 )u . u!

This shows that the sum of two independent Poisson random variables with parameters λ 1 and λ2 is Poisson with parameters λ 1 + λ2 . We will prove this important fact again using a different method in Section 11.2 of the book (see Theorem 11.5). =

Example 8i Let X and Y be two independent beta random variables both with parameters (2, 2). Find the joint probability density function of U = X/Y and V = XY .

Solution: Let f (x, y) be the joint probability density function of X and Y . We are given that ⎧ ⎨6x(1 − x) 0<x<1 fX (x) = ⎩ 0 otherwise, and fY (y) = Thus f (x, y) = fX (x)fY (y) =

⎧ ⎨6y(1 − y)

0<y<1

⎩ 0

otherwise.

⎧ ⎨36xy(1 − x)(1 − y)

0 < x < 1, 0 < y < 1

⎩ 0

otherwise.

Consider the system of two equations in two unknowns ⎧ ⎨x/y = u ⎩ xy = v.

Since x > 0 and y > 0, this system has the unique solution ⎧ √ ⎨x = uv ⎩ y=

v/u.

Now 0 < x < 1 implies that v < 1/u, and 0 < y < 1 implies that v < u. So this system transforms R = (x, y) : 0 < x < 1, 0 < y < 1 to the shaded region of Figure 8h, which is Q = (u, v) : 0 < u ≤ 1, 0 < v < u ∪ (u, v) : 1 < u < ∞, 0 < v < 1/u .

Chapter 8

Additional Examples

11

1 1

Figure 8h

Since

The shaded region is Q in Example 8i.

1

2 v/u J = − 1 v/u3

2

1 u/v 2 = 1 = 0, 2u 1 1/uv 2

by Theorem 8.8 of the book, g(u, v), the joint probability density function of U and V, is given by ⎧

√ v 18v ⎪ ⎪ 1 − uv 1 − if (u, v) ∈ Q ⎨ u u g(u, v) = f (u, v)|J| = ⎪ ⎪ ⎩ 0 elsewhere.

Example 8j From a certain airport, starting midnight, Airline A has flights every 4 hours every day. From the same airport, starting Y hours after midnight, Airline B has flights every 6 hours every day. Suppose that Y is a random number from the interval (0, 4); that is, the first flight of the Airline B is at a random time between midnight and 4:00 A . M .

Find the probability that a passenger who arrives at this airport at a random time finds that the next departure is from Airline A. Solution: On a given day, the number of hours after midnight that Airline A has departures are 0, 4, 8, 12, 16, and 20, and the number of hours after midnight that Airline B has departures are Y, Y + 6, Y + 12, and Y + 18, where Y is a uniform random variable over the interval (0, 4). By looking at these departure times carefully, we observe that, after every 12 hours, the system restarts probabilistically all over again. That is, if the 12th hour

Chapter 8

Additional Examples

12

is taken as 0, then again the departure times of Airline A will be 0, 4, 8, 12, 16, and 20, and the departure times of Airline B will be Y, Y + 6, Y + 12, and Y + 18, which is a replica of the original situation. By this observation, the problem can be rephrased as follows: In the interval [0, 12], Airline A has flights at 0, 4, 8, and 12. Airline B has flights at Y and Y + 6, where Y is a random number from the interval (0, 4). If the time interval [0, 12] is observed at a random time X, what is the probability that the first departure after X is from Airline A? To answer this, note that, if 0 < Y ≤ 2, then in the period [0, 12], the departure times for flights from Airlines A and B, combined, are 0, Y, 4, Y + 6, 8, and 12. Hence the first departure after X is from A if and only if Y ≤ X < 4 or

Y + 6 ≤ X < 12.

If Y > 2, then, in [0, 12], the departure times of flights A and B, combined, are 0, Y, 4, 8, Y + 6, and 12. Therefore, in this case, the first departure after X is from A if and only if Y ≤ X < 8 or

Y + 6 ≤ X < 12.

Let R1 = (x, y) : 0 < y < 2 ∩ (x, y) : y ≤ x < 4 or y + 6 ≤ x < 12 , and R2 = (x, y) : y > 2 ∩ (x, y) : y ≤ x < 8 or y + 6 ≤ x < 12 .

Then the set R 1 ∪ R2 is the shaded region of Figure 8i.

Figure 8i

Geometric Model of Example 8j.

Chapter 8

Additional Examples

13

Clearly, the desired probability is the area of R 1 ∪R2 divided by 48, the area of the rectangle OMNS, which is the sample space. But the area of R 1 ∪ R2 is the summation of the areas of the three shaded trapezoids, which is easily calculated and found to be 32. Hence the desired probability is 32/48=2/3.

Example 8k Let X and Y be two independent random variables with the same probability density function, given by ⎧ if x ≥ 1 ⎨1/x2 f (x) = ⎩ 0 elsewhere. (i)

Calculate the probability density function of X/Y directly, without using Theorem 8.8 of the book.

(ii)

Find P (X < 3Y ).

Solution:

Since X and Y are independent and have the same probability density function, f (x, y) = fX (x)fY (y),

where, fX (x) = and fY (y) =

⎧ ⎨1/x2

if x ≥ 1

⎩

0

elsewhere,

⎧ ⎨1/y 2

if y ≥ 1

⎩ 0

elsewhere.

Thus f (x, y) = fX (x)fY (y) =

⎧ ⎨1/(x2 y 2 )

if x ≥ 1, y ≥ 1

⎩ 0

elsewhere.

Let G be the distribution function of X/Y . We will first calculate G and then, by differentiation, find g, the probability density function of X/Y . Clearly, for t ≤ 0, G(t) = 0. If 0 < t < 1, then the event X/Y ≤ t occurs if and only if X ≤ tY . Since X ≥ 1 implies that Y ≥ 1/t, we have that X ≤ tY is equivalent to Y ≥ 1/t and 1 ≤ X ≤ tY. As Figure 8j shows, for 0 < t < 1,

1 G(t) = P (X/Y ≤ t) = P (X ≤ tY ) = P Y ≥ and 1 ≤ X ≤ tY t

1 = P (X, Y ) ∈ (x, y) : 1 ≤ x ≤ ty, y ≥ t

∞ ∞ ty 1 1 ty dx dy = dy − = 2 2 xy 2 1 1/t 1 x y 1/t

Chapter 8

=

∞ 1/t

1 1 − 3+ 2 ty y

14

Additional Examples

1 ∞ 1 − = t/2. dy = 2ty 2 y 1/t

(1 , 1/ t) 1

1

Figure 8j

The shaded area is the region of integration of Example 8k, 0 < t < 1.

As Figure 8k shows, for t ≥ 1,

G(t) = P (X ≤ tY ) = P (X, Y ) ∈ (x, y) : 1 < x ≤ ty, y > 1

∞ 1 1 1 dx dy = − 3 + 2 dy = 2 2 ty y 1 1 x y 1

1 ∞ 1 1 − =1− . = 2 2ty y 1 2t ∞ ty

Thus,

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨t G(t) = 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩1 − 1 2t

t<0 0≤t<1 t ≥ 1,

Chapter 8

and g(t) = G (t) is given by

(ii)

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨1 g(t) = 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 2t2

Additional Examples

15

t<0 0≤t<1 t ≥ 1.

P (X < 3Y ) = P (X/Y < 3) = G(3) = 1 −

5 1 = .

6 6

1

(t , 1)

1 Figure 8k

The shaded area is the region of integration of Example 8k, t ≥ 1.

Example 8 For an integer n ≥ 3, let X be a standard normal random variable and Y be a gamma random variable with parameters (n/2, 1/2). Suppose that X and Y are independent and U = X/ Y /n. The random variable U, called t-distributed with n degrees of freedom, has significant applications in statistics. Find the probability density function of U. Solution: Let f (x, y) be the joint probability density function of X and Y . Since X and Y are independent, f (x, y) = fX (x)fY (y), where 1 2 fX (x) = √ e−x /2 , 2π fY (y) =

y (n/2)−1 e−y/2 , 2n/2 Γ(n/2)

−∞ < x < ∞ y > 0.

Chapter 8

Additional Examples

16

Thus f (x, y) = √

1

2

2π · 2n/2 · Γ(n/2)

· y (n/2)−1 · e−(x +y)/2 ,

−∞ < x < ∞, y > 0.

Let V = Y . We will find the joint probability density function of U and V . The desired function is the marginal probability density function of U . The system of two equations in two unknowns ⎧ x ⎪ =u ⎪ ⎨ y/n ⎪ ⎪ ⎩y = v defines a one-to-one transformation of R = (x, y) : − ∞ < x < ∞, y > 0 onto the region Q = (u, v) : − ∞ < u < ∞, v > 0 . It has the unique solution x = u v/n, y = v. Hence

u v/n √

2 nv = v/n = 0.

J =

0 1

By Theorem 8.8 of the book, g(u, v), the joint probability density function of U and V is given by g(u, v) = f u v/n, v =√

v/n

v u2 1 1+ , v [(n+1)/2]−1 exp − 2 n 2πn Γ(n/2) 2n/2

−∞ < u < ∞, v > 0.

1 . Then the probability density function of U is For simplicity, let k = √ 2πn Γ(n/2) 2n/2 gU (u), where ∞ v u2 1+ dv. v [(n+1)/2]−1 exp − gU (u) = k 2 n 0 Let

u2 2 2 v 1+ = w. Then v = w and dv = dw. Thus 2 2 n 1 + (u /n) 1 + (u2 /n) gU (u) = k

∞ 0

2 2[(n+1)/2]−1 [(n+1)/2]−1 −w dw e · [(n+1)/2]−1 w

1 + (u2 /n) 1 + (u2 /n)

2(n+1)/2 =k· (n+1)/2 1 + (u2 /n) =k·

∞

w [(n+1)/2]−1 e−w dw

0

n + 1 2(n+1)/2 (n+1)/2 Γ 2 1 + (u2 /n)

Chapter 8

=√

Additional Examples

n + 1 2(n+1)/2 1 . · Γ

2 2πn Γ(n/2) 2n/2 1 + (u2 /n) (n+1)/2

Simplifying the last term, we have that the probability density function of U is given by n + 1 Γ 1 2 · , gU (u) = √

πn Γ(n/2) 1 + (u2 /n) (n+1)/2

−∞ < u < ∞.

17

COMPANION FOR CHAPTER 9 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 9 Multivariate Distributions 9A 9B

Additional Examples 3 Transformations of n > 2 Random Variables

3

9

Section 9A

Additional Examples

3

Chapter 9

Multivariate Distributions 9A

ADDITIONAL EXAMPLES

Example 9a Suppose that mine explosions in a certain country occur at X 1 , X1 + X2 , X1 +X2 +X3 , . . . , where Xi’s are identically distributed and independent random variables with probability density function. f (x) = λ2 xe−λx

x ≥ 0.

Let N (t) be the total number of mine explosions in that country by time t. Find the probability mass function of N (t).

Solution: Note that for i ≥ 1, X i is a gamma random variable with parameters 2 and λ. two independent exponential random variables. Consider a Poisson Hence X i is the sum of process N1 (t) : t ≥ 0 with parameter λ. The times between consequent events of this process are exponential with parameter λ. Thus N (t) = n if and only if N 1 (t) = 2n or 2n + 1. This observation yields

e−λt (λt)2n e−λt (λt)2n+1 + , P N (t) = n = (2n)!

(2n)!

n = 0, 1, 2, . . ..

Example 9b Let X1 , X2 , . . . , Xn be independent exponential random variables with parameters λ1 , λ2 , . . . , λn , respectively. There are n! possible rank orderings of X 1 , X2 , . . . , Xn , one of them being X 1 < X2 < X3 < · · · < Xn . Show that the rank ordering of {X1 , X2 , . . ., Xn } is independent of min(X 1 , X2, . . . , Xn ). That is, if (i1 , i2, . . . , in) is a permutation of (1, 2, . . ., n), then P (Xi1 < Xi2 < · · · < Xin | min Xi > t) = P (Xi1 < Xi2 < · · · < Xin ).

1≤i≤n

Solution: Given that min 1≤i≤n Xi > t, each Xi is greater than t. Due to the memoryless property of exponential random variables, after t units of time, X i1 −t, Xi2 −t, . . . , Xin −t have the same distributions as X i1 , Xi2 , . . . , Xin , respectively. So P (Xi1 < Xi2 < · · · < Xin | min Xi > t) = P (Xi1 − t < Xi2 − t < · · · < Xin − t) 1≤i≤n

= P (Xi1 < Xi2 < · · · < Xin ).

Section 9A

Additional Examples

4

Example 9c At an institution, the registration process consists of three steps. First, students discuss their options for courses with their faculty advisors. Second, students check in at the Administration Services to receive clearance on their financial and health status. Third, they meet with a registrar who inputs their courses into the system. Suppose that the amounts of time spent at each step are independent exponential random variables with parameters λ1 , λ2 , and λ3 , respectively. Find the probability density function of the total time a student spends on the entire registration process if (a)

λi = λj , if i = j;

(b)

λ1 = λ2 = λ3 = λ.

Solution: Let X 1 , X2 , and X3 be the amounts of time the student spends at steps 1, 2, and 3, respectively. We are given that X 1 , X2 , and X3 are independent exponential random variables with parameters λ 1 , λ2 , and λ3 , respectively. Let f1 , f2 , and f3 be the probability density functions of X 1 , X2 , and X3 , respectively. We will now find the probability density function of the desired quantity, X 1 + X2 + X3 , in cases (a) and (b). (a)

Using the convolution theorem, we will first find, g, the probability density function of X1 + X2 . We have ∞ f1 (x)f2 (t − x) dx g(t) = 0

=

t 0

λ1 e−λ1 x λ2 e−λ2 (t−x) dx −λ2 t

= λ1 λ2 e

= λ1 λ2 e−λ2 t =

t

e−(λ1 −λ2 )x dx

0

1 1 − e−(λ1 −λ2 )t λ1 − λ2

λ2 λ1 λ2 e−λ2 t + λ1 e−λ1 t . λ1 − λ2 λ2 − λ1

Let h be the probability density function of X 1 + X2 + X3 ; applying the convolution theorem again, straightforward calculations give ∞ g(x)f3 (t − x) dx h(t) = 0

=

λ1 λ3 λ2 λ3 λ1 eλ1 t + λ2 e−λ2 t (λ2 − λ1 )(λ3 − λ1 ) (λ1 − λ2 )(λ3 − λ2 ) +

(b)

λ1 λ2 λ3 e−λ3 t . (λ1 − λ3 )(λ2 − λ3 )

In this case also, by the convolution theorem, g, the probability density function of X1 + X2 , is given by t ∞ f1 (x)f2 (t − x) dx = λe−λx · λe−λ(t−x)dx = λ2 te−λt . g(t) = −∞

0

Section 9A

Additional Examples

5

This is the density function of a gamma random variable with parameters 2 and λ.

Hence X1 + X2 is gamma with parameters 2 and λ. Let h be the probability density function of X 1 + X2 + X3 ; applying the convolution theorem again, straightforward calculations give ∞ g(x)f3 (t − x) dx = (1/2)λ3t2 e−λt . h(t) = 0

Hence if λ1 = λ2 = λ3 = λ, then X1 + X2 + X3 is gamma with parameters 3 and λ.

Remark Let X1 , X2 , . . . , Xn be n independent exponential random variables with parameters λ1 , λ2 , . . . , λn , respectively. Using induction and the result of this example, we can show that, (a)

if λi = λj , for i = j, then, f, the probability density function of X 1 + X2 + · · ·+ Xn is given by n λj λi e−λi t ; f (t) = λj − λi i=1

(b)

j =i

if λi = λ, for 1 ≤ i ≤ n, then X1 + X2 + · · · + Xn is gamma with parameters n and λ.

Example 9d Michelle and Fred are shopping at a mall. Michelle decides to buy a leather coat and Fred, knowing that it will take her a while, goes to buy an ice cream. At the ice cream parlor, there is only one employee behind the counter. When Fred arrives, there is one customer currently being waited on and 6 more are in line. Suppose that the time that it takes to buy an ice cream is exponentially distributed with mean 3 minutes. Furthermore, the time it takes to choose and buy a leather coat is also exponentially distributed, but with mean 28 minutes. Also suppose that Michelle is the only customer in the coat shop, and, lastly, it takes 2 minutes to travel from the coat shop to the ice cream parlor. What is the probability that Fred will return before Michelle has finished buying her coat?

Solution: When Fred enters the ice cream parlor, there are 7 customers ahead of him. Let us call the customer who is currently being waited on Customer 1, and the 6 customers in line Customers 2 to 7, respectively. Fred is Customer 8. Let X 1 be the period starting when Fred enters the ice cream parlor and ending when Customer 1 leaves. By the memoryless property of the exponential random variables, X 1 is exponential with mean 3 minutes. For 2 ≤ i ≤ 8, let Xi be the time that it takes for Customer i to buy an ice cream; X i is exponential with mean 3 minutes. Finally, let Y be the time until Michelle buys her favorite leather coat. Then, it is reasonable to assume that {Y, X 1, X2, . . . , X8 } is a set of independent random variables. To find the desired probability, P (4 + X1 + X2 + · · · + X8 < Y ) = P (X1 + X2 + · · · + X8 < Y − 4), we will successively calculate

Section 9A

Additional Examples

6

P (X1 < Y − 4), P (X1 + X2 < Y − 4), P (X1 + X2 + X3 < Y − 4), .. .

P (X1 + X2 + · · · + X8 < Y − 4).

To do so, note that since the random variables X 1 and Y are independent, f (x, y), the joint probability density function of these two random variables is given by 1 1 f (x, y) = e−x/3 · e−y/28 , x ≥ 0, y ≥ 0.

3 28 Therefore,

1 −x/3 1 −y/28 e · e dy dx P (X1 < Y − 4) = 28 0 x+4 3

∞ ∞ 1 −x/3 1 −y/28 e e dy dx = 3 0 x+4 28 ∞ ∞ 1 1 −x/3 −(x+4)/28 e ·e dx = e−4/28 e−31x/84 dx = 3 3 0 0 ∞ ∞

=

84 −4/28 e = 0.783.

93

Thus, P (X1 < Y − 4) = P (2 + X1 < Y − 2), the probability that Customer 1 leaves at least 2 minutes before Michelle has bought her leather coat is 0.783. Next, we calculate, P (X1 + X2 < Y − 4), the probability that Customer 2 also leaves at least 2 minutes before Michelle is done. To do so, note that by the memoryless property of the exponential, the period starting when Customer 1 leaves, and ending when Michelle is done, is still exponentially distributed with mean 28 minutes. Therefore repeating the same argument yields P (X1 + X2 < Y − 4 | X1 < Y − 4) = 0.783.

Thus P (X1 + X2 < Y − 4) =

P (X1 + X2 < Y − 4 | X1 < Y − 4)P (X1 < Y − 4) = (0.783)2.

Similarly, P (X1 + X2 + X3 < Y − 4 | X1 + X2 < Y − 4) = 0.783.

Hence, P (X1 + X2 + X3 < Y − 4) = P (X1 + X2 + X3 < Y − 4 | X1 + X2 < Y − 4) ×

P (X1 + X2 < Y − 4) = (0.783)(0.783)2 = (0.783)3.

Section 9A

Additional Examples

7

Continuing this argument, we have that the probability that Fred (Customer 8) leaves the ice cream parlor at least 2 minutes before Michelle has bought her leather coat is P (X1 + X2 + · · · + X8 < Y − 4) = (0.783)8 = 0.141.

Example 9e Suppose that shocks occur to a system according to a Poisson process with parameter λ. Furthermore, suppose that k types of shocks are identified to occur and, independently form other shocks, the probability that a shock is of type i, 1 ≤ i ≤ k, is p i .

For 1 ≤ i ≤ k, let Xi be the number of shocks of type i occurring between 0 and t. Find the joint probability mass function of X 1 , X2 , . . . , Xk .

Solution: Let j 1 , j2 , . . . , jk be non-negative integers. For 1 ≤ i ≤ k, let q i = P (Xi = ji ).

Then P (X1 = j1 , X2 = j2 , . . ., Xk = jk ) =

(j1 + j2 + · · · + jk )! j1 j2 q1 q2 · · · qkjk . j1 ! j2 ! · · · , jk !

To find qi = P (Xi = ji ), let N (t) be the number of shocks occurring to the system between 0 and t. We have ∞

P Xi = ji | N (t) = n P N (t) = n

qi = P (Xi = ji ) = n=0 ∞

= n=ji

e−λt (λt)n n ji pi (1 − pi )n−ji · n! ji

= pji i e−λt

∞ n=ji

(pi λt)ji e−λt = ji ! =

(λt)n−ji (λt)ji n!

(1 − pi )n−ji · ji ! (n − ji )! n! ∞ n=ji

n−ji λt(1 − pi ) (n − ji )!

(piλt)ji e−pi λt (pi λt)ji e−λt λt(1−pi) ·e . = ji ! ji !

Therefore, the joint probability mass function of X 1 , X2 , . . . , Xk is P (X1 = j1 , X2 = j2 , . . . , Xk = jk ) =

(pk λt)jk e−pk λt (j1 + j2 + · · · + jk )! (p1 λt)j1 e−p1 λt (p2 λt)j2 e−p2 λt ··· , j1 ! j2 ! · · · , jk ! j1 ! j2 ! jk !

where j1 , j2 , . . . , jk are non-negative integers.

Section 9A

Additional Examples

8

Example 9f (Genetics) As we know, in humans, for blood type, there are three alleles A, B, and O. The alleles A and B are codominant to each other and dominant to O. A man of genotype AB marries a woman of genotype BO. If they have six children, what is the probability that three will have type B blood, two will have type A blood, and one will have type AB blood?

Solution: The probability is 1/4 that the blood type of a child of this man and this woman is AB. The probability is 1/4 that it is A, and the probability is 1/2 that it is B. The desired probability is equal to 1 6!

3! 2! 1! 2

3

1 4

2

1 4

1

=

15 = 0.117.

128

Example 9g (Genetics) Let p and q be positive numbers with p + q = 1. For a gene with dominant allele A and recessive allele a, let p 2 , 2pq, and q 2 be the probabilities that a randomly selected person from a population has genotype AA, Aa, and aa, respectively. A group of six members of the population is selected randomly. Determine the value of p that maximizes the probability of the event of obtaining two AA’s, two Aa’s, and two aa’s.

Solution: The probability of two AA’s, two Aa’s, and two aa’s is g(p) =

2 2 6! (p2 )2 2p(1 − p) (1 − p)2 = 360p6 (1 − p)6 .

2! 2! 2!

To find the maximum of this function, set g (p) = 0 to obtain p = 1/2.

Section 9B

9B

TRANSFORMATIONS OF

Transformations of n > 2 Random Variables

9

n > 2 RANDOM VARIABLES

Theorem 8.8 of the book can be generalized in the following obvious way for functions of more than two random variables.

Theorem 9a Let X1 , X2 , . . . , Xn be n random variables with the joint probability density function f (x 1 , x2 , . . . , xn ). Let g1 , g2 , . . . , gn be n real valued functions of n variables and Y1 = g1 (X1 , X2 , . . ., Xn ), Y2 = g2 (X1 , X2, . . . , Xn ), . . . , Yn = gn (X1 , X2 , . . ., Xn ).

If (i) the system of n equations in n unknowns ⎧ ⎪ ⎪ ⎪g1 (x1 , x2 , . . . , xn) = y1 ⎪ ⎪ ⎨g2 (x1 , x2 , . . . , xn) = y2 .. ⎪ ⎪ . ⎪ ⎪ ⎪ ⎩g (x , x , . . . , x ) = y n 1 2 n n

(1)

has a unique solution for x 1 , x2 , . . . , xn in terms of y 1 , y2 , . . . , yn , and (ii) the functions g 1 , g2 , . . . , gn have continuous partial derivatives and the Jacobian of the transformation ⎧ ⎪ y1 = g1 (x1 , x2 , . . . , xn ) ⎪ ⎪ ⎪ ⎪ ⎨y2 = g2 (x1 , x2 , . . . , xn ) .. ⎪ ⎪ . ⎪ ⎪ ⎪ ⎩y = g (x , x , . . . , x ), n n 1 2 n is nonzero at all points (x 1 , x2 , . . ., xn ); that is, the following n × n determinant is nonzero

∂g1 ∂g1 ∂g1

∂x1 ∂x2 · · · ∂xn

∂g

∂g ∂g

2 2 2 ···

∂x1 ∂x2 ∂xn ∂(g1 , g2, . . . , gn )

= 0, =

∂(x1 , x2 , . . ., xn ) .. .. ... . .

∂gn ∂gn

∂g n

···

∂x1 ∂x2 ∂xn then, the random variables Y 1 , Y2 , . . . , Yn are jointly continuous with the joint probability density function h(y 1 , y2 , . . ., yn ) given by

∂(g1 , g2 , . . . , gn) −1 , h(y1 , y2 , . . ., yn ) = f (x1 , x2 , . . . , xn ) ∂(x1 , x2 , . . . , xn) where in this formula, for given (y 1 , y2 , . . ., yn ), we have that (x1 , x2 , . . . , xn) is the unique solution of the system (1).

Transformations of n > 2 Random Variables

Section 9B

10

Example 9h Let f (x1 , x2 , x3 ) be the joint probability density function of X 1 , X2 , and X3 . Find the joint probability density function of Y 1 = X1 +X2 +X3 , Y2 = X1 −X2 +X3 , and Y3 = X1 − X2 − X3 .

Solution: Let g 1 (x1 , x2 , x3 ) = x1 + x2 + x3 , g2 (x1 , x2 , x3 ) = x1 − x2 + x3 , and g3 (x1 , x2 , x3 ) = x1 − x2 − x3 . Then the system of equations ⎧ ⎪ ⎪ ⎨x1 + x2 + x3 = y1 x1 − x2 + x3 = y2 ⎪ ⎪ ⎩x − x − x = y 1

2

3

3

has the unique solution x 1 = (y1 + y3 )/2, x2 = (y1 − y2 )/2, x3 = (y2 − y3 )/2. Also

1 1 1

∂(g1 , g2, g3 ) = 1 −1 1 = 4 = 0. ∂(x1 , x2 , x3 ) 1 −1 −1 Hence, by Theorem 9a, h(y1 , y2 , y3 ), the joint probability density function of Y 1 , Y2 , Y3 is given by

y1 + y3 y1 − y2 y2 − y3 1 1 , , . h(y1 , y2 , y3 ) = f (x1 , x2 , x3 ) = f 4 4 2 2 2

COMPANION FOR CHAPTER 10 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 10 More Expectations and Variances 10A Additional Examples 3 10B Pattern Appearance 16 10C More on Memoryless Property of Exponential Random Variables 10D Conditioning on More than One Random Variable 25 Answers to Exercises of Section 10B 31

3

20

Section 10A

Additional Examples

3

Chapter 10

More E xpectations and Variances 10A

ADDITIONAL EXAMPLES

Example 10a Three points A, B, and C are randomly placed on a circle. What is the probability that they lie on a semicircle?

H K

O

A B Figure 10c

The circle of Example 10a.

Solution: Let O be the center of the circle. After A and B are placed, draw the diameter through A intersecting the circle at K (see Figure 10c). Also draw the diameter through B intersecting the circle at H. Let the measure of the acute angle AOB be X (note that the probability is 0 that the angle AOB is straight). The random variable X is uniform over the interval (0, π). So f, the probability density function of X, is given by ⎧ ⎨1/π 0<x<π f (x) = ⎩0 otherwise.

Let E be the event that C is placed on the arc HAK. Clearly, A, B, and C lie on a

Section 10A

Additional Examples

4

semicircle if and only if E occurs. Thus the desired probability is π π 1 P (E | X = x)f (x) dx = P (E | X = x) dx P (E) = π 0 0 π 3 2π − x 1 = · dx = .

2π π 4 0 Remark: Note that A, B and C lie on a semicircle if and only if the triangle ABC is obtuse. This example implies the fact that 75% of “all triangles” are obtuse. That is, if from the class of all triangles a triangle is selected at random, then the probability is 3/4 that it is obtuse. (See Exercise 8, Section 7.1 of the book.)

Example 10b The lifetimes of batteries (in hours) manufactured by a certain company are identically distributed with distribution and probability density functions F and f, respectively. (a)

In terms of F, f, and s, find the expected value of the lifetime of an s-hour-old battery.

(b)

Suppose that the lifetimes of the batteries are exponential random variables with identical parameter λ. Find the expected value of the lifetime of a s-hour-old device, and the expected value of the lifetime of a battery, which is not yet s hours old.

Solution: (a) Let X be the lifetime of the s-hour-old device. We want to calculate E(X | X > s). Let FX|X>s (t) = P (X ≤ t | X > s),

and fX|X>s (t) = FX|X>s (t). Then E(X | X > s) =

∞ 0

tfX|X>s (t) dt.

Now P (X ≤ t, X > s) FX|X>s (t) = P (X ≤ t | X > s) = P (X > s) ⎧ ⎪ if t ≤ s ⎪ ⎨0 = P (s < X ≤ t) ⎪ ⎪ ⎩ if t > s.

P (X > s) Therefore, FX|X>s (t) =

⎧ ⎪ 0 ⎪ ⎨

if t ≤ s

F (t) − F (s) ⎪ ⎪ ⎩ 1 − F (s)

if t > s.

Section 10A

Additional Examples

5

Differentiating F X|X>s (t) with respect to t, we obtain

fX|X>s (t) =

⎧ ⎪ 0 ⎪ ⎨

if t ≤ s

f (t) 1 − F (s)

if t > s.

tfX|X>s (t) dt =

1 1 − F (s)

⎪ ⎪ ⎩

This yields E(X | X > s) =

∞ 0

∞ s

tf (t) dt.

(b) Given that the battery is s hours old, by part (a), its probability density function, fX|X>s , is given by

0 if t ≤ s fX|X>s (t) = −λ(t−s) if t > s, λe and E(X | X > s) = eλs

∞ s

λte−λt dt = s +

1 , λ

a result expected because of the memoryless propoerty of exponential random variables.

Now, by Theorem 10.10 of the book, E(X) = E(X | X < s)P (X < s) + E(X | X > s)P (X > s), which implies that

1 1 = E(X | X < s) · (1 − e−λs ) + s + · e−λs . λ λ Solving this equation for E(X | X < s), we have E(X | X < s) =

se−λs 1 − . λ 1 − e−λs

Note that, in this example, we have established another important property of exponential random variables: Let X be an exponential random variable with parmeter λ. For s > 0, E(X | X > s) = s + E(X | X < s) =

1 , λ

se−λs 1 − . λ 1 − e−λs

Section 10A

Additional Examples

6

Example 10c A spice company distributes cinnamon in one-pound bags. Suppose that the Food and Drug Administration (FDA) considers more than 500 insect fragments in one bag excessive and hence unacceptable. To meet the standards of the FDA, the quality control division of the company begins inspecting the bags of the cinnamon at time t = 0 according to the following scheme. It inspects each bag with probability α until it encounters an unacceptable bag. At that point, the division inspects every single bag until it finds m consecutive acceptable bags. When this happens, the division has completed one inspection cycle. It then resumes its normal inspection process. Inspected bags that are found with excessive numbers of insect fragments are sent back for further cleaning. Let p be the probability that a bag is acceptable, independent of the number of insect fragments in other bags. Find the expected value of the number of bags inspected in one inspection cycle.

Solution: Let X be the number of bags inspected until an unacceptable bag is found. Let Kn be the number of consequent bags inspected until n consecutive acceptable bags are found. The number of bags inspected in one inspection cycle is X + K m. We are interested in E(X + Km ) = E(X) + E(Km). Clearly, X is a geometric random variable with parameter α(1 − p). So E(X) = 1/ α(1 − p) . To find E(Km), note that ∀n, E(Kn) = E E(Kn | Kn−1 ) .

Now E(Kn | Kn−1 = i) = (i + 1)p + i + 1 + E(Kn) (1 − p) = (i + 1) + (1 − p)E(Kn).

(1)

To derive this relation, we noted the following. It took i inspections to find n−1 consecutive acceptable bags. If the next bag inspected is also acceptable, we have the n consecutive acceptable bags required in i + 1 inspections. This occurs with probability p. However, if the next bag inspected is unacceptable, then, on the average, we need an additional E(K n) inspections a total of i + 1 + E(K n) inspections until we get n consecutive acceptable bags of cinnamon. This happens with probability 1 − p.

From (1), we have E(Kn | Kn−1 ) = (Kn−1 + 1) + (1 − p)E(Kn).

Finding the expected values of both sides of this relation gives E(Kn) = E(Kn−1 ) + 1 + (1 − p)E(Kn).

Solving for E(K n), we obtain E(Kn) =

1 E(Kn−1 ) + . p p

Noting that E(K 1) = 1/p and solving recursively, we find that E(Kn) =

1 1 1 + 2 + ···+ n. p p p

Section 10A

Additional Examples

7

Therefore, the desired quantity is E(X + Km ) = E(X) + E(Km) =

=

1 1 1 1 + 1 + + · · · + m−1 α(1 − p) p p p 1 1 + · α(1 − p) p

1

m

−1 (1 − α)pm + α p = .

1 αpm (1 − p) −1 p

Example 10d A system has two components, the lives of which are independent exponential random variables with means 1/λ 1 and 1/λ2, respectively. The system functions if and only if at least one of the components operates. For i = 1, 2, when component i fails, it is replaced with a new component that has the same lifetime distribution as the old one.

The time it takes to replace component i is exponential with mean 1/μ i independently of the lifetimes of the components. If the system is presently functioning with both components operative, what is the expected value of the time until the system fails?

Solution: For i = 1, 2, let X i be the lifetime of component i, and let T i be the time it takes to replace component i when it fails. We say that the system is in state 0 if none of its components operates, it is in state 1 if its first component operates and its second component does not, it is in state 2 if the second component operates but its first component does not, and, finally, it is in state 3 if both of its components operate. For i = 1, 2, 3, let W i be the time until the system fails given that it is currently in state i. We are interested in E(W 3 ).

Given that the system is in state 3, let N be the next state of the system. By Theorem 10.6 of the book, E(W3) = E E(W3 | N ) = E(W3 | N = 1)P (N = 1) + E(W3 | N = 2)P (N = 2).

(2)

Considering the memoryless property of exponential random variables, by Remark 10.3 of the book, E(W3 | N = 1) = E(X2 | X2 < X1 ) + E(W1 ) =

1 + E(W1 ) λ1 + λ2

E(W3 | N = 2) = E(X1 | X1 < X2 ) + E(W2 ) =

1 + E(W2 ). λ1 + λ2

Now, by Remark 10.2 of the book, P (N = 1) = P (X2 < X1 ) =

λ2 , λ1 + λ2

P (N = 2) = P (X1 < X2 ) =

λ1 . λ1 + λ2

Section 10A

Additional Examples

8

Thus relation (2) gives E(W3 ) = =

1 λ2 1 λ1 + E(W1) + + E(W2 ) λ1 + λ2 λ1 + λ2 λ1 + λ2 λ1 + λ2 1 λ2 λ1 + E(W1) + E(W2 ). λ1 + λ2 λ1 + λ2 λ1 + λ2

(3)

Now suppose that the system is in state 1. Letting N be the next state of the system, by Theorem 10.6 of the book, E(W1) = E E(W1 | N ) = E(W1 | N = 0)P (N = 0) + E(W2 | N = 3)P (N = 3).

(4)

Taking the memoryless property of exponential random variables into account, we have E(W1 | N = 0) = E(X1 | X1 < T2 ) =

1 , λ1 + μ2

E(W1 | N = 3) = E(T2 | T2 < X1 ) + E(W3) =

1 + E(W3). λ1 + μ2

Noting that λ1 , λ1 + μ2 μ2 , P (N = 3) = P (T2 < X1 ) = λ1 + μ2 P (N = 0) = P (X1 < T2 ) =

relation (4) gives E(W1) = =

1 λ1 1 μ2 · + + E(W3) · λ1 + μ2 λ1 + μ2 λ1 + μ2 λ1 + μ2 1 μ2 + E(W3 ). λ1 + μ2 λ1 + μ2

(5)

By symmetry, we also have E(W2) =

1 μ1 + E(W3). λ2 + μ1 λ2 + μ1

(6)

Substituting (5) and (6) in (3) and solving for E(W 3) yields E(W3 ) =

λ1 (λ1 + μ2 ) + λ2 (λ2 + μ1 ) + (λ1 + μ2 )(λ2 + μ1 ) . λ1 λ2 (λ1 + λ2 + μ1 + μ2 )

Example 10e Let X be a binomial random variable with parameters n and p, where p itself is a random variable, which has a beta distribution with parameters α and β. Find the probability mass function of X.

Section 10A

Additional Examples

9

Solution: Let f be the probability density function of p. By conditioning on p, for 0 ≤ i ≤ n, we have 1 P (X = i | p = x)f (x) dx P (X = i) = 0

1 n i Γ(α + β) α−1 x = (1 − x)β−1 dx x (1 − x)n−i Γ(α)Γ(β) i 0

n Γ(α + β) 1 α+i−1 x (1 − x)n−i+β−1 dx = i Γ(α)Γ(β) 0

n Γ(α + β) Γ(α + i)Γ(β + n − i) · . = Γ(α + β + n) i Γ(α)Γ(β)

n Γ(n + 1) . = Γ(i + 1)Γ(n − i + 1) i

Note that

So P (X = i) =

Γ(n + 1)Γ(α + β)Γ(α + i)Γ(β + n − i) , Γ(i + 1)Γ(n − i + 1)Γ(α)Γ(β)Γ(α + β + n)

i = 0, 1, . . .n.

Example 10f An Alternative Proof for Cauchy-Schwarz Inequality dom variables X and Y with finite variances,

E(XY ) ≤ E(X 2)E(Y 2 ).

Proof:

For ran-

For all real numbers t, (X − tY ) 2 ≥ 0. Hence, for all values of t, X 2 − 2XY t + t2 Y 2 ≥ 0.

Since nonnegative random variables have nonnegative expected values, E(X 2 − 2XY t + t2 Y 2 ) ≥ 0, which implies that E(X 2) − 2E(XY )t + t2 E(Y 2 ) ≥ 0.

Rewriting this as a polynomial in t of degree 2, we get E(Y 2 )t2 − 2E(XY )t + E(X 2) ≥ 0.

It is a well-known fact that if a polynomial of degree 2 is nonnegative, its discriminant is nonpositive. Therefore, 4 E(XY )

2

− 4E(X 2)E(Y 2 ) ≤ 0

Section 10A

Additional Examples

10

or E(XY )

2

≤ E(X 2)E(Y 2 ).

This gives

E(XY ) ≤ E(X 2)E(Y 2 ).

Example 10g A device has n components the lifetimes of which are independent exponential random variables with parameters λ 1 , λ2 , . . . , λn , respectively. What is the probability that the ith component is the first to fail?

Solution: Let X 1 , X2 , . . . , Xn be the lifetimes of the components. Let Y = min(X1 , . . . , Xi−1 , Xi+1 , . . . , Xn).

By Remark 9.2 of the book, Y is exponential with parameter λ 1 +· · ·+λi−1 +λi+1 +· · ·+λn .

By Remark 10.2of the book, the desired probability is

P min(X1 , . . . , Xn) = Xi = P (Xi < Y ) =

λi . λ1 + λ2 + · · · + λn

Remark: Note that this example implies the following important property of the exponential random variables.

Let X1 , X2 , . . . , Xn be n > 1 independent exponential random variables with parameters λ1 , λ2 , . . . , λn , respectively. Then

P min(X1 , X2, . . . , Xn) = Xi =

λi . λ1 + λ2 + · · · + λn

Example 10h Let X1 , X2 , . . . , Xn be n independent exponential random variables with parameters λ1 , λ2 , . . . , λn , respectively. Let X = min(X 1 , X2, . . . , Xn), and define N to be i if and only if X = X i . Therefore, N is a discrete random variable with the set of possible values {1, 2, . . ., n} and probability mass function given by P (N = i) = P (X = Xi) =

λi . λ1 + λ2 + · · · + λn

(See Example 10g.) Show that X and N are independent random variables.

Solution: It suffices to show that P (X > t, N = i) = P (X > t)P (N = i).

Section 10A

Additional Examples

11

We have P (X > t, N = i) = P (X > t, X = Xi) ∞ P (X > t, X = Xi | Xi = x)λie−λi x dx = 0

= = =

∞ t

∞ t

∞ t

= λi =

P (X = Xi | Xi = x)λie−λi x dx P (X1 ≥ x, . . . , Xi−1 ≥ x, Xi+1 ≥ x, . . . , Xn ≥ x | Xi = x)λie−λi x dx P (X1 ≥ x) · · · P (Xi−1 ≥ x)P (Xi+1 > x) · · ·P (Xn ≥ x)λi e−λi x dx

∞ t

e−λ1 x · e−λ2 x · · · e−λi−1 x · e−λi+1 x · · · e−λn x · e−λi x dx

λi · e−(λ1 +λ2 +···+λn )t λ1 + λ2 + · · · + λn

= P (X = Xi )P (X > t) = P (N = i)P (X > t).

Example 10i Suppose that there are k, k > 1, types of defects, identified by a wire manufacturing company, that occur independently over the cable as it comes out of a machine.

For 1 ≤ i ≤ k, the number of defects of type i is distributed according to a Poisson process with parameter λi. Find the probability that the nth defect occurring over the cable is of type i, 1 ≤ i ≤ k.

Solution: For 1 ≤ j ≤ k, let V j be the number of meters on the cable until the first defect of type i occurs. Note that P Vi = min(V1 , V2 , . . . , Vk ) is the probability that the first defect occurring over the cable is of type i. Suppose that the first defect occurs when exactly u meters of the cable are produced. That point labeled t = 0, from there on, by stationarity and independent-increment property, probabilistically, the behaviors of the k given Poisson processes are identical to the original processes when the first u meters of cable wire were being produced. So the probability that the second defect is of type i is identical to the probability that the first defect is of type i and so on. Hence, by Example 10g, the desired quantity is

λi , P Vi = min(V1 , V2 , . . ., Vk ) = λ1 + λ2 + · · · + λk since Vi ’s are independent exponential random variables with parameters λ 1 , λ2 , · · · , λk , respectively.

Example 10j (Investment) Dr. Caprio has invested money in three uncorrelated assets; 25% in the first one, 43% in the second one, and 32% in the third one. The means of the annual rate of returns for these assets, respectively, are 10%, 15%, and 13%. Their standard deviations are 8%, 12%, and 10%, respectively. Find the mean and standard deviation of the annual rate of return for Dr. Caprio’s total investment.

Section 10A

Additional Examples

12

Solution: Let r be the annual rate of return for Dr. Caprio’s total investment. Let r 1 , r2 , and r3 be the annual rate of returns for the first, second, and third assets, respectively. By relation (4) of Section 4B, r = 0.25r1 + 0.43r2 + 0.32r3.

Thus E(r) = 0.25E(r1) + 0.43E(r2) + 0.32E(r3) = (0.25)(0.10) + (0.43)(0.15) + (0.32)(0.13) = 0.1311.

Since the assets are uncorrelated, by relation (10.11) of the book, Var(r) = (0.25)2 Var(r1 ) + (0.43)2 Var(r2 ) + (0.32)2 Var(r3 ) = (0.25)2(0.08)2 + (0.43)2(0.12)2 + (0.32)2(0.10)2 = 0.004087. √ Therefore, σr = 0.004087 = 0.064. Hence Dr. Caprio should expect an annual rate of return of 13.11% with standard deviation 6.4%. Note that Dr. Caprio has reduced the standard deviation of his investments considerably by diversifying his investment; that is, by not putting all of his eggs in one basket.

In general, If Dr. Caprio divides his money into n equal portions and invests them in n uncorrelated financial assets with identically distributed rates of returns r1 , r2 , . . . , rn , each with variance σ2 , then by (10.11), 1 1 1 Var(r) = Var r1 + r2 + · · · + rn n n n =

1 1 1 σ2 1 2 .

Var(r ) + Var(r ) + · · · + Var(r ) = · nσ = 1 2 n n2 n2 n2 n2 n

This shows that the more financial assets Dr. Caprio invests in, the less the variance of the return will be—an indication that diversification reduces the investment risks. .

Example 10k (Investment) Dr. Siciliano has invested money in two financial assets.

Let r be the annual rate of return for his total investment. Let r 1 and r2 be the annual rates of return for the first and second assets, respectively. Let σ 2 = Var(r), σ12 = Var(r1 ), and σ22 = Var(r2 ). Prove that

σ 2 ≤ max σ12 , σ22 .

In particular, by this inequality, if r 1 and r2 are identically distributed, then σ 12 = σ22 implies that σ 2 ≤ σ12 = σ22 .

This is an extension of the result discussed in Example 10j. It shows that, even if the financial assets are correlated, still diversification reduces the investment risk.

Section 10A

Additional Examples

13

Proof: We will show that if σ 22 ≤ σ12 , then σ 2 ≤ σ12 . By symmetry, if σ12 ≤ σ22 , then σ 2 ≤ σ22 . Therefore, σ2 is either less than or equal to σ 12 , or it is less than or equal to σ 22 , implying that

σ 2 ≤ max σ12 , σ22 .

To show that σ 22 ≤ σ12 implies that σ 2 ≤ σ12 , let w1 and w2 be the fractions of Dr. Siciliano’s investment in the first and second financial assets, respectively. Then w 1 + w2 = 1 and r = w1 r1 + w2 r2 .

Let ρ be the correlation coefficient of r 1 and r2 . Then Cov(r1 , r2 ) = ρσ1 σ2 .

We have Var(r) = Var(w1 r1 + w2 r2 ) = Var(w1 r1 ) + Var(w2 r2 ) + 2Cov(w1 r1 , w2 r2 ) = w12 Var(r1 ) + w22 Var(r2 ) + 2w1 w2 Cov(r1 , r2 ).

Noting that w 1 + w2 = 1, −1 ≤ ρ ≤ 1, and σ22 ≤ σ12 , this relation implies that σ 2 = w12 σ12 + w22 σ22 + 2w1 w2 ρσ1 σ2 ≤ w12 σ12 + w22 σ12 + 2w1 w2 σ12 = (w12 + w22 + 2w1 w2 )σ12 = (w1 + w2 )2 σ12 = σ12 .

Example 10 (Investment) Mr. Ingham has invested money in three assets; 18% in the first asset, 40% in the second one, and 42% in the third one. Let r 1 , r2 , and r3 be the annual rate of returns for these three investments, respectively. For 1 ≤ i, j ≤ 3, Cov(r i, rj ) is the ith element in the jth row of the following table. Note that Var(r i) = Cov(ri, ri). r1 r2 r3

r1

r2

r3

0.064 0.03 0.015

0.03 0.0144 0.021

0.015 0.021 0.01

Find the standard deviation of the annual rate of return for Mr. Ingham’s total investment.

Solution: Let r be the annual rate of return for Mr. Ingham’s total investment. We have Var(r) = Var(0.18r1 + 0.40r2 + 0.42r3) = (0.18)2 Var(r1 ) + (0.40)2 Var(r2 ) + (0.42)2 Var(r3 ) + 2(0.18)(0.40)Cov(r1 , r2) + 2(0.18)(0.42)Cov(r1 , r3 ) + 2(0.40)(0.42)Cov(r2 , r3) = (0.18)2(0.064) + (0.40)2(0.0144) + (0.42)2(0.01) + 2(0.18)(0.40)(0.03) + 2(0.18)(0.42)(0.015) + 2(0.40)(0.42)(0.021) = 0.01979.

Section 10A

Additional Examples

14

Hence √ the standard deviation of the annual rate of return for Mr. Ingham’s total investment is 0.01979 = 0.14.

Example 10m (Investment) Mr. Kowalski has invested $50,000 in three uncorrelated financial assets: 25% in the first financial asset, 40% in the second one, and 35% in the third one. The annual rates of return for these assets, respectively, are normal random variables with means 12%, 15%, and 18%, and standard deviations 8%, 12%, and 15%.

What is the probability that, after a year, Mr. Kowalski’s net profit from this investment is at least $10,000? Ignore transaction costs and assume that there is no annual dividend.

Solution: Let r 1 , r2 , and r3 be the annual rates of return for Mr. Kowalski’s investments in financial assets 1, 2, and 3, respectively. Let r be the annual rate of return for his total investment. Then, by relation (4) of Section 4B, r = 0.25r1 + 0.40r2 + 0.35r3.

Since the assets are uncorrelated, we have E(r) = (0.25)(0.12) + (0.40)(0.15) + (0.35)(0.18) = 0.153, Var(r) = (0.25)2(0.08)2 + (0.40)2(0.12)2 + (0.35)2(0.15)2 = 0.00546,

σr = Var(r) = 0.074.

Hence r ∼ N (0.153, 0.00546). Let X be the total investment of Mr. Kowalski. We are given that X = 50, 000. Let Y be the total return of Mr. Kowalski’s investment next year.

The desired probability is Y − X 10, 000 ≥ P (Y − X ≥ 10, 000) = P X 50, 000

0.2 − 0.153 = P (r ≥ 0.2) = P Z ≥ 0.074 = P (Z ≥ 0.64) = 1 − Φ(0.64) = 1 − 0.7389 = 0.2611.

Example 10n (Genetics) Hemophilia is a sex-linked disease with normal allele H dominant to the mutant allele h. Kim and John are married, and John is phenotypically normal. Suppose that, in the entire population, the frequencies of H and h are 0.98 and 0.02, respectively. If Kim and John have four sons and three daughters, what is the expected number of their hemophilic children?

Solution: Since hemophilia is a sex-linked disease, and John is phenotypically normal, John is H. Therefore, no matter what Kim’s genotype is, none of the daughters has hemophilia. Whether a boy has hemophilia or not depends solely on the genotype of Kim. Let X be the number of the boys who have hemophilia. To find, E(X), the expected number of the boys who have hemophilia, let ⎧ ⎪ 0 if Kim is hh ⎪ ⎨ Z = 1 if Kim is Hh ⎪ ⎪ ⎩ 2 if Kim is HH.

Section 10A

Additional Examples

15

Then E(X) = E E(X | Z) = E(X | Z = 0)P (Z = 0) + E(X | Z = 1)P (Z = 1) + E(X | Z = 2)P (Z = 2) = 4(0.02)(0.02) + 4(1/2) 2(0.98)(0.02) + 0 0.98)(0.98) = 0.08.

Therefore, on average, 0.08 of the boys and hence 0.08 of the children are expected to have hemophilia.

Section 10B

10B

Pattern Appearance

16

PATTERN APPEARANCE

Suppose that a coin is tossed independently and successively. We are interested in the expected number of tosses until a specific pattern is first obtained. For example, the expected number of tosses until the first appearance of HT, the first appearance of HHH, or the first appearance of, say, HHTHH. Similarly, suppose that in generating random numbers from the set {0, 1, 2, . . ., 9} independently and successively, we are interested in the expected number of digits to be generated until the first appearance of 453, the first appearance of 353, or the first appearance of, say, 88588.

In coin tossing, for example, we can readily calculate the expected number of tosses after the first appearance of a pattern until its second appearance. The same is true in generating random numbers and other similar experiments. As an example, shortly, we will show that it is not difficult to calculate the expected number of digits to be generated after the first appearance of a pattern until its second appearance. We use this to calculate the expected number of digits to be generated until the first appearance of the pattern. First, we introduce some notation and make some definitions: Suppose that A is a pattern of certain characters. Let the number of random characters generated until A appears for the first time be denoted by → A. Let the number of random characters generated after the appearance of a pattern A until the next appearance of a pattern B be denoted by A → B. Definition 10a Suppose that A is a pattern of length n of certain characters. For 1 ≤ i < n, let A(i) and A(i) be the first i and the last i characters of A, respectively. If A(i) = A(i) for some i (1 ≤ i < n), we say that the pattern is overlapping. Otherwise, we say it is a pattern with no self-overlap. For an overlapping pattern A, if A (i) = A(i) , then i is said to be an overlap number.

For example, in successive and independent tosses of a coin, let A =HHTHH; we have A(1) = H, A(2) = HH, A(3) = HHT, A(4) = HHTH; A(1) = H, A(2) = HH, A(3) = THH, A(4) = HTHH. We see that A is overlapping with overlap numbers 1 and 2. As another example, in generating random digits from the set {0, 1, 2, . . . , 9} independently and successively, consider the pattern A = 45345; we have A(1) = 4, A(2) = 45, A(3) = 453, A(4) = 4534; A(1) = 5, A(2) = 45, A(3) = 345, A(4) = 5345.

Since A(2) = A(2), the pattern A is overlapping, and 2 is its only overlap number. As a third example, the pattern 333333 is overlapping. Its overlap numbers are 1, 2, 3, 4, and 5.

The patterns 3334 and 44565 are patterns with no self-overlap.

Section 10B

Pattern Appearance

17

Next, we make the following observations: For a pattern with no self-overlap such as 453, E(→ 453) = E(453 → 453). That is, the expected number of random numbers to be generated until the first 453 appears is the same as the expected number of random numbers to be generated after the first appearance of 453 until its second appearance. However, this does not apply to an overlapping pattern such as 353. To our surprise, E(→ 353) is larger than E(353 → 353). This is because E(353 → 353) = E(3 → 353) and E(→ 353) = E(→ 3) + E(3 → 353) = E(→ 3) + E(353 → 353).

That is, the expected number of digits to be generated until the pattern 353 appears is the expected number of digits to be generated until the first appearance of 3, plus the expected number of additional digits after that until 353 is obtained for the first time. In general, for patterns of the same length, on average, overlapping ones occur later than those with no self-overlap. For example, when random digits are generated one after another independently and successively, on average, 453 appears sooner than 353, and 353 appears sooner than 333. Note that E(→ 333) = E(→ 3) + E(3 → 33) + E(33 → 333) = E(→ 3) + E(33 → 33) + E(333 → 333).

In general, let A be an overlapping pattern. Let i 1 , i2 , i3 , . . . , im be the overlap numbers of A arranged in (strictly) ascending order. Then E(→ A) = E(→ A(i1 ) ) + E(A(i1) → A(i2 ) ) + E(A(i2) → A(i3 )) + · · · + E(A(im) → A).

Clearly, the number of digits to be generated to obtain, say, the first 3, the first 5, or the first 8 are all geometric random variables with parameter p = 1/10. So E(→ 3) = E(→ 5) = E(→ 8) =

1 = 10. p

In general, for a pattern A, this and E(A → A) enable us to find E(→ A). So all that remains is to find the expected number of digits to be generated after the first appearance of a pattern until its second appearance. To demonstrate one method of calculation, we will find E(353 → 353) using an intuitive probabilistic argument. To do so, in the process of generating random digits, let D 1 be the first random digit and D 2 be the first two random digits generated. For i ≥ 3, let D i be the (i − 2)nd, (i − 1)st, and ith random digits generated. For example, if the first 6 random digits generated are 507229, then D 1 = 5, D2 = 50, D3 = 507, D4 = 072, D5 = 722, and D6 = 229. Let X1 = X2 = 0. For i ≥ 3, let

1 if Di = 353 Xi = 0 if Di = 353.

Then the number of appearances of 353 among the first n random ndigits generated is

n i=1 Xi. Hence the proportion of the D i’s that are 353 is (1/n) i=1 Xi when n random digits are generated. Now E(X 1) = E(X2) = 0, and for i ≥ 3, E(Xi) = E(X3)

Section 10B

Pattern Appearance

18

since the X i ’s are identically distributed. Thus the expected value of the fraction of D i ’s that are 353 is E

n 1

n

i=1

Xi

n n 1 1 1 Xi = E(Xi) = · (n − 2)E(X3) = E n n n i=1 i=1

3 n−2 1 n−2 n−2 P (D3 = 353) = . = = n n 10 1000 n

Thus the expected value of the fraction of D i ’s in n random digits that are 353 is (n − 2)/(1000n). Now as n → ∞, this expected value approaches 1/1000, implying that in the long-run, on average, the fraction of D i ’s that are 353 is 1/1000. This and the fact that the expected number of random digits between any two consecutive 353’s is the same, imply that the average number of random digits between two consecutive 353’s is 1000.

Hence E(353 → 353) = 1000. A rigorous proof for this fact needs certain theorems of renewal theory, a branch of stochastic processes. See, for example, Stochastic Modeling and the Theory of Queues, by Ronald W. Wolff, Prentice Hall, 1989. We can use the argument above to show that, for example, E(23 → 23) = 100, E(453 → 453) = 1000, E(333 → 333) = 1000, E(5732 → 5732) = 10, 000.

Therefore, E(→ 453) = E(453 → 453) = 1000, E(→ 353) = E(→ 3) + E(3 → 353) = E(→ 3) + E(353 → 353) = 10 + 1000 = 1010, E(→ 333) = E(→ 3) + E(3 → 33) + E(33 → 333) = E(→ 3) + E(33 → 33) + E(333 → 333) = 10 + 100 + 1000 = 1110, and E(→ 88588) = E(→ 8) + E(8 → 88) + E(88 → 88588) = E(→ 8) + E(88 → 88) + E(88588 → 88588) = 10 + 100 + 100000 = 100110.

An argument similar to the preceding will establish the following generalization.

Section 10B

Pattern Appearance

19

Suppose that an results in one of the outcomes a1 , a2 , experiment

k . . . , ak , and P {ai } = pi , i = 1, 2, . . ., k; i=1 pi = 1. Let ai1 , ai2 , . . . , ai , be (not necessarily distinct) elements of {a1 , a2 , . . ., ak }.

Then in successive and independent performances of this experiment, the expected number of trials after the first appearance of the pattern ai1 ai2 · · · ai until its second appearances is 1/(pi1 pi2 · · · pi ).

By this generalization, for example, in successive independent flips of a fair coin, E(→ HH) = E(→ H) + E(H → HH) = E(→ H) + E(HH → HH) 1 1 + = 6, = 1/2 (1/2)(1/2) whereas E(→ HT) = E(HT → HT) =

1 = 4.

(1/2)(1/2)

Similarly, E(→ HHHH) = E(→ H) + E(H → HH) + E(HH → HHH) + E(HHH → HHHH) = E(→ H) + E(HH → HH) + E(HHH → HHH) + E(HHHH → HHHH) = 2 + 4 + 8 + 16 = 30, whereas E(→ THHH) = E(THHH → THHH) = 16.

At first glance, it seems paradoxical that, on the average, it takes nearly twice as many flips of a fair coin to obtain the first HHHH as to encounter THHH for the first time. However, THHH is a pattern with no self-overlap, whereas HHHH is an overlapping pattern.

EXERCISES FOR SECTION 10B

1.

Suppose that random digits are generated from the set {0, 1, . . ., 9} independently and successively. Find the expected number of digits to be generated until the pattern (a) 007 appears, (b) 156156 appears, (c) 575757 appears.

2.

In successive independent flips of a fair coin, what is the expected number of trials until the pattern THTHTTHTHT appears?

Section 10C

10C

More on Memoryless Properties of Exponential Random Variables

20

MORE ON MEMORYLESS PROPERTY OF EXPONENTIAL RANDOM VARIABLES

In this section, we will discuss an aspect of the exponential random variables, which demonstrates the validity of memoryless property even in more complicated situations. Suppose that a system has two devices, the lifetime of which, X 1 and X2 , are independent exponential random variables with means 1/λ 1 and 1/λ2 , respectively. We will show that, given device 1 fails before device 2, the remaining lifetime of device 2 is still exponential with the mean 1/λ2 independently of the lifetime of device 1. This property has many useful applications and is the subject of the following theorem.

Theorem 10a Let X1 and X2 be independent exponential random variables with parameters λ 1 and λ2 , respectively. Then, for t ≥ 0, P (X2 − X1 > t | X2 > X1 ) = e−λ2 t; That is, given X 2 > X1 , the random variable X 2 − X1 is exponential with parameter λ 2 .

Furthermore, X 2 − X1 and X1 are independent random variables.

Proof: Using the definition of conditional probability and Remark 10.2 of the book yields P (X2 > X1 + t) P (X2 − X1 > t, X2 > X1 )

= P (X2 > X1 ) λ1 (λ1 + λ2 )

λ1 + λ2 ∞ P (X2 > X1 + t | X1 = x)λ1 e−λ1 x dx = λ1 0 ∞ λ1 + λ2 P (X2 > x + t)λ1 e−λ1 x dx = λ1 0

λ1 + λ2 ∞ −λ2 (x+t) e · λ1 e−λ1 x dx = λ1 0

P (X2 − X1 > t | X2 > X1 ) =

=

λ1 λ1 + λ2 · e−λ2 t = e−λ2 t . λ1 λ1 + λ2

The following shows that given X 2 > X1 , the random variables X 2 − X1 and X1 are independent.

P (X2 − X1 > t2 | X2 > X1 , X1 = t1 ) = =

P (X2 − X1 > t2 , X2 > X1 | X1 = t1 ) P (X2 > X1 | X1 = t1 )

e−λ2 (t1 +t2 ) P (X2 > t1 + t2 ) = = e−λ2 t2 P (X2 > t1 ) e−λ2 t1

= P (X2 − X1 > t1 | X2 > X1 ).

Theorem 10a can be generalized in the following way. Suppose that a system has n (n > 2) devices, the lifetime of which, X 1 , X2 , . . . , Xn , are independent exponential random variables with means 1/λ 1, 1/λ2 , . . . , 1/λn, respectively. Suppose that we are given that device 1 is the first device to fail. Given this information, X 2 − X1 , X3 − X1 ,

Section 10C

More on Memoryless Properties of Exponential Random Variables

21

. . . , Xn − X1 are n − 1 independent exponential random variables with means 1/λ 2 , 1/λ3, . . . , 1/λn , respectively. Furthermore, Xi − X1 is independent of X 1 for 2 ≤ i ≤ n.

Example 10o Upon arrival at a shoe store, the times it will take Sara and Martha to buy a pair of shoes are independent exponential random variables with means 1/λ and 1/μ, respectively. If Sara and Martha enter the shoe store at the same time, find the expected length of time until they have both purchased pairs of shoes if (a) 1/λ = 1/μ = 30 minutes; (b) 1/λ = 30 minutes, but 1/μ = 45 minutes.

Solution: Let X be the time it will take for Sara to buy a pair of shoes, and let Y be the time it will take for Martha to buy a pair of shoes. Let S 1 be the time when exactly one of Sara or Martha has purchased a pair of shoes, and S 2 be the time when both of them have purchased shoes. (a)

1 = 15 (1/30) + (1/30) minutes. At S 1 , by Theorem 10a, for whoever is not done buying shoes yet, the remaining time to complete her purchase is exponential with mean 30 minutes. Hence E(S2 − S1 ) = 30 minutes and thus

By Remark 9.2 of the book, E(S1) = E min(X, Y ) =

E(S2 ) = E(S1 ) + E(S2 − S1 ) = 45 minutes. (b)

1 = 18 minutes. To find (1/30) + (1/45) E(S2 − S1 ), by Theorem 10.10 of the book, In this case, E(S1 ) = E min(X, Y ) =

E(S2 − S1 ) = E(S2 − S1 | X < Y )P (X < Y ) + E(S2 − S1 | Y < X)P (Y < X) = E(Y − X | X < Y )P (X < Y ) + E(X − Y | Y < X)P (Y < X) = 45 ·

(1/45) 1/30 + 30 · = 39 minutes.

(1/30) + (1/45) (1/30) + (1/45)

Hence E(S2) = E(S1) + E(S2 − S1 ) = 18 + 39 = 57 minutes.

Remark: An alternative solution to this problem is to note that S 2 = max(X, Y ). Hence,

by Remark 6.4of the book, ∞

P max(X, Y ) > t dt E(S2 ) = E max(X, Y ) = = =

∞ 0

∞ 0

0

1 − P (max(X, Y ) ≤ t) dt = 1 − P (X ≤ t)P (Y ≤ t) dt =

= 57 minutes.

∞ 0

∞ 0

1 − P (X ≤ t, Y ≤ t) dt (1 − e−t/30 )(1 − e−t/45 ) dt

Section 10C

More on Memoryless Properties of Exponential Random Variables

22

Example 10p To ship merchandise abroad, by freight, customers must go through two stations sequentially. At the first station, it takes a customer an exponential time with mean 1/α to get clearance from the customs. At the second station, it takes a customer an exponential time with mean 1/β to repackage the merchandise and deliver it to the carrier for shipment. The service times at stations 1 and 2 are independent sequences of independent random variables. Suppose that when Mary arrives to ship her antique chair, Alan is being served in the customs, but there are no customers in the queue or in the second station. Find Mary’s expected waiting time.

Solution: Let X 1 and X2 be Mary’s service times at stations 1 and 2, respectively. Let R be the remaining service time of Alan at station 1 when Mary arrives, and Y be Alan’s service time at station 2. Due to memoryless property of exponential random variables, R is exponential with mean 1/α. Let V be the period between completion times of Mary’s service at stations 1 and 2. Let W be Mary’s total waiting time. Clearly, E(W ) = E(R + X1 + V ) = E(R) + E(X1 ) + E(V ) =

1 2 1 + + E(V ) = + E(V ) α α α

To find E(V ), note that, by Theorem 10.10 of the book, E(V ) = E(V | Y < X1 )P (Y < X1 ) + E(V | Y > X1 )P (Y > X1 ) = E(X2)P (Y < X1 ) + E(Y − X1 | Y > X1 ) + E(X2) P (Y > X1 ).

By Theorem 10a, given Y > X1 , Y − X1 is exponential with mean 1/β. Hence E(V ) =

1 β 1 α β + 2α 1 · + + = . β α+β β β α+β β(α + β)

The final answer is E(W ) =

2 β + 2α 2 + E(V ) = + . α α β(α + β)

Example 10q After a long day of work, the tellers of a seven-teller bank are anxious to leave for the day. However, at 5:00 P.M. when the bank closes, there are seven customers still in the bank being served. Luckily, there is no customer in the queue waiting to be served.

If the service times are identically distributed, independent exponential random variables, each with mean 10 minutes, at what time should we expect the last teller to leave the bank?

Assume that a teller who has served his or her last customer leaves for the day.

Solution: For the tellers’ departure times, suppose that the ith departure occurs Y i minutes past 5:00 P.M. Clearly, Y1 is the minimum of seven independent exponential random variables each with parameter λ = 1/10. Thus, by Remark 9.2 of the book, Y 1 is exponential with parameter 7/10, and hence E(Y1 ) = 10/7 minutes. By Theorem 10a, at Y1 minutes past 5:00 P.M., the remaining service times of the remaining six customers are independent

Section 10C

More on Memoryless Properties of Exponential Random Variables

23

exponential random variables each with parameter 1/10. Therefore, Y 2 − Y1 is the minimum of six independent exponential random variables each with parameter 1/10. Hence Y2 − Y1 is exponential with parameter 6/10, and hence E(Y 2 − Y1 ) = 10/6 minutes. Applying Theorem 10a again, at Y 2 minutes past 5:00 P.M., the remaining service times of the remaining five customers are independent exponential random variables each with parameter 1/10. Thus Y 3 − Y2 is the minimum of five independent exponential random variables each with parameter 1/10, implying that Y 3 − Y2 is exponential with parameter 5/10. Hence E(Y3 − Y2 ) = 10/5 minutes. Continuing this argument, we have that, for 1 ≤ i ≤ 6, E(Yi+1 − Yi ) = 10/(7 − i). Thus E(Y7 ) = E(Y1 ) + E(Y2 − Y1 ) + · · · + E(Y6 − Y5 ) + E(Y7 − Y6 ) =

10 10 10 363 + +···+ + 10 = , 7 6 2 14

which shows that the expected departure time for the last teller for the day is approximately 25 minutes and 56 seconds past 5:00 P.M. Example 10r The Dean of a college buys an 8-sheet crosscut paper shredder for his own office, a 12-sheet crosscut paper shredder for the Center for Writing and Reading (CWR), and a 15-sheet crosscut paper shredder for the main office. Suppose that, under normal usage, which is the case at this College, the lifetimes of these shredders are independent exponential random variables with means 6, 8, and 10 years, respectively. What is the probability that the Dean’s own shredder dies first, the CWR’s shredder dies next, and the main office’s shredder dies last?

Solution: Let X 1 be the lifetime of the Dean’s own shredder, X 2 be the lifetime of the CWR’s shredder, and X3 be the lifetime of the main office’s shredder. We are given that X1 , X2 and X3 are independent exponential random variables with parameters λ 1 = 1/6, λ2 = 1/8, and λ3 = 1/10, respectively. To calculate the desired probability P (X1 < X2 < X3 ), note that, by Example 10g and Theorem 10a, P (X1 < X2 < X3 ) =

P X2 < X3 | min(X1 , X2 , X3) = X1 P min(X1 , X2 , X3) = X1 = P (X2 < X3 | X1 < X2 , X1 < X3 )

λ1 = λ1 + λ2 + λ3

P (X2 − X1 < X3 − X1 | X2 − X1 > 0, X3 − X1 > 0)

λ1 . λ1 + λ2 + λ3

We know that given X 2 −X1 > 0 and X3 −X1 > 0, X2 −X1 and X3 −X1 are independent exponential random variables with parameters λ 2 and λ3 , respectively. Hence P (X2 − X1 < X3 − X1 | X2 − X1 > 0, X3 − X1 > 0) =

λ2 . λ2 + λ3

Section 10C

More on Memoryless Properties of Exponential Random Variables

24

Therefore, P (X1 < X2 < X3 ) = =

λ2 λ1 · λ2 + λ3 λ1 + λ2 + λ3 1/6 100 1/8 · = = 0.236.

(1/8) + (1/10) (1/6) + (1/8) + (1/10) 423

Section 10D

10D

Conditioning on More Than One Random Variable

25

CONDITIONING ON MORE THAN ONE RANDOM VARIABLE

We will now generalize the definition of E(X|Y ) to the multivariate case.

Definition 10b Let X, Y, and Z be three random variables. By the random variable E(X | Y, Z) we mean a function of the random variables Y and Z that is defined to be E(X | Y = y, Z = z) when Y = y and Z = z.

Let p be the joint probability mass function of three discrete random variables X, Y and Z. Let pX|Y,Z (x|y, z) be the conditional probablilty mass function of X given that Y = y and Z = z. Let pY,Z be the joint probability mass function marginalized over Y and Z; that is,

pY,Z (y, z) = p(x, y, z). x

For all y, z for which p Y,Z (y, z) > 0, pX|Y,Z (x|y, z) =

p(x, y, z) . pY,Z (y, z)

Therefore, E(X | Y = y, Z = z) =

x

xpX|Y,Z (x|y, z) =

xp(x, y, z) x

pY,Z (y, z)

.

(7)

This shows that, as a function of Y and Z, at all sample points where Y = y and Z = z, E(X | Y, Z) is defined by relation (7).

Similarly, let f be the joint probability density function of three continuous random variables X, Y, and Z. Let fX|Y,Z (x|y, z) be the conditional probability density function of X given that Y = y and Z = z. Let f Y,Z be the joint probability density function marginalized over Y and Z; then ∞ f (x, y, z) dx. f Y,Z (y, z) = −∞

For all y, z, for which f Y,Z (y, z) > 0, fX|Y,Z (x|y, z) =

f (x, y, z) . f Y,Z (y, z)

Therefore, E(X | Y = y, Z = z) =

∞ −∞

xfX|Y,Z (x|y, z) dx =

∞

xf (x, y, z) dx. −∞ f Y,Z (y, z)

(8)

As in discrete case, this shows that, as a function of Y and Z, at all sample points where Y = y and Z = z, E(X | Y, Z) is defined by relation (8).

Section 10D

Conditioning on More Than One Random Variable

26

z (0,0,3)

,0)

(0,3

y

(2,0,0)

x Figure 10d

The tetrahedron of Example 10s.

Example 10s Let X, Y, and Z be the coordinates of a point selected randomly from the inside of the tetrahedron of Figure10d. (a)

Find E(Z | X, Y ).

(b)

Verify that E E(Z | X, Y ) = E(Z).

Solution: (a)

The equation of the plane passing through the points (2, 0, 0), (0, 3, 0), and (0, 0, 3) is 3x + 2y + 2z = 6. For 0 < x < 2 and 0 < y < (6 − 3x)/2, given that X = x 6 − 3x − 2y . and Y = y, the random variable Z is uniform over the interval 0 , 2 Hence 6 − 3x − 2y .

E(Z | X = x, Y = y) = 4 This implies that 6 − 3X − 2Y , E(Z | X, Y ) = 4 where X and Y are jointly distributed random variables and their

joint distribution is uniform over the set (x, y) : 0 < x < 2, 0 < y < (6 − 3x)/2 .

(b)

The volume of the terahedron in Figure10d is the area of its base, which is (2 × 3)/2 = 3, times one-third of its height, which is 1. So the volume is 3. Let f be the joint probability density function of X, Y, and Z, the coordinates of a point randomly selected from the tetrahedron. Then, by (9.10), ⎧ (x, y, z) ∈ inside the tetrahedron ⎨1/3 f (x, y, z) = ⎩ 0 otherwise.

Section 10D

Conditioning on More Than One Random Variable

27

Clearly, E(Z) = E(X) = E(Y ) =

2 (6−3x)/2 (6−3x−2y)/2 0

0

0

2 (6−3x)/2 (6−3x−2y)/2 0

0

0

2 (6−3x)/2 (6−3x−2y)/2 0

0

0

3 z dz dy dx = , 3 4 1 x dz dy dx = , 3 2 3 y dz dy dx = .

3 4

We see that 6 − 3X − 2Y

1 6 − 3E(X) − 2E(Y ) 4 4 3 1 = 6 − 3(1/2) − 2(3/4) = = E(Z).

4 4

E E(Z | X, Y ) = E

Example 10t

=

Let the joint probability density function of X, Y, and Z be given by f (x, y, z) =

Then f Y,Z (y, z) =

1 0

4 (xy + yz + zx), 3

0 < x, y, z < 1.

2 4 (xy + yz + zx) dx = (2yz + y + z).

3 3

Hence E(X | Y = y, Z = z) =

1

=2

0

xfX|Y,Z (x|y, z) dx =

1 0

1 0

x

f (x, y, z) dx f Y,Z (y, z)

3yz + 2y + 2z x(xy + yz + zx) dx = .

2yz + y + z 3(2yz + y + z)

Thus the random variable E(X | Y, Z) is given by E(X | Y, Z) =

3Y Z + 2Y + 2Z .

3(2Y Z + Y + Z)

The following is a natural extension of the definition of E(X | Y, Z).

Definition 10c Let X, Y1 , . . . , Yn be n + 1 random variables. By the random variable E(X | Y1 , . . . , Yn), we mean a function of the random variables Y 1 , . . . , Yn that is defined to be E(X | Y1 = y1 , . . . , Yn = yn ) when Y1 = y1 , . . . , and Yn = yn .

Section 10D

28

Conditioning on More Than One Random Variable

Let p be the joint probability mass function of the discrete random variables X, Y 1 , Y2 , . . . , Yn . Let pX|Y1 ,Y2 ,...,Yn (x|y1 , y2 , . . . , yn ) be the conditional probability mass function of X given that Y 1 = y1 , Y2 = y2 , . . . , Yn = yn . Let pY1 ,Y2 ,...,Yn be the joint probability mass function marginalized over Y 1 , Y2 , . . . , Yn ; that is,

pY1 ,Y2 ,...,Yn (y1 , y2 , . . . , yn ) =

p(x, y1, y2 , . . . , yn ).

x

For all y1 , y2 , . . . , yn for which pY1 ,Y2 ,...,Yn (y1 , y2 , . . . , yn ) > 0, pX|Y1 ,Y2 ,...,Yn (x | y1 , y2 , . . ., yn ) =

p(x, y1, y2 , . . ., yn ) . pY1 ,Y2 ,...,Yn (y1 , y2 , . . . , yn )

Therefore, E(X | Y1 = y1 , Y2 = y2 , . . . , Yn = yn ) =

x

=

x

xpX|Y1 ,Y2 ,...,Yn (x|y1 , y2 , . . . , yn ) xp(x, y1, y2 , . . ., yn ) . pY1 ,Y2 ,...,Yn (y1 , y2 . . . , yn )

(9)

This shows that, as a function of Y 1 , Y2 , . . . , Yn , at all sample points where Y 1 = y1 , Y2 = y2 . . . , and Yn = yn , E(X | Y1 , Y2 , . . . , Yn ) is defined by relation (9).

Similarly, let f be the joint probability density function of the continuous random variables X, Y1 , Y2 , . . . , Yn . Let fX|Y1 ,Y2 ,...,Yn (x|y1 , y2 , . . . , yn ) be the conditional probability density function of X given that Y 1 = y1 , Y2 = y2 , . . . , Yn = yn . Let f Y1 ,Y2 ,...,Yn be the joint probability density function marginalized over Y 1 , Y2 , . . . , and Yn ; then ∞ f (x, y1 , y2 , . . . , yn ) dx. f Y1 ,Y2 ,...,Yn (y1 , y2 , . . . , yn ) = −∞

For all y1 , y2 , . . . , yn for which f Y1 ,Y2 ,...,Yn (y1 , y2 , . . . , yn ) > 0, fX|Y1 ,Y2 ,...,Yn (x|y1 , y2 , . . . , yn ) =

f (x, y1 , y2 , . . . , yn) . f Y1 ,Y2 ,...,Yn (y1 , y2 , . . . , yn )

Therefore, E(X | Y1 = y1 , Y2 = y2 , . . . , Yn = yn ) = =

∞ −∞

xfX|Y1 ,Y2 ,...,Yn (x|y1 , y2 , . . . , yn ) dx

∞

xf (x, y1 , y2 , . . . , yn ) dx. −∞ f Y1 ,Y2 ,...,Yn (y1 , y2 , . . . yn )

(10)

As in other cases discussed above, this shows that, as a function of Y 1 , Y2 , . . . , Yn , at all sample points where Y 1 = y1 , Y2 = y2 , . . . , and Yn = yn , E(X | Y1 = y1 , Y2 = y2 , . . . , Yn = yn ) is defined by relation (10).

It is important to note that, as in Theorem 10.6 of the book, the expected value of the random variable E(X | Y1 , Y2 , . . . , Yn ) is equal to the expected value of X. To show this

Section 10D

29

Conditioning on More Than One Random Variable

fact, say, for continuous random variables, since E(X | Y 1 = y1 , Y2 = y2 , . . . Yn = yn ) is a function of Y1 , Y2 , . . . , Yn , by Theorem 9.4 of the book, E E(X | Y1 , Y2 , . . . , Yn ) ∞ ∞ ··· E(X | Y1 = y1 , . . ., Yn = yn )f Y1 ,...,Yn (y1 , . . . , yn ) dy1 · · · dyn = −∞

−∞

xf (x, y1 , . . . , yn ) dx f Y1 ,...,Yn (y1 , . . . , yn ) dy1 · · · dyn = ··· −∞ −∞ −∞ f Y1 ,...,Yn (y1 , . . . , yn )

∞ ∞ ∞ = x ··· f (x, y1 , y2 , . . . , yn ) dy1 dy2 · · · dyn dx ∞ ∞

∞

−∞

=

∞

−∞

−∞

−∞

xfX (x) dx = E(X).

In particular, this proof implies that E(X) =

∞ −∞

···

∞ −∞

E(X | Y1 = y1 , . . . , Yn = yn )f Y1 ,...,Yn (y1, . . . , yn ) dy1 · · · dyn .

(11)

The proof of E(X) = E E(X | Y1 , Y2 , . . . , Yn ) in the discrete case is similar to the continuous case. For discrete random variables, the relation analogous to (11) is E(X) =

y1

···

E(X | Y1 = y1, . . . , Yn = yn)P (Y1 = y1, . . . , Yn = yn).

(12)

yn

Note that by (11) and (12), E(X) = E E(X | Y1 , Y2 , . . ., Yn ) is a condensed way to say that E(X) is the weighted average of the conditional expectations of X (given Y 1 = y1 , Y2 = y2 , . . . , Yn = yn ) over all possible values of Y 1 , Y2 , . . . , and Yn . We have established the following important generalization of Theorem 10.6 of the book, both in the continuous and in the discrete cases.

Theorem 10b Let X, Y1 , Y2 , . . . , Yn be n + 1 random variables. Then E E(X | Y1 , Y2 , . . . , Yn ) = E(X).

We will now prove two theorems, which in the study of stochastic processes, especially martingales, are proved to be very useful.

Theorem 10c Let X, Y1 , Y2 , . . . , Yn be n + 1 random variables and h be a real valued function from Rn to R. Then E h(Y1 , Y2 , . . . , Yn )X | Y1 , Y2 , . . . , Yn = h(Y1 , Y2 , . . ., Yn )E(X | Y1 , Y2 , . . . , Yn ).

Proof: We will show that both the right and left sides are the same for all sample points.

Suppose that at a certain sample point we have that Y 1 = y1 , Y2 = y2 , . . . , and Yn = yn .

At that point the right side of the quation is h(y1 , . . . , yn )E(X | Y1 = y1 , . . . , Yn = yn ).

Section 10D

Conditioning on More Than One Random Variable

30

The following shows that the left side is also the same.

E h(Y1 , . . . , Yn )X | Y1 = y1 , . . . , Yn = yn = E h(y1 , . . . , yn)X | Y1 = y1 , . . . , Yn = yn = h(y1 , . . ., yn )E(X | Y1 = y1 , . . . , Yn = yn ).

Theorem 10d

For n ≥ 1, let X, W, Y1 , . . . , Yn be n + 2 random variables. Then

E(W + X | Y1 , Y2 , . . . , Yn ) = E(W | Y1 , Y2 , . . . , Yn ) + E(X | Y1 , Y2 , . . ., Yn ).

Proof:

At a sample point where Y 1 = y1 , . . . , Yn = yn , the right side is E(W |Y1 = y1 , . . . , Yn = yn ) + E(X|Y1 = y1 , . . . , Yn = yn ).

Let f be the joint probability density function of W, X, Y 1 , . . . , Yn . We will show that the left side is also the same thing.

E(W + X | Y1 = y1 , . . . , Yn = yn ) ∞ ∞ f (w, x, y1, . . . , yn ) dw dx (w + x) = fY1 ,...,Yn (y1 , . . . , yn ) −∞ −∞ ∞ ∞ ∞ ∞ f (w, x, y1, . . . , yn ) f (w, x, y1, . . . , yn ) dw dx + dw dx w x = −∞ −∞ fY1 ,...,Yn (y1 , . . . , yn ) −∞ −∞ fY1 ,...,Yn (y1 , . . . , yn ) = E(W |Y1 = y1 , . . . , Yn = yn ) + E(X|Y1 = y1 , . . . , Yn = yn ).

Answers to Exercises of Section 10B

Answers to Exercises of Section 10B 1. (a) 1000; (b) 1,001,000; 2. 1066.

(c) 1,010,1000.

31

COMPANION FOR CHAPTER 11 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 11

Sums of Independent Random Variables and Limit Theorems

11A Additional Examples 3 11B Proportion versus Difference in Coin Tossing

3

10

11A

Additional Examples

3

Chapter 11

S ums of I ndependent R andom Variables and L imit Theorems 11A

ADDITIONAL EXAMPLES

Example 11a Let X be a uniform random variable over the interval (a, b). For n = 1, 2, . . ., calculate E(X n).

Solution: M X (t), the moment generating function of X is given by ⎧ ⎪ 1 etb − eta ⎪ ⎨ if t = 0 t MX (t) = b − a ⎪ ⎪ ⎩1 if t = 0.

(See Exercise 10, Section 11.1 of the book.) To find E(X n), we will use relation (11.2) of the book, ∞ ∞

tn E(X n) n t =1+ (1) E(X n) · .

MX (t) = n! n! n=0

n=1

To do so, note that, for t = 0, ∞ ∞

1 (tb)n (ta)n − MX (t) = t(b − a) n! n! n=0

n=0

∞

∞

n=1

n=1

(tb)n (ta)n 1 1+ −1− = t(b − a) n! n! ∞

∞

n=1

n=1

(tb)n − (ta)n 1 bn − an n−1 1 = t = t(b − a) n! b−a n!

11A

Additional Examples

∞

∞

n=2

n=1

4

1 bn+1 − an+1 n 1 bn − an n−1 t t =1+ = 1+ b−a n! b−a (n + 1)! ∞

bn+1 − an+1 tn · . = 1+ (b − a)(n + 1) n! n=1

Comparing this with relation (1), we have E(X n) =

bn+1 − an+1 . (b − a)(n + 1)

Example 11b The service times at a two-teller bank are independent exponential random variables each with parameter λ. At a random time Jim is being served by teller 2, Jeff is being served by teller 1, Judy and Julie are in the queue waiting to be served, and Judy is ahead of Julie. What is the probability that when Julie’s service is completed Jim is still being served?

Solution: Let A 1 be the event that Jeff leaves the bank before Jim, A 2 be the event that Judy leaves the bank before Jim, and A 3 be the event that Julie leaves the bank before Jim. Using the memoryless property of exponential random variables, by symmetry, the desired probability is P (A1 A2 A3 ) = P (A1 )P (A2 | A1 )P (A3 | A1 A2 ) =

1 1 1 1 · · = .

2 2 2 8

An Alternative Solution: At the random time, let R be the remaining service time of Jim, let X1 be the remaining service time of Jeff, and let X 2 and X3 be the service times of Judy and Julie, respectively. Due the memoryless property of exponential random variables, R and X1 are also exponential, each with parameter λ. The desired probability is P (X1 + X2 + X3 < R). Let Z = X1 + X2 + X3 ; then Z is gamma with parameters 3 and λ. By (10.21), P (Z < R) = = =

= =

∞ 0 ∞ 0

P (Z < R | R = x)λe−λx dx P (Z < x)λe−λx dx

∞

x

0

0 ∞

1 22 (λ t + 2λt + 2)e−λx 2

0 ∞ 0

1 −λt λe (λt)2 dt λe−λx dx 2 x

λe−λx dx 0

1 1 − e−λx − λxe−λx − λ2 x2 e−λx λe−λx dx 2

11A

=

−λx

−e

Additional Examples

1 −2λx 4 2 + e 4λ x + 12λ3 x + 14λ2 16λ2

= 0− −1+

5

∞ 0

1 1 14 = .

(14λ2) = 1 − 2 16λ 16 8

Example 11c The service times at a two-teller bank are independent exponential random variables each with parameter λ. At a random time, Jim is being served by teller 2, Jeff is being served by teller 1, Judy and Julie are in the queue waiting to be served, and Judy is ahead of Julie. What is the probability that when Julie’s service is completed Jim is still being served?

Solution: At the random time, let R be the remaining service time of Jim, let X 1 be the remaining service time of Jeff, and let X 2 and X3 be the service times of Judy and Julie, respectively. Due the memoryless property of exponential random variables, R and X 1 are also exponential, each with parameter λ. The desired probability is P (X 1 +X2 +X3 < R).

Let Z = X1 + X2 + X3 ; then Z is gamma with parameters 3 and λ. By relation (10.21) of the book, P (Z < R) = = =

= = =

∞ 0 ∞ 0

P (Z < R | R = x)λe−λx dx P (Z < x)λe−λx dx

∞

x

0

0 ∞

1 −λt λe (λt)2 dt λe−λx dx 2

1 22 (λ t + 2λt + 2)e−λx 2

0 ∞

x

λe−λx dx 0

1 1 − e−λx − λxe−λx − λ2 x2 e−λx λe−λx dx 2

0

−λx

−e

1 −2λx 4 2 + e 4λ x + 12λ3 x + 14λ2 16λ2

= 0− −1+

∞ 0

1 1 14 = .

(14λ2) = 1 − 2 16λ 16 8

An Alternative Solution: Let A 1 be the event that Jeff leaves the bank before Jim, A 2 be the event that Judy leaves the bank before Jim, and A 3 be the event that Julie leaves the bank before Jim. Using the memoryless property of exponential random variables, by symmetry, the desired probability is P (A1 A2 A3 ) = P (A1 )P (A2 | A1 )P (A3 | A1 A2 ) =

1 1 1 1 · · = .

2 2 2 8

11A

Additional Examples

6

Example 11d A random variable X is called chi-squared with n degrees of freedom if its probability density function is given by f (x) =

x(n/2)−1 e−x/2 , Γ(n/2) 2n/2

x > 0.

In Example 11.11 of the book, we showed that a chi-squared random variable with n degrees of freedom is gamma with parameters (n/2, 1/2). By Exercise 22, Section 11.1 of the book, the moment generating function of a gamma random variable Y with parameters r and λ is MY (t) =

λ r , λ−t

t < λ.

Therefore, the moment generating function of a chi-squared random variable X with n degrees of freedom is MX (t) =

1/2 n/2 = (1 − 2t)−n/2 , 1/2 − t

t < 1/2.

Let X1 , X2 , . . . , Xk be independent chi-squared random variables with n 1 , n2 , . . . , nk degrees of freedom, respectively. Find the distribution function of S = X 1 +X2 +· · ·+Xk .

Solution: By Theorem 11.3 of the book , for t < 1/2, MS (t) = MX1 (t)MX2 (t) · · · MXk (t) = (1 − 2t)−n1 /2 (1 − 2t)−n2 /2 · · · (1 − 2t)−nk /2 = (1 − 2t)−(n1 +n2 +···+nk )/2 .

Since this is the moment generating function of a chi-squared random variable with n 1 + n2 + · · · + nk degrees of freedom, by Theorem 11.2 of the book, S is chi-squared with n1 + n2 + · · · + nk degrees of freedom. We have shown that the sum of independent chi-squared random variables with n 1 , n2 , . . . , nk degrees of freedom is chi-squared with n1 + n2 + · · · + nk degrees of freedom.

Example 11e Suppose that, with equal probabilities, the value of a specific stock on any trading day increases 40% or decreases 30%, independently of the fluctuations of the stock value on the past and future trading days. Suppose that currently the stock price is X0 = 250, and let Xn be the stock price after n trading days. (i)

Find the distribution of ln X n for large n.

(ii)

By finding X n for large n, determine whether or not owning this specific stock over the long-term will turn out to be a mistake.

(iii)

Find limn→∞ E(Xn).

11A

Additional Examples

7

Solution: (i) In the first n trading days, let Y n be the number of the days in which the price of the stock increases. Then Y n is a binomial random variable with parameters n and 1/2. Clearly, Xn = 250(1.4)Yn (0.7)n−Yn .

This gives ln Xn = ln 250 + Yn ln 1.4 + (n − Yn ) ln 0.7 = ln 250 + Yn ln

1.4 + n ln 0.7 0.7

= ln 250 + Yn ln 2 + n ln 0.7.

Now, as n → ∞, by the central limit theorem, Y n is approximately N n/2, n/4 . (See self-test problem 8, Chapter 11 of the book.) So ln X n is also approximately normal with mean n E(ln Xn ) = ln 250 + ln 2 + n ln 0.7 ≈ 5.52 − (0.0101)n, 2 and variance n Var(ln Xn ) = (ln 2)2 Var(Yn ) = (ln 2)2 · ≈ (0.12)n.

4 (ii) Note that Xn = 250

1.4 Yn

n (0.7)n = 250 2Yn /n (0.7)n.

0.7 By the strong law of large numbers, as n → ∞, Y n /n → 1/2 and (0.7)n → 0. Therefore, for large n, Xn → 0 and owning this specific stock over the long-term will turn out to be a 1/2 n n disaster. For large n, Xn → 250(2 · 0.7) ≈ 250(0.9899) , which goes to 0 as n goes to infinity. (iii) Note that E(Xn+1 |Xn ) = (1.4Xn) · Thus

1 1 + (0.7Xn) · = (1.05)Xn. 2 2

E(Xn+1 ) = E E(Xn+1 |Xn) = (1.05)E(Xn).

This implies that E(X1 ) = (1.05)E(X0) = (1.05)(250) E(X2 ) = (1.05)E(X1) = (1.05)2(250) E(X3 ) = (1.05)E(X2) = (1.05)3(250) .. .

E(Xn) = (1.05)E(Xn−1) = (1.05)n(250)

11A

So, as n → ∞, E(Xn) → ∞.

Additional Examples

8

Example 11f For n ≥ 1, let Xn be a gamma random variable with parameters n and λ.

Let Fn be the distribution function of Y n = Xn /n. Let F be the distribution function of the constant random variable X = 1/λ. Show that, for all t, t = 1/λ, lim n→∞ Fn (x) = F (x).

Solution: Since

MX (t) = E(etX ) = E(et/λ) = et/λ ,

by the Lévy Continuity Theorem, it suffices to show that lim MYn (t) = et/λ .

n→∞

Note that since F (t) =

⎧ ⎨0

t < 1/λ

⎩ 1

t ≥ 1/λ

is continuous at all points except t = 1/λ, the relation lim n→∞ MYn (t) = et/λ implies that, for all t, t = 1/λ, limn→∞ Fn (x) = F (x). We will now establish the convergence of MYn (t) to et/λ . Noting that MXn (t) =

λ n , λ−t

t < λ,

it follows that

MYn (t) = E(etYn ) = E etXn /n = MXn (t/n) =

λ n nλ n = , λ − t/n nλ − t

t < nλ.

Now nλ n nλ − t −n = lim n→∞ nλ − t n→∞ nλ

lim MYn (t) = lim

n→∞

1 1 t/λ −n = lim = −t/λ = et/λ , n→∞ n→∞ t/λ n n e 1− n

x n = ex . where the penultimate equality follows from the fact that lim 1 + n→∞ n = lim

1−

Example 11g Convergence in Distribution Let X 1 , X2 , . . . be a sequence of random variables with distribution functions F 1 , F2 , . . . , respectively. Let X be a random variable with distribution function F . If lim Fn (x) = F (x)

n→∞

11A

Additional Examples

9

at all points in which F is continuous, then we say that X n converges in distribution to the random variable X. Note that, with this definition, the Lévy Continuity Theorem states that, for a sequence of random variables X 1 , X2 , . . . , if MXn (t) converges to MX (t), the moment generating function of a random variable X, then X n converges to X in distribution.

Let U1 , U2 , . . . be a sequence of independently selected random numbers from the interval (0, 1). For n ≥ 1, let Xn = min(U1 , U2 , . . . , Un). Let X be the constant random variable X = 0. Show that Xn converges to X in distribution.

Solution: The distribution function of X is given by ⎧ x<0 ⎨0 F (x) = ⎩ 1 x≥0 Since F is continuous everywhere except at x = 0, we must show that at all points where x = 0, lim Fn (x) = F (x), n→∞

where, for n ≥ 1, Fn is the distribution function of X n . Clearly, for x < 0, Fn (x) = 0, and limn→∞ Fn (x) = 0 = F (x). Similarly, for x ≥ 1, Fn (x) = 1 and limn→∞ Fn (x) = 1 = F (x). For 0 < x < 1, it follows that

Fn (x) = P (Xn ≤ x) = P min(U1 , U2 , . . . , Un ) ≤ x

= 1 − P min(U1 , U2 , . . ., Un ) > x = 1 − P (U1 > x, U2 > x, . . . , Un > x) = 1 − P (U1 > x)P (U2 > x) · · · P (Un > x) = 1 − Clearly, lim Fn (x) =

n→∞

⎧ ⎨0

x<0

⎩ 1

x>0

This shows that except at x = 0, lim n→∞ Fn (x) = F (x).

1 − x n 1−0

= 1 − (1 − x)n .

11B

11B

Proportion versus Difference in Coin Tossing

10

PROPORTION VERSES DIFFERENCE IN COIN TOSSING

Suppose that a fair coin is flipped independently and successively. Let n(H) be the number of times that heads occurs in the first n flips of the coin. Let n(T ) be the number of times that tails occurs in the first n flips. By the strong law of large numbers, with probability 1, 1 n(H) n(T ) = lim = . n→∞ n→∞ n n 2 That is, with probability 1, the proportion of heads and the proportion of tails, as n → ∞, is 1/2. This result is subject to misinterpretation. Some might think that if a fair coin is tossed a very large number of times, then heads occurs as often as tails. For example, Bertrand Russell (1872–1970), in his Religion and Science (Oxford University Press, 1947, page 158), writes lim

It is said (though I have never seen any good experimental evidence) that if you toss a penny a great many times, it will come heads about as often as tails. It is further said that this is not certain, but only extremely probable.

None of these statements is false. But they are subject to interpretations.

In Chapter 12, when discussing recurrent states of symmetric simple random walks, we have shown that, in successive and independent flips of a fair coin, it happens that at times the number of heads obtained is equal to the number of tails obtained. In that chapter, we will show that, with probability 1, this happens infinitely often, but the expected number of tosses between two such consecutive times is ∞. Hence, for some large number of tosses, what Bertrand Russell quotes is accurate. However, in general, when we say heads occurs “about as often” as tails, it is important to note that the fuzzy term about as often is a relative term and should be interpreted only relative to the number of times the coin is flipped. Frequently, when a fair coin is flipped a large number of times, in the usual sense, one side will not come “about as often” as the other side. But relatively speaking, it does. For example, we will show below that if a fair coin is flipped independently and successively 10 20 times, then the probability exceeds 0.96 that the absolute value of the difference between the number of heads and the number of tails obtained is at least half a billion. This is a very large difference. Nevertheless when compared with the huge number of tosses, 10 20 , it is insignificant. In fact, it is straightforward to show that in 10 20 tosses of the coin, almost certainly, the difference between the number of heads and the number of tails is less than 40 billion. Considering the fact that 10 20 is one hundred billion billion, 1 th of the number of times relative to 10 20 , 40 billion is really a small number. It is 25 × 109 the coin is flipped. Therefore, relatively speaking, when the coin is flipped 10 20 times, still “it will come heads about as often as tails.”

We will end this section by showing that in 10 20 flips of a fair coin, successively and independently, the probability exceeds 0.96 that the absolute value of the difference between the number of heads and the number of tails obtained is at least half a billion. Let X be the number of heads obtained. Clearly, X is a binomial random variable with parameters 10 20 and 1/2. Therefore,

1 1 1 1 20 1− = · 1010 and σX = 1020 E(X) = · 10 2 2 2 2

11B

Proportion versus Difference in Coin Tossing

To show that

11

1

P X − (1020 − X) ≥ · 109 ≥ 0.96, 2 we will use the De Moivre-Laplace theorem, and we ignore the correction for continuity since its effect, when dealing with large numbers, is negligible. We have

1 P X−(1020 − X) ≥ · 109 2

1 = P 2X − 1020 ≥ · 109 2

1 1 = P 2X − 1020 ≥ · 109 or 2X − 1020 ≤ − · 109 2 2

1 1 1 1 = P X ≥ · 1020 + · 109 + P X ≤ · 1020 − · 109 2 4 2 4

X − 1 · 1020

X − 1 · 1020 2 2 ≥ 0.05 + P ≤ −0.05 =P 1 1 · 1010 · 1010 2 2 ∞ −0.05 1 −x2 /2 1 −x2 /2 √ e √ e dx + dx ≈ 2π 2π 0.05 −∞

= 1 − Φ(0.05) + Φ(−0.05) = 2 1 − Φ(0.05) ≈ 2(1 − 0.5199) = 0.9602.

COMPANION FOR CHAPTER 12 OF

FUNDAMENTALS OF PROBABILITY WITH

STOCHASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

C ontents 12 Stochastic Processes 12A Additional Examples 3 12B Brownian Motion 12 First Passage Time Distribution 18 The Maximum of a Brownian Motion 19 The Zeros of Brownian Motion 19 Brownian Motion with Drift 22 Geometric Brownian Motion 23 Self-Quiz on Section 12B 27 Section 12B Summary 27 Answers to Odd-Numbered Exercises of Section 12B

3

30

12A

Additional Examples

3

Chapter 12

Stochastic Processes 12A

ADDITIONAL EXAMPLES

Example 12a Buses of tourists arrive at a museum at a Poisson rate of λ per hour. For

15 ≤ i ≤ 34, each bus has i passengers with probability p i , 34 i=15 pi = 1. The time each tourist spends at the museum is a random period with distribution function G independent of all other tourists. Find the expected value of the total number of tourists who have visited and left the museum by time t.

Solution: Let X(t) be the total number of tourists who have visited and left the museum by time t. Let Xi (t) be the total number of tourists from the ith bus who have visited and left the museum by time t. Let S ij be the time that the jth tourist on the ith bus spends in the museum, and define Yij = 1, if the jth tourist on the ith bus has visited and left the museum by time t; Yij = 0, otherwise. Conditioning on N (t) yields ∞

E X(t) | N (t) = n P N (t) = n E X(t) = n=0 ∞ n

E Xi (t) | N (t) = n P N (t) = n = n=0

i=1

∞ n

E Xi (t) P N (t) = n . = n=0

i=1

Let Ni be the number of tourists on the bus i. Let T i be the arrival time of the ith bus.

Conditioning on T i , which is a uniform random variable over the interval (0, t), yields

E Xi(t) =

t

1 E Xi(t) | Ti = x dx. t 0

12A

Additional Examples

4

But, by Wald’s equation, Ni

Yij (t) | Ti = x E Xi (t) | Ti = x = E j=1

= E(Ni)E Yij (t) | Ti = x 34

=

jpj P (Sij < t − x)

j=15 34

=

jpj G(t − x).

j=15

Therefore, letting u = t − x, we have

E Xi (t) =

34

t 0

j=15

34

=

1 jpj G(t − x) dx t 1 t

jpj

j=15

t

G(u) du.

0

Putting all these together, we obtain ∞ 34

1 jpj n E X(t) = t n=0

=

j=15

34

jpj

j=15

=

34

jpj

1 t

jpj

1 t

j=15

=

34

j=15

=λ

34

j=15

t

1 t

0 t 0 t 0 t

jpj

0

t 0

G(u) du P N (t) = n

G(u) du ·

∞

n P N (t) = n

n=0

G(u) du · E N (t)

G(u) du · λt

G(u) du.

Example 12b (An Inventory System with Back Orders Allowed) Suppose that for a certain product sold by a company, customer demands occur at a Poisson rate of λ per unit of time. Suppose that in the inventory of the product, whether there is shortage or there are

12A

Additional Examples

5

no shortages, for each customer demand, a replenishment order of the same size is placed so that the inventory level remains the same. When customer demands occur, if there is no shortage, the demand is met immediately. However, if there is shortage, the demand will be back-ordered and will be met when the replenishment order arrives. Such inventory models in which back orders are allowed and all customers accept late delivery can be modeled as an M/G/∞ queueing system. In this model, “customers” are the replenishment orders and the service time of each “customer”, a replenishment order, is the lead time of the order.

That is, the time interval which begins when the replenishment order is placed and ends when the replenishment order will arrive. We assume that the lead times are independent variables identically distributed each with distribution function G. Furthermore, we assume that the lead times are independent of the times customer demands occur.

Example 12c Recall that an M/G/∞ queueing system is an infinite-server system in which customers arrive according to a Poisson process with rate λ, service times are independent random variables with distribution function G, and service times and the arrival process are independent. Suppose that at time 0 there are no customers in an M/G/∞ queueing system. Find the probability mass function of the number of customers present in the system at time t.

Solution: Let X(t) be the number of customers in the system at time 0. Let N (t) be the total number of customers who have arrived at the system at or prior to t. Note that N (t) is the sum of X(t) and the number of customers in [0, t] who have already left the system.

We are given that X(0) = 0. Conditioning on N (t), we have ∞

P X(t) = i | N (t) = n P N (t) = n P X(t) = i = n=i ∞

e−λt (λt)n .

P X(t) = i | N (t) = n = n! n=i

Given that N (t) = n, the arrival times of the n customers who arrived in [0, t], T 1 , T2 , . . . , Tn , are independent uniform random variables over [0, t]. For 1 ≤ k ≤ n, let S k be the service time of the kth customer entered the system. Let A k be the event that the kth customer is still in the system at t. By conditioning on T k , P (Ak ) = =

t

1 P (Ak | Tk = x) dx = t 0 1 t

t 0

1 1 − G(t − x) dx = t

t

1 P (Sk > t − x) dx t 0 t 0

1 − G(u) du,

where the last equality follows by making the substitution u = t − x. As expected, P (A k ) does not depend on k. Let p(t) be the probability that any one of the customers who entered the system in [0, t] is still in the system at t. We have p(t) =

1 t

t 0

1 − G(u) du,

12A

which yields

P X(t) = i | N (t) = n =

Additional Examples

6

n−i i n . p(t) 1 − p(t) i

Therefore,

∞

n−i e−λt (λt)n i n P X(t) = i = p(t) 1 − p(t) n! i

n=i

n−i i ∞ λtp(t) −λt λt(1 − p(t)) = e i! (n − i)!

n=i

j ∞ λtp(t) −λt λt(1 − p(t)) e = i! j!

i

i

j=0

λtp(t) −λt λt[1−p(t)] e ·e i!

i e−λtp(t) λtp(t) . = i

=

λ t 1 − G(U ) du. As t 0 an application, in the inventory model of Example 12b, this result implies that the number of outstanding replenishment orders at time t is Poisson with rate

λ t 1 − G(U ) du. t 0 So X(t) is a Poisson random variable with parameter λp(t) =

Example 12d At a higher education institution, students are caught plagiarizing at a Poisson rate of λ per month. For each case of plagiarism, a hearing process begins when the student is caught and ends when a hearing is convened in the Dean’s office and an appropriate punishment is determined. Suppose that the duration of hearing processes are identically distributed and independent with distribution function G. Furthermore, suppose that the duration of hearing processes and the Poisson process according which the students are caught plagiarizing are independent. At this institution, what is the probability that, at time t, there are exactly n unsettled plagiarism cases?

Solution: This problem can be modeled as an M/G/∞ queueing system in which arrivals are plagiarizing cases caught and service times are the durations of hearing processes. Let X(t) be the number of unsettled plagiarism cases by time t. Then, by Example 12c, X(t) is a Poisson process with parameter μ= Therefore,

λ t

t 0

1 − G(u) du.

e−μt (μt)n .

P X(t) = n = n!

12A

Additional Examples

7

Example 12e Passengers arrive at a certain train station according to a Poisson process with rate λ and wait for a train to arrive. Trains arrive at the same station, beginning at time 0, every T minutes. Each time a train arrives, all the passengers waiting at the station will board the train. The train then immediately leaves the station. If there are no passengers waiting at the station, the train will not wait until passengers arrive. For i ≥ 1, find the expected value of the sum of the waiting times of all of the customers who will take the train arriving at time iT .

Solution: Let W be the sum of the waiting times of all of the customers who will take the train arriving at time T . It should be clear that, for i ≥ 1, E(W ) is the desired quantity. For t ≥ 0, let N (t) be the number of customers who have arrived at the station at or prior to t.

Given N (T ) = n, for 1 ≤ i ≤ n, let S i be the time of the ith arrival. We have ∞

E W | N (T ) = n P N (T ) = n E(W ) =

=

n=0 ∞

E W | N (T ) = n P N (T ) = n ,

n=1

where

E W | N (T ) = n = E (T − S1 ) + (T − S2 ) + · · · + (T − Sn )

= E nT − (S1 + S2 + · · · + Sn ) = nT − E(S1 + S2 + · · · + Sn ).

Now let X 1 , X2 , . . . , Xn be n randomly and independently selected points from the interval [0, T ]. Let X(1), X(2), . . . , X(n) be the order statistics of X 1 , X2 , . . . , Xn . Then E(S1 + S2 + · · · + Sn ) = E[X(1) + X(2) + · · · + X(n)] = E(X1 + X2 + · · · + Xn ) T = nE(X1) = n .

2 Hence ∞

E W | N (T ) = n P N (T ) = n E(W ) = n=1

=

∞

n=1

=

nT −

nT e−λT (λT )n 2 n!

T λT 2 T E N (T ) = · λT = .

2 2 2

12A

Additional Examples

8

Example 12f At a winery in Napa Valley, California, starting at 10:00 A.M., tours are given every T units of time. Visitors arrive to take a tour according to a Poisson process with rate λ, independently of the times and duration of the tours. Upon arrival, visitors wait in the wine tasting room for the next tour. While waiting, they engage in wine tasting, they order hors d’oeuvres, purchase wine and wine accessories, and buy books on wine and winery guides. As a result, independently of his or her waiting time, a random visitor spends, on average, d dollars per unit of time while waiting for the next tour. Furthermore, each time a tour begins, all the visitors waiting at the wine tasting room will take the tour.

What is the expected value of the total sales from the wine tasting room to the tour-taking visitors between two consecutive tours?

Solution: Label the time point 10:00 A.M. as t = 0. Let W be the total sale of the wine tasting room to the tour-taking visitors between 0 and T . Since the behavior of the Poisson process between any two consecutive tours is a probabilistic replica of its behavior between 0 and T, the desired quantity is E(W ). For t ≥ 0, let N (t) be the number of visitors who arrived at the wine tasting room at or prior to t. Given N (T ) = n, for 1 ≤ i ≤ n, let S i be the arrival time of the ith visitor, and let Y i be the amount in dollars, per unit of time, the ith visitor spends while waiting for the tour. We are given that, for all i, E(Y i) = d. We have E(W ) =

∞

E W | N (T ) = n P N (T ) = n n=0

=

∞

E W | N (T ) = n P N (T ) = n . n=1

For n ≥ 1,

E W | N (T ) = n = E Y1 (T − S1 ) + Y2 (T − S2 ) + · · · + Yn (T − Sn )

= E (Y1 + Y2 + · · · + Yn )T − (Y1 S1 + Y2 S2 + · · · + Yn Sn )

= E(Y1 ) + E(Y2 ) + · · · + E(Yn ) T

− E(Y1 )E(S1) + E(Y2 )E(S2) + · · · + E(Yn)E(Sn)

= ndT − dE(S1) + dE(S2) + · · · + dE(Sn) = ndT − dE(S1 + S2 + · · · + Sn ).

Now, let X1 , X2 , . . . , Xn be n randomly selected points from the interval [0, T ]. Let X (1), X(2), . . . , X(n) be the order statistics of X 1 , X2 , . . . , Xn . Then, by Theorem 12.4 of the book, E(S1 + S2 + · · · + Sn ) = E(X(1) + X(2) + · · · + X(n) ) = E(X1 + X1 + X2 + · · · + Xn ) = nE(X1) =

nT .

2

12A

Additional Examples

9

Hence E(W ) =

∞

E W | N (T ) = n P N (T ) = n n=1

=

∞

n=1

=

nT ndT − d 2

∞

e−λT (λT )n dT e−λT (λT )n · = n n!

2 n! n=1

aT aT aλT 2 E N (T ) = · λT = .

2 2 2

Example 12g (Genetics) Recall that in organisms having two sets of chromosomes, called diploid organisms, which we all are, each hereditary character of each individual is carried by a pair of genes. A gene has alternate alleles which usually are dominant A or recessive a, so that the possible pairs of genes are AA, Aa (same as aA), and aa. Suppose that the zeroth generation of a diploid organism consists of two individuals of opposite sex of the entire population who are randomly mated. Let the first generation be two opposite sex offspring of the zeroth generation who are randomly mated, the second generation be two opposite sex offspring of the first generation who are randomly mated, and so on. For n ≥ 0, define Xn to be aa × aa if the two individuals of the nth generation both are aa.

Define Xn to be Aa × aa if one individual of the nth generation is AA and the other one is Aa. Define Xn to be Aa × aa, Aa × Aa, AA × Aa, and AA × AA similarly. Let State 0 = aa × aa

State 1 = AA × aa

State 2 = Aa × aa

State 3 = Aa × Aa

State 4 = AA × Aa

State 5 = AA × AA.

The following transition probability matrix shows that {X n : n ≥ 0} is a Markov chain.

That is, given the present state, we can determine the transition probabilities to the next step no matter what the course of transitions in the past was. ⎞ ⎛ 1 0 0 0 0 0 ⎜ 0 0 0 1 0 0 ⎟ ⎜ ⎟ ⎜ 1/4 0 1/2 1/4 0 0 ⎟ ⎟ P =⎜ ⎜1/16 1/8 1/4 1/4 1/4 1/16⎟ . ⎜ ⎟ ⎝ 0 0 0 1/4 1/2 1/4 ⎠ 0 0 0 0 0 1 To see that, for example, the probabilities of one-step transitions from step 3 to various steps are given by the numbers in the fourth row, suppose that X n is Aa × Aa. Then an offspring of the nth generation is AA with probability 1/4, Aa with probability 1/2, and aa with probability 1/4. Therefore, X n+1 is aa × aa with probability (1/4)(1/4) = 1/16. It is AA × aa if the father is AA and the mother is aa or vice versa. So X n+1 is AA × aa with probability 2(1/4)(1/4) = 1/8. Similarly, X n+1 is Aa × aa with probability 2(1/2)(1/4) = 1/4, it is Aa × Aa with probability (1/2)(1/2) = 1/4, AA × Aa with probability 2(1/4)(1/2) = 1/4, and AA × AA with probability (1/4)(1/4) = 1/16.

12A

Additional Examples

10

It should be clear that the finite-state Markov chain above is reducible, and its communication classes are {0}, {1, 2, 3, 4} and {5}. Clearly, states 0 and 5 are absorbing. Therefore, the classes {0} and {5} are positive recurrent. The class {1, 2, 3, 4} is transient.

For the Markov chain {Xn : n ≥ 0}, starting from state i ∈ {1, 2, 3, 4}, absorption to state 0 (aa × aa) will occur at the nth generation if the state at the (n − 1)st generation is either 2 (Aa × aa) or 3 (Aa × Aa), and the next transition is into state 0. Therefore, the probability of absorption to aa × aa at the nth generation is n−1 pn−1 i2 p20 + pi3 p30 =

1 n−1 1 + pn−1 , p 4 i2 16 i3

where pn−1 and pn−1 are the i2-entry and i3-entry of the matrix P n−1 . Similarly, starting i2 i3 from i ∈ {1, 2, 3, 4}, the probability of absorption to state AA × AA at the nth generation is 1 n−1 1 n−1 n−1 p + pi4 . pn−1 i3 p35 + pi4 p45 = 16 i3 4 Example 12h (Genetics–The Wright-Fisher Model) Suppose that, in a population of N diploid organisms with alternate dominant allele A and recessive allele a, the population size remains constant for all generations.

Under random mating, for n = 0, 1, . . ., let Xn be the number of A alleles in the nth generation. Clearly, {Xn : n = 0, 1, . . .} is a Markov chain with state space {0, 1, . . ., 2N } and, given X n = i, Xn+1 is a binomial random variable with parameters 2N and i/2N . (Note that there are 2N alleles altogether.) Hence, for 0 ≤ i, j ≤ 2N, the transition probabilities are given by

2N i j i 2N −j .

1− pij = 2N 2N j Furthermore, E(Xn+1 | Xn = i) = 2N ·

i = i.

2N

Hence E(Xn+1 | Xn ) = Xn .

This implies that

E(Xn+1 ) = E E(Xn+1 | Xn ) = E(Xn),

showing that the expected value of the A alleles is the same for all generations.

The Markov chain {Xn : n = 0, 1, . . .} has two absorbing states, 0 and 2N . Absorbing to 0 means permanent loss of allele A from the population. Obviously, if for some generation, all individuals are aa, then, from that generation on, the individuals of all future generations will also be aa. Similarly, absorbing to 2N means permanent loss of allele a from the population and, hence, all of its future generations. Let x i be the probability that the Markov chain will be absorbed into 0 or 2N given that, initially, the number of A alleles is i. By the law of total probability, ⎧

2N ⎪

i 2N −j 2N i j ⎪ ⎨ xj , 1 ≤ i ≤ 2N − 1, 1− xi = 2N 2N j j=0 ⎪ ⎪ ⎩x = x = 1.

0

2N

12A

Additional Examples

11

It is readily seen that the solution to this system of 2N + 1 equations in 2N + 1 unknowns is x0 = x1 = · · · = x2N = 1.

This is true since by the binomial expansion,

2N

2N i 2N j

2N

1−

j=0

i 2N

2N −j

2N

i i +1− 2N 2N

=

= 1.

Having x0 = x1 = · · · = x2N = 1 shows that no matter what the number of A alleles in the initial population is, the Markov chain will eventually be absorbed into state 0 or state 2N with probability 1. We will now find y i , the probability that the Markov chain will be absorbed into state 2N given that X 0 = i, 0 ≤ i ≤ 2N . Note that (y0 , y1 , . . . , y2N ) is the solution to the following system of linear equations obtained by the law of total probability: ⎧ ⎪ y0 = 0 ⎪ ⎪ ⎪ ⎪

⎨ 2N

i 2N −j 2N i j = yj , 1 ≤ i ≤ 2N − 1 1 − y i ⎪ 2N 2N j ⎪ ⎪ j=0 ⎪ ⎪ ⎩ y2N = 1.

If we show that, for 0 ≤ i ≤ N, yi = i/2N satisfy the equations of this system, we are done. The fact that, for i = 0 and 2N, yi = i/2N is trivial. For 1 ≤ i ≤ 2N − 1,

2N

2N i 2N j

j

j=0

i 1− 2N

2N −j

2N

2N i yj = 2N j 2N

j=0

= 2N

j=0

2N j j

1−

j=0

1 2N j = 2N j

where

j

i 2N

j

i 2N

j

i 2N 1−

2N −j

i 2N

·

j 2N

2N −j

i i 1 · 2N · = = yi , 2N 2N 2N

1−

i 2N

2N −j

= 2N ·

i 2N

follows since it is the expected value of a binomial random variable with parameters 2N and i/2N . Hence, if initially the number of A alleles is i, the probability that eventually all of the individuals in the population are AA is i/2N . The probability that eventually they are all aa is 1 − (i/2N ).

12B

12B

Brownian Motion

12

BROWNIAN MOTION

To show that all material is made up from molecules, in 1827, the English Botanist Robert Brown (1773–1851) studied the motion of pollen particles in a container of water. He observed that, even though liquid in a container may appear to be motionless, particles suspended in the liquid, under constant and incessant random collisions with nearby particles, undergo unceasing motions in a totally erratic way. In 1905, unaware of Robert Brown’s work, Albert Einstein (1879–1955) presented the first mathematical description of this phenomenon using laws of physics. After Einstein, much work was done by many scientists to advance the physical theory of the motion of particles in liquid. However, it was in 1923 that Norbert Wiener (1894–1964) was able to formulate Robert Brown’s observations with mathematical rigor.

Suppose that liquid in a cubic container is placed in a coordinate system, and at time 0, a particle is at (0, 0, 0), the origin. Unless otherwise specified, for mathematical simplicity, we assume that the container is unbounded from all sides and is full of liquid. Let

X(t), Y (t), Z(t) be the position of the particle after t units of time. We will find the distribution functions of X(t), Y (t), and Z(t). To do so, we will find the distribution function of X(t). It should be clear that the distribution functions of Y (t) and Z(t) are similarly calculated and are identical to that of the distribution function of X(t). Observe that, in infinitesimal lengths of times during the time period (0, t), X(t) is a sum of infinitely many small movements in the direction of the x-coordinate. One way to find it is to divide the time interval t into n = [t/h] subintervals, where the length of each subinterval is h, for some infinitesimal h, and [t/h] is the greatest integer less than or equal to t/h. Suppose that, for infinitesimal δ > 0, in each of these time subintervals, the x-coordinate of the particle moves to the right δ units with probability 1/3, moves to the left δ units with probability 1/3, and does not move with probability 1/3. For i ≥ 1, let ⎧ ⎪ δ with probability 1/3 ⎪ ⎨ Xi = −δ with probability 1/3 ⎪ ⎪ ⎩ 0 with probability 1/3.

Then {Xi : i ≥ 1} is a random walk with E(X i) = 0 and 1 2 1 1 2δ 2 · δ + · (−δ)2 + · 02 = .

3 3 3 3

For large n, it should be clear that X(t) and ni=1 Xi have approximately the same distribution. Since the motions of the particle are totally erratic, X 1 , X2 , . . . are independent random variables. Hence, by the central limit theorem, for large n, the distribution of n 2δ 2 2δ 2 n or N 0, t . For the limiting behavior of the X is approximately N 0, i i=1 3 3h random walk {Xi : i ≥ 1} to closely approximate the x-coordinate, X(t), of the position of the particle after t units of time, we need to let h → 0 and δ → 0 in a way that (2δ 2 )/(3h) remains a constant σ 2 . Doing this, we obtain that, for all t > 0, X(t) is a normal random variable with mean 0 and variance σ 2 t. The same argument can be used to show that Y (t), the y-coordinate, and Z(t), the z-coordinate of the position of the particle after t units of time, are also normal with mean 0 and variance σ 2 t. It can be shown that Var(Xi ) = E(Xi2) =

12B

Brownian Motion

13

X(t), Y (t), and Z(t) are independent random variables.

Moreover, physical observations show that

they possess stationary and independent increments. For the process X(t) : t ≥ 0 , by stationary increments we mean that, for s < t and h ∈ (−∞, ∞), the random variables X(t) − X(s) and X(t + h) − X(s + h)

are identically distributed. Similarly, as previously defined, we say that X(t) : t ≥ 0 possesses independent increments if for s ≤ t ≤ u ≤ v, we have that X(t) − X(s) random variables. and X(v) − X(u) are independent

The properties of the processes X(t) : t ≥ 0 , Y (t) : t ≥ 0 and Z(t) : t ≥ 0 obtained by studying the x, y, and z coordinates of the positions of a particles’s motion in a liquid motivate the following definition.

Definition 12a A stochastic process X(t) : t ≥ 0 with state space S = (−∞, ∞) that possesses stationary and independent increments is said to be a Brownian motion if X(0) = 0 and, for t > 0, X(t) is a normal random variable with mean 0 and variance σ 2 t, for some σ > 0.

Brownian motions are also called Wiener processes. σ 2 is called the variance parameter of the Brownian motion. For t > 0, the probability density function of X(t) is denoted by φt (x). Therefore,

x2 1 exp − 2 .

(1) φt (x) = √ 2σ t σ 2πt

For a Brownian motion X(t) : t ≥ 0 , note that by stationarity of the increments: For s, t ≥ 0, X(t + s) − X(s) is N (0, σ 2 t).

Brownian motion is an important area in stochastic processes. It has diverse applications in various fields. Even though its origin is in the studies of motions of particles in liquid or gas, it is used in quantum mechanics, stock market fluctuations, and testing of goodness of fit in statistics.

2 Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ . Let W (t) = X(t)/σ. It is straightforward to show that W (t) : t ≥ 0 is a Brownian motion with variance parameter 1. A Brownian motion with variance parameter 1 is called a standard Brownian motion. For simplicity, we sometimes prove theorems about a Brownian motion with variance parameter 1 and then use the transformation W (t) = X(t)/σ to find the more general theorem.

Intuitively, it is not difficult to see that the graph of the x-coordinates of the positions of a particle’s motion in a liquid, as a function of t, is continuous. This is true for the y- and zcoordinates as well. It is interesting to know that these functions are everywhere continuous but nowhere differentiable. Therefore, being at no point smooth, they are extremely kinky.

The study of nowhere differentiability of such graphs is difficult and beyond the scope of this book. However, from a mechanical point of view, nowhere differentiability is a result of the assumption that the mass of the particle in motion is 0.

We will now discuss, step by step, some of the most important properties of Brownian motions.

12B

Brownian Motion

14

The Conditional Probability Density Function of X(t) Given that X(0) = x 0 In Section 12.3 of the book, we showed that for a discrete-time Markov chain, p nij , the probability of moving from state i to state j in n steps satisfies the Chapman-Kolmogorov equations. In Section 12.4 of the book, we proved that p ij (t), the probability that a continuoustime Markov chain moves from state i to state j in t units of time satisfies the Kolmogorov forward and backward equations. To point out the parallel between these and analogous results in Brownian motions, let f t|0 (x|x0 ) be the conditional probability density function of X(t) given that X(0) = x 0 .† This is the function analogous to p ij (t) for continuoustime Markov chain, and pnij for the discrete-time Markov chain. The functions f t|0 (x|x0 ) are called transition probability density functions of the Brownian motion. They should not be confused with f X|Y (x|y), the conditional probability density function of a random variable X given that Y = y.

Note that, by definition of a probability density function, for u ∈ (−∞, ∞),

P X(t) ≤ u | X(0) = x0 =

u −∞

ft|0 (x|x0 ) dx.

Also note that, since Brownian motions possess stationary increments, the probability density function of X(t + t 0 ) given that X(t 0 ) = x0 is ft|0 (x|x0 ) as well. By selecting appropriate scales, Albert Einstein showed that f t|0 (x|x0 ) satisfies the partial differential equation 1 ∂ 2f ∂f = σ2 , (2) ∂t 2 ∂x2 the backward diffusion equation. Considering the facts that f t|0 (x|x0 ) ≥ 0, called ∞ f −∞ t|0 (x|x0 ) dx = 1, and limt→0 ft|0 (x|x0 ) = 0 if x = x0 , it is straightforward to see that the unique solution to the backward diffusion equation under these conditions is ft|0 (x|x0 ) =

(x − x )2 1 0 exp − , √ 2 2σ t σ 2πt

(3)

which is the probability √ density function of a normal random variable with mean x 0 and standard deviation σ t.

The Joint Probability Density Function of X(t 1 ), X(t2 ), . . . , X(tn) Next, for t1 < t2 , let f (x1 , x2 ) be the joint probability density function of X(t 1 ) and X(t2 ). To find this important function, let U = X(t 1 ) and V = X(t2 ) − X(t1 ). Applying Theorem 8.8 of the book, we will first find the joint probability density function of U and V . Observe that the system of equations

x1 = u x2 − x1 = v †

Even though, in general, we adopt the convention of X(0) = 0, this choice is not necessary. It is not contradictory to assume that X(0) = x 0 , for some point x0 = 0.

12B

Brownian Motion

15

has the unique solution x 1 = u, x2 = u + v. Since

∂x1 ∂x1

∂u ∂v 1 0

= 1 = 0, J = = 1 1

∂x ∂x 2 2

∂u ∂v we have that, g(u, v), the joint probability density function of U and V is given by g(u, v) = f (u, u + v)|J| = f (u, u + v).

By this relation, f (x 1 , x2 ) = g(x1, x2 − x1 ). Since U and V are independent random variables, g(x1, x2 − x1 ) is the product of the marginal probability density functions of X(t1 ) and X(t2 ) − X(t1 ), which by the stationarity and independence of the increments of Brownian motion processes, are N (0, σ 2t1 ) and N 0, σ 2(t2 − t1 ) , respectively. Hence f (x1 , x2 ) =

(x2 − x1 )2 1 x2 1 √ exp − 2 , exp − 21 · 2σ t1 2σ (t2 − t1 ) σ 2πt1 σ 2π(t2 − t1 )

or, equivalently, f(x1 , x2 ) =

1 x21 (x2 − x1 )2 1

exp − + 2σ 2 t1 t2 − t1 2σ 2 π t1 (t2 − t1 )

,

−∞ < x1 , x2 < ∞.

(4)

For t1 < t2 < · · · < tn , let f (x1 , x2 , . . . , xn ) be the joint probability density function of X(t1 ), X(t2 ), . . . , X(tn ). Let g(x1, x2 , . . . , xn ) be the joint probability density function of X(t1 ), X(t2 ) − X(t1 ), . . . , X(tn ) − X(tn−1 ). An argument similar to the one presented for the case n = 2, shows that f (x1 , x2, . . . , xn ) = g(x1 , x2 − x1 , . . . , xn − xn−1 ).

Since the random variables X(t1 ), X(t2 ) − X(t1 ), . . . , X(tn) − X(t2n−1 ) are independent,

2 X(t1 ) ∼ N (0, σ t) and, for 2 ≤ i ≤ n, X(ti) − X(ti−1 ) ∼ N 0, σ (ti − ti−1 ) . Thus f (x1 ,x2 , . . . , xn) = n−1

(xi+1 − xi)2 1 x21 + exp − 2σ 2 t1 ti+1 − ti i=1

, σ n (2π)nt1 (t2 − t1 ) · · · (tn − tn−1 )

−∞ < x1 , x2 , . . . , xn < ∞.

(5)

12B

Brownian Motion

16

For t1 < t < t2 , the Conditional Probability Density Function of X(t) Given That X(t1 ) = x1 and X(t2 ) = x2 Let us begin with the special case in which t 1 = 0, x1 = 0, t2 = u1 , and x2 = 0. In such a case, the x-coordinate of the particle under consideration in the liquid, which at time 0 was at 0, is again at 0 at time u 1 . For 0 < u < u1 , we want to find the conditional probability density function of X(u) given that X(u 1) = 0. The condition X(0) = 0 is automatically assumed for all Brownian motions unless otherwise stated. Let f (x, y) be the joint probability density function of X(u) and X(u 1 ). Then, by (4), f (x, y) =

1 x2 (y − x)2 1

exp − 2 + 2σ u u1 − u 2σ 2 π u(u1 − u)

,

−∞ < x, y < ∞.

(6)

Let fX(u)|X(u1) (x|0) be the conditional probability density function of X(u) given that X(u1 ) = 0. Then, by relation (8.21) of the book and (1), fX(u)|X(u1) (x|0) =

√ f (x, 0) f (x, 0) = = σ 2πu1 f (x, 0). fX(u1 )(0) φu1 (0)

Thus, by (6),

1 x2 x2 1

exp − 2 + 2σ u u1 − u 2σ 2 π u(u1 − u)

1 u1 u1 1 · exp − 2 · x2 , −∞ < x < ∞, = √ 2σ u(u1 − u) σ 2π u(u1 − u)

√ fX(u)|X(u1 )(x|0) = σ 2πu1 ·

which is the probability density function of a normal random variable with mean 0 and σ 2 u(u1 − u) . Therefore, for 0 < u < u1 , variance u1

(7) E X(u) | X(0) = 0 and X(u1 ) = 0 = 0 and

σ 2 u(u1 − u) .

Var X(u) | X(0) = 0 and X(u1 ) = 0 = u1

(8)

Now, for t1 < t < t2 , to find the conditional probability density function of X(t) given that X(t1 ) = x1 and X(t2 ) = x2 , define X(t) = X(t + t1 ) − x1 −

t (x2 − x1 ). t2 − t1

X(t − t1 ) = X(t) − x1 −

t − t1 (x2 − x1 ). t2 − t1

This is equivalent to (9)

12B

Brownian Motion

17

Clearly, for t = t1 , we obtain X(0) = X(t1 ) − x1 = 0, and X(t2 − t1 ) = X(t2 ) − x1 − (x2 − x1 ) = 0.

Since t1 < t < t2 implies that 0 < t − t 1 < t2 − t1 , in (7) and (8) letting u 1 = t2 − t1 and u = t − t1 , for the process X(t), we have

E X(t − t1 ) | X(0) = 0 and X(t2 − t1 ) = 0 = 0 and

Var X(t − t1 ) | X(0) = 0 and X(t2 − t1 ) = 0

σ 2 (t − t1 ) (t2 − t1 ) − (t − t1 ) σ 2 (t2 − t)(t − t1 ) = . = t2 − t1 t2 − t1 Now, by (9),

0 = E X(t − t1 ) | X(0) = 0 and X(t2 − t1 ) = 0

t − t1 (x2 − x1 ). = E X(t) | X(t1 ) = x1 and X(t2 ) = x2 − x1 − t2 − t1

Hence

x2 − x1 (t − t1 ), E X(t) | X(t1 ) = x1 and X(t2 ) = x2 = x1 + t2 − t1 which is the equation of a line passing through (t 1 , x1) and (t2 , x2 ). Similarly, by (9),

Var X(t) | X(t1 ) = x1 and X(t2 ) = x2

(t2 − t)(t − t1 ) . = Var X(t − t1 ) | X(0) = 0 and X(t2 − t1 ) = 0 = σ 2 t2 − t1

We have shown the following important theorem: Theorem 12a

For t1 < t < t2 , the conditional probability density function of X(t) x2 − x1 (t − t1 ) and given that X(t 1 ) = x1 and X(t2 ) = x2 is normal with mean x 1 + t2 − t1 (t2 − t)(t − t1 ) . variance σ 2 t2 − t1 Example 12i Suppose that liquid in a container is placed in a coordinate system, and at time 0, a pollen particle suspended in the liquid is at (0, 0, 0), the origin.

Let X(t) be the x-coordinate of the position of the pollen after t minutes. Suppose that X(t) : t ≥ 0 is a Brownian motion with variance parameter 4 and, after one minute, the x-coordinate of the pollen’s position is 2. (a)

What is the probability that after 2 minutes it is between 0 and 1?

(b)

What is the expected value and variance of the x-coordinate of the position of pollen after 30 seconds?

12B

Brownian Motion

18

Solution: (a)

The desired probability is

P 0 < X(2) < 1 | X(1) = 2 = P − 2 < X(2) − X(1) < −1 | X(1) = 2

= P − 2 < X(2) − X(1) < −1 , by the independent-increments property of Brownian motion. Since X(2) − X(1) is normal with mean 0 and variance (2 − 1)σ 2 = 4, letting Z ∼ N (0, 1), we have

−2 − 0 −1 − 0 P − 2 < X(2) − X(1) < −1 = P <Z< 2 2 = P (−1 < Z < −0.5) = Φ(−0.5) − Φ(−1) = 0.1523.

(b)

In Theorem 12a, let t1 = 0, t2 = 1, x1 = 0, x2 = 2. We have

2−0 1 −0 =1 E X(1/2) | X(0) = 0 and X(1) = 2 = 0 + 1−0 2 1 1 −0 1−

2 2 = 1.

Var X(1/2) | X(0) = 0 and X(1) = 2 = 4 · 1−0

First Passage Time Distribution

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . The first passage time to a point α, also called the time of hitting α first, denoted by T α , is the minimum of the set t : X(t) = α . We will first find P (T α ≤ t), −∞ < t < ∞, for α > 0. Then we will calculate it for α < 0. For α > 0, note that, by Theorem 3.4 of the book,

P X(t) ≥ α = P X(t) ≥ α | Tα < t P (Tα < t) + P X(t) ≥ α | Tα = t P (Tα = t)

(10) + P X(t) ≥ α | Tα > t P (Tα > t).

Since α > 0, and the graph of X(t), as a function of t, is continuous, it follows that P X(t) ≥ α | Tα > t = 0. Since Tα is a continuous random variable, P (T α = t) = 0.

Observe that after the process hits α, for t > T α, by symmetry, it is equally likely that X(t) ≥ α and X(t) ≤ α. Hence 1

P X(t) ≥ α | Tα < t = .

2 Incorporating these facts into (10), we obtain

1 P X(t) ≥ α = P (Tα < t).

2 Since X(t) ∼ N (0, σ 2t), we have

2 P (Tα ≤ t) = P (Tα < t) = 2P X(t) ≥ α = √ σ 2πt

∞ α

2

2

e−x /(2tσ ) dx.

12B

Brownian Motion

19

x Letting √ = y, this reduces to σ t ∞

2 P (Tα ≤ t) = √ 2π

√ α/(σ t)

2 e−y /2 dy = 2 1 − Φ

α √

.

σ t

(11)

To find P (Tα ≤ t) for α < 0, note that T α and T−α are identically distributed. Hence

α .

(12) P (Tα ≤ t) = P (T−α ≤ t) = 2 1 − Φ − √ σ t Putting (11) and (12) together, we have that for all α ∈ (−∞, ∞), the distribution function of the first passage time to α is given by

P (Tα ≤ t) = 2 1 − Φ

|α| √ σ t

.

(13)

As in the case of symmetric random walks, it can be shown that, even though, with proba bility 1, X(t) : t ≥ 0 eventually will hit α, none of the moments of the first passage time to α is finite.

The Maximum of a Brownian Motion As before, let Tα be the first passage time to α. By the continuity of the paths of Brownian motions, for α > 0, the event max X(s) ≥ α occurs if and only if T α ≤ t. Therefore, 0≤s≤t

P max X(s) ≥ α = P (Tα ≤ t) = 2 1 − Φ

α √ σ t

0≤s≤t

,

or, equivalently,

P max X(s) ≤ α = 1 − 2 1 − Φ 0≤s≤t

α √ σ t

= 2Φ

α √

σ t

− 1.

Thus the distribution function of max X(s) is given by 0≤s≤t

P

max X(s) ≤ x =

0≤s≤t

⎧ ⎨2Φ ⎩0

x √ −1 σ t

if x ≥ 0 if x < 0.

(14)

The Zeros of Brownian Motion

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . In this subsection, we will find the probability that X(t) = 0 at least once in the time interval (t 1 , t2 ), 0 < t1 < t2 . The event that X(t) = 0 for at least one t in this interval is !

X(t) = 0 . t1 <t<t2

12B

20

Brownian Motion

To begin! with, note that if X(t 1 ) = a > 0, then, by the continuity of the Brownian motion’s

X(t) = 0 occurs if and only if min X(t) ≤ 0. Hence, for a > 0, paths, t1 <t<t2

t1 <t<t2

P

!

X(t) = 0 X(t1 ) = a = P min X(t) ≤ 0 | X(t1 ) = a . t1 <t<t2

t1 <t<t2

(15)

For a < 0, we consider the Brownian motion − X(t) : t ≥ 0 . Since, for all t ≥ 0, X(t) and −X(t) are identically distributed, we have !

P

!

X(t) = 0 X(t1 ) = a = P

t1 <t<t2

− X(t) = 0 − X(t1 ) = a

X(t) = 0 X(t1 ) = −a ,

t1 <t<t2

!

=P

t1 <t<t2

which, by (15), is equal to P

min X(t) ≤ 0 | X(t1 ) = −a since −a > 0. This and

t1 <t<t2

(15) imply that, for all a ∈ (−∞, ∞), P

!

X(t) = 0 X(t1 ) = a = P min X(t) ≤ 0 X(t1 ) = |a| . t1 <t<t2

t1 <t<t2

(16)

Now, again, using the fact that X(t) and −X(t) are identically distributed, and − min −X(t) = max X(t), t1 <t<t2

t1 <t<t2

we have

P min X(t) ≤ 0 X(t1 ) = |a| = 1 − P min X(t) > 0 X(t1 ) = |a| t1 <t<t2 t1 <t<t2

= 1 − P min −X(t) > 0 − X(t1 ) = |a| t1 <t<t2

= 1 − P − min −X(t) < 0 − X(t1 ) = |a| t1 <t<t2

= 1 − P max X(t) < 0 X(t1 ) = −|a| t1 <t<t2

= P max X(t) ≥ 0 X(t1 ) = −|a| t1 <t<t2

= P max X(t) ≥ |a| X(t1 ) = 0 t1 <t<t2

=P max X(t) ≥ |a| X(0) = 0 0<t<t2 −t1

= 2 1−Φ

|a| σ t2 − t1 √

,

(17)

where the last equality follows from (14). Now, on the one hand, by conditioning on X(t 1 )

12B

and by relations (16) and (17), we obtain ∞ !

P X(t) = 0 = P −∞

t1 <t<t2

= =

P

X(t) = 0 X(t1 ) = a φt1 (a) da

min X(t) ≤ 0 X(t1 ) = |a| φt1 (a) da

t1 <t<t2

−∞

−∞

21

t1 <t<t2

∞

∞

!

Brownian Motion

2 1−Φ

|a| √ σ t2 − t1

φt1 (a) da.

(18)

On the other hand, letting Z ∼ N (0, 1), we get ∞

P X(t2 ) − X(t1 ) > |X(t1 )| = P X(t2 ) − X(t1 ) > |X(t1 )| X(t1 ) = a φt1 (a) da −∞

= = =

∞ −∞

P X(t2 ) − X(t1 ) > |a| φt1 (a) da

∞

P Z> −∞ ∞ −∞

1−Φ

|a| √ φt1 (a) da σ t2 − t1 |a| √ σ t2 − t1

φt1 (a) da.

(19)

Comparing (18) and (19) yields !

X(t) = 0 = 2P X(t2 ) − X(t1 ) > |X(t1 )| .

P t1 <t<t2

Then Let Z1 and Z2 be two independent standard normal random variables. √ X(t2 ) − X(t1 ) is identically distributed with σ t2 − t1 Z1 and X(t1 ) is identically dis√ tributed with σ t1 Z2 . Hence

! √

√ t1 Z1

> X(t) = 0 = 2P σ t2 − t1 Z1 > σ t1 Z2 = 2P .

P |Z2 | t2 − t1 t <t<t 1

2

By Example 10.24 of the book, Z 1 /|Z2 | is a Cauchy random variable. Therefore, ∞ !

1 dx X(t) = 0 = 2 √ P π(1 + x2 ) t1 /(t2 −t1 ) t1 <t<t2

"

∞ 2 π t1 2 arctan x √ − arctan = = π t1 /(t2−t1 ) π 2 t2 − t1

t1 2 .

(20) = 1 − arctan π t2 − t1 √ Now draw a right triangle, such as the one in Figure 12a, with hypotenuse of length t2 √ √ √ and sides of lengths t1 and t2 − t1 . Let the angle adjacent to the side of length t1 be θ radians. Then √ t1 π −θ = √ , tan 2 t2 − t1

12B

Brownian Motion

22

2 1

_ 2

_

1

Figure 12a Triangle of Theorem 12b.

which implies that

arctan

π t1 = − θ, t2 − t1 2

2 2 t1 t1 2 = θ = arccos .

1 − arctan π t2 − t1 π π t2 Considering this identity and (20), we have shown the following theorem:

Theorem 12b For a Brownian motion X(t) : t ≥ 0 with variance parameter σ 2 , the probability that X(t) = 0 at least once in the time interval (t 1 , t2 ), 0 < t1 < t2 , is t1 2 arccos . π t2 or

Brownian Motion with Drift

Let W (t) : t ≥ 0 be a standard Brownian motion. That is, a Brownian motion with variance parameter 1. For μ ∈ (−∞, ∞) and σ > 0, the process X(t) = μt + σW (t),

t ≥ 0,

is called the Brownian motion with drift parameter μ and variance parameter σ2.

Note that for such a process, it is still the case that X(0) = 0 and X(t) : t ≥ 0 possesses stationary and independent increments. However, for t > 0, X(t) is normal with mean μt and variance σ 2 t.

For a Brownian motion X(t) : t ≥ 0 with drift parameter μ and variance parameter σ 2 , let ft|0 (x|x0 ) be the conditional probability density function of X(t) given that X(0) = x0 , or, equivalently, the conditional probability density function of X(t + t 0 ) given that X(t0 ) = x0 . To find ft|0 (x|x0 ), note that

P X(t) ≤ x | X(0) = x0 = P μt + σW (t) ≤ x | σW (0) = x0 x0 x − μt W (0) = σ σ x0 x − x0 − μt W (0) = . = P W (t) − W (0) ≤ σ σ

= P W (t) ≤

12B

Brownian Motion

23

Let Z ∼ N (0, 1); since W (t) − W (0) and W (0) are independent random variables, and W (t) − W (0) is N (0, t), this equation gives

x − x0 − μt P X(t) ≤ x | X(0) = x0 = P W (t) − W (0) ≤ σ x − x0 − μt # −0 $ x − x0 − μt σ √ =Φ . =P Z≤ √ σ t t Differentiating this equation with respect to x yields

x − x0 − μt 1 1 √ exp − ft|0 (x|x0 ) = √ · √ σ t σ t 2π −(x − x − μt)2 1 0 exp , = √ 2σ 2 t σ 2πt

2%

2

which is the probability density function of a normal random variable with mean x 0 + μt and variance σ 2 t.

Geometric Brownian Motion As mentioned before, particles suspended in a liquid are constantly and incessantly under random collisions with nearby particles. As a result, they undergo unceasing motions in a totally erratic way, and the changes in each coordinate of the positions of a particle, in infinitesimal time periods, form a Brownian motion. In finance, random fluctuations of prices of shares of stocks, or other assets, are a result of an enormity of political, social, economical, and financial factors independently assessed by a large number of traders. For that reason financial scholars and consultants use Brownian motions as models for prices of assets in general and shares of stocks in particular. However, since such prices are nonnegative and in the long run follow exponential growth or decay curves, they use a type of a stochastic process that is called geometric Brownian motion.

Let W (t) : t ≥ 0 be a standard Brownian motion. Let X(t) = μt + σW (t) so that

X(t) : t ≥ 0 is a Brownian motion with drift parameter μ and variance parameter σ 2 .

The process V (t) : t ≥ 0 defined by

V (t) = v0 exp X(t) = v0 exp μt + σW (t) is said to be a geometric Brownian motion with drift parameter μ and variance parameter σ 2 . Clearly, W (0) = 0 implies that V (0) = v0 .

Stock prices are often assumed to be Markov processes, since, usually, the price of a stock in the future depends on its present value only. The past history of the value of a stock and how it evolved to form the current price does not play a major role on the price of the stock in the future. In the preceding geometric model, if V (t) is the price of some stock at t, then, for t0 < t1 < · · · < tn , the relations

V (ti ) = exp μ(ti − ti−1 ) · exp σ[W (ti) − W (ti−1 )] V (ti−1 )

12B

show that

Brownian Motion

24

V (tn ) V (t1 ) V (t2 ) , , . . ., , V (t0 ) V (t1 ) V (tn−1 )

the ratio changes of the price of the stock over nonoverlapping time intervals, are independent random variables.

We will now calculate the expected value and variance of V (t) given that V (0) = v 0 .

To do so, let Z ∼ N (0, 1). We will first show that, for α ∈ (−∞, ∞), 2

E(eαZ ) = eα /2 .

(21)

This follows from the following calculations: E(eαZ ) =

∞

1 2 eαx · √ e−x /2 dx 2π −∞ 2

= eα /2 2

= eα /2

∞

1 2 1 2 1 √ e− 2 α +αx− 2 x dx 2π −∞

∞

1 1 2 2 √ e− 2 (x−α) dx = eα /2 , 2π −∞

∞

1 1 1 1 2 2 √ e− 2 (x−α) dx = 1, since √ e− 2 (x−α) is the probability density func2π 2π −∞ tion of a normal random variable with mean α√ and variance 1.

t ∼ N (0, 1), calculation of E V (t) and Using (21) and the fact that Z = W (t)/

Var V (t) is straightforward. We have

where

√

1 2 E V (t) = E v0 eμt+σW (t) = v0 eμt E eσ tZ = v0 eμt+ 2 σ t .

Also, √

2 E V (t)2 = E v02 e2μt+2σW (t) = v02 e2μt E e2σ tZ = v02 e2μt+2σ t .

Therefore,

2

2 2 Var V (t) = E V (t)2 − E V (t) = v02 e2μt+σ t(eσ t − 1).

Example 12j Let V (t) be the price of a stock, per share, at time t. Suppose that the stock’s current value, per share, is $40 with drift parameter $2 per year and variance parameter 4.41. If V (t) : t ≥ 0 is a geometric Brownian motion, what is the probability that after 18 months the stock price, per share, exceeds $65? √ Solution: Noting that 4.41 = 2.1, we have V (t) = 40e2t+2.1W (t),

12B

Brownian Motion

25

where W (t) : t ≥ 0 is a standard Brownian motion. Hence W (t) ∼ N (0, t). The desired probability is

P V (1.5) > 65 = P 40e2(1.5)+2.1W (1.5) > 65

= P e2.1W (1.5) > 0.0809 = P W (1.5) > −1.197 =P

W (1.5) − 0 1.197 √ > −√ = P (Z > −0.98) 1.5 1.5

= 1 − Φ(−0.98) = Φ(0.98) = 0.8365.

EXERCISES 1.

Suppose that liquid in a container is placed in a coordinate system, and at time 0, a pollen particle suspended in the liquid is at (0, 0, 0), the origin. Let Z(t) be the z-

coordinate of the position of the pollen after t minutes. Suppose that Z(t) : t ≥ 0 is a Brownian motion with variance parameter 9. Suppose that after 5 minutes the z-coordinate of the pollen’s position is 0 again.

What is the probability that after 10 minutes it is between −1/2 and 1/2?

(a)

If after seven minutes the z-coordinate of the pollen’s position is −1, find the expected value and variance of the z-coordinate of the position of pollen after six minutes.

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . Show that, for all t > 0, |X(t)| and max X(s) are identically distributed. (b)

2.

3.

4.

5.

6.

7.

0≤s≤t

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . For ε > 0, |X(t)| |X(t)| > ε = 1, whereas lim P > ε = 0. show that lim P t→∞ t→0 t t

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . Let Tα be the time of hitting α first. Let Y ∼ N (0, σ 2/α2 ). Show that, for α > 0, Tα and 1/Y 2 are identically distributed.

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . For a fixed t > 0, let T be the smallest zero greater than t. Find the distribution function of T .

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . As we know, for t1 and t2 , t1 < t2 , the random variables X(t1 ) and X(t2 ) are not independent.

Find the distribution of X(t 1 ) + X(t2 ).

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . For u > 0, show that

E X(t + u) | X(t) = X(t).

Therefore, for s > t, E X(s) | X(t) = X(t).

12B

8.

Brownian Motion

26

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . For u > 0,

t ≥ 0, find E X(t)X(t + u) .

9.

(Reflected Brownian Motion) Suppose that liquid in a cubic container is placed in a coordinate system in such a way that the bottom of the container is placed on the xy-plane. Therefore, whenever a particle reaches the xy-plane, it cannot cross the bottom of the container. So it reverberates back to the nonnegative side of the z-axis. Suppose that at time 0, a particle is at (0, 0, 0), the Let origin.

V (t) be the z-coordinate of the particle after t units of time. Find E V (t) , Var V (t) , and

P V (t) ≤ z | V (0) = z0 .

Hint: Let Z(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . Note that

Z(t) if Z(t) ≥ 0 V (t) = −Z(t) if Z(t) < 0.

The process V (t) : t ≥ 0 is called reflected Brownian motion.

10.

Suppose that liquid in a cubic container is placed in a coordinate system. Suppose

that at time 0, a particle is at (0, 0, 0), the origin. Let X(t), Y (t), Z(t) be the coordinates of the particle after t units of time, and assume that X(t), Y (t), and Z(t) are independent Brownian motions, each with variance parameter σ 2 . Let D(t)

be the distance of the particle from the origin after t units of time. Find E D(t) .

11.

Let V (t) be the price of a stock, per share, at time t. Suppose that the stock’s current value, per share, is $95.00 with drift parameter −$2 per year and variance parameter 5.29. If V (t) : t ≥ 0 is a geometric Brownian motion, what is the probability that after 9 months the stock price, per share, is below $80?

2 Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ . For t, s > 0, find E X(s)X(t) . Use the fact that, by Exercise 7, for s < t,

E X(t) | X(s) = X(s).

12.

12B

†

Section 12B Summary

27

Self-Quiz on Section 12B

Time allotted: 15 Minutes 1.

2.

3.

Suppose that liquid in a cubic container is placed in a coordinate system, and at time 0, a pollen particle suspended in the liquid is at (0, 0, 0), the origin. Let Y (t) be the that the process y-coordinate of the position of the pollen after t minutes. Suppose 2 . For 0 < s < t, Y (t) : t ≥ 0 is a Brownian motion with variance parameter σ

find E Y (s) | Y (t) = y and Var Y (s) | Y (t) = y . (3 points)

Let V (t) be the price of a stock, per share, at time t. Suppose that V (t) : t ≥ 0 is a geometric Brownian motion with drift parameter $3 per year and variance parameter 27.04. What is the probability that the price of this stock is at least twice its current price after two years? (4 Points)

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . For t > 0, let T be the smallest zero greater than t, and let U be the largest zero smaller than t. For x < t < y, find P (U < x and T > y). (3 points)

SECTION 12B SUMMARY

A stochastic process X(t) : t ≥ 0 with state space S = (−∞, ∞) that possesses stationary and independent increments is said to be a Brownian motion with variance parameter σ 2 if X(0) = 0 and, for t > 0, X(t) is a normal random variable with mean 0 and variance σ 2 t, for some σ > 0. For such a process, by stationarity of the increments, for s, t ≥ 0, X(t + s) − X(s) is N (0, σ 2t). A Brownian motion with variance parameter 1 is called a standard Brownian motion.

For a Brownian motion X(t) : t ≥ 0 , ft|0 (x|x0), the conditional probability density

function of X(t) given that X(0) = x 0 is normal with mean x 0 and standard deviation √ σ t.

For a Brownian motion X(t) : t ≥ 0 , for t1 < t2 , f (x1 , x2) the joint probability

density function of X(t 1 ) and X(t2 ) given by f (x1 , x2 ) =

1 x2 (x2 − x1 )2 1

exp − 2 1 + 2σ t1 t2 − t1 2σ 2 π t1 (t2 − t1 )

,

−∞ < x1 , x2 < ∞.

† For the solutions to these Self-Quiz problems, see Solutions to Self-Quizzes and Self-Tests, Chapter 12, of this Companion Website.

12B

Section 12B Summary

28

For t1 < t2 < · · · < tn , f (x1 , x2 , . . . , xn), the joint probability density function of X(t 1 ), X(t2 ), . . . , X(tn ), is f (x1 , x2 , . . . , xn ) = n−1

1 x2 (xi+1 − xi )2 exp − 2 1 + 2σ t1 ti+1 − ti i=1

, σ n (2π)nt1 (t2 − t1 ) · · · (tn − tn−1 )

−∞ < x1 , x2 , . . . , xn < ∞.

For a Brownian motion X(t) : t ≥ 0 ,

E X(u) | X(0) = 0 and X(u1 ) = 0 = 0, σ 2 u(u1 − u)

.

Var X(u) | X(0) = 0 and X(u1 ) = 0 = u1 Furthermore, for t1 < t < t2 , the conditional probability density function of X(t) given x2 − x1 (t − t1 ) and variance that X(t1 ) = x1 and X(t2 ) = x2 is normal with mean x 1 + t2 − t1 (t2 − t)(t − t1 ) . σ2 t2 − t1

Let X(t) : t ≥ 0 be a Brownian motion with variance parameter σ 2 . The first passage time to a point α, also called the time of hitting α first, denoted by T α , is the minimum of the set t : X(t) = α . We have that for all α ∈ (−∞, ∞), the distribution function of the first passage time to α is given by

|α| .

P (Tα ≤ t) = 2 1 − Φ √ σ t

Note that, even though, with probability 1, X(t) : t ≥ 0 eventually will hit α, none of the moments of the first passage time to α is finite.

As before, let Tα be the first passage time to α. By the continuity of the paths of Brownian motions, for α > 0, the event max X(s) ≥ α occurs if and only if T α ≤ t. This 0≤s≤t

observation enables us to calculate the distribution function of max X(s): 0≤s≤t

P max X(s) ≤ x =

0≤s≤t

⎧ ⎨2Φ ⎩0

x √

σ t

− 1 if x ≥ 0 if x < 0.

The Zeros of Brownian Motion For a Brownian motion X(t) : t ≥ 0 with vari-

that X(t) = 0 at least once in the time interval (t 1 , t2 ), ance parameter σ2 , the probability

2 t1 .

0 < t1 < t2 , is arccos π t2

12B

Section 12B Summary

29

Brownian Motion with Drift Let W (t) : t ≥ 0 be a standard Brownian motion.

That is, a Brownian motion with variance parameter 1. For μ ∈ (−∞, ∞) and σ > 0, the process X(t) = μt + σW (t), t ≥ 0,

is called the Brownian motion with drift parameter μ and variance parameter σ 2 . Note

that for such a process, it is still the case that X(0) = 0 and X(t) : t ≥ 0 possesses stationary and independent increments. However, for t > 0, X(t) is normal with mean μt and variance σ 2 t. For a Brownian motion X(t) : t ≥ 0 with drift parameter μ and variance parameter σ 2 , let ft|0 (x|x0 ) be the conditional probability density function of X(t) given that X(0) = x 0 , or, equivalently, the conditional probability density function of X(t + t0 ) given that X(t 0 ) = x0 . Then ft|0 (x|x0 ) is the probability density function of a normal random variable with mean x 0 + μt and variance σ 2 t.

Geometric Brownian Motion Let W (t)

: t ≥ 0 be a standard Brownian motion.

Let X(t) = μt + σW (t) so that X(t) : t ≥ motion with drift parameter 0 is a Brownian

2 μ and variance parameter σ . The process V (t) : t ≥ 0 defined by

V (t) = v0 exp X(t) = v0 exp μt + σW (t) is said to be a geometric Brownian motion with drift parameter μ and variance parameter σ 2 . We have

1 2 2 2 E V (t) = v0 eμt+ 2 σ t and Var V (t) = v02 e2μt+σ t(eσ t − 1).

Answers to Odd-Numbered Exercises of Section 12B

30

Answers to Odd-Numbered Exercises of Section 12B 1. (a) 0.056; (b) −0.5 and 4.5, respectively.

5. 0, if x ≤ t; (2/π) arccos t/x, if x ≥ t. √ √ 9. σ 2t/π, σ 2 t[1 − (2/π)], and Φ[(z + z0 )/σ t] + Φ[(z − z0 )/σ t] − 1, respectively.

11. 0.7486.