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Statistical Method In Economics

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PROBLEMS

Problem 1 Suppose t hat X , Y , and o are random variables with Y =

+ βX + o.

Assume E (X ) = 0, V ar (X ) = σ , o has a uniform distribution over , and Cov (X , ) = 0.

2 a.

What is the dist ribut ion (pdf) of Y |X ?

b. c.

What is E (Y |X )? What is V ar (Y |X )? What is

, —1

1

2

2

E (Y )? What is V ar (Y )?

Problem 2 a.Assume that X is distributed uniformly over the interval [0, 4]. Calculate the moment generating function of X . Use the MGF to find the mean and variance of X . b.Use the Chebyshev inequality to calculate an upper bound on the probability that X is outside the interval [0.5, 3.5]. c.Now use the fact that X is distributed uniformly over the interval [0, 4] to calculate the probability that X is outside the interval [0.5, 3.5]. Is it higher or lower than the answer you got in part b? d.Based on your answers in parts b and c, comment on the usefulness of the Chebyshev inequality.

Problem 3 Let X be distributed uniformly on the interval [0, 1]. e.

Let Y = X 2. Use the 1-step and 2-step methods to find the pdf of Y. b.

Let Z = —Z log(X), for Z > 0. Calculate the pdf of Z. 1

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Problem 4 Let X have the following distribution. ƒ (x) x Find the pdf of Y =

E

.

1 2 x 3 3 ∈ ( 0, 1, 2, ...}

=

E +1

Problem 5 1

Let X have the pdf ƒ (x) = (x + 1)2 for — a.

Y . b.

1 ≤ x ≤ 1.

Let Y = X 2. Use the 1-step and 2-step methods to find the pdf of Find the MGF of Y .

c.

What is E (Y )? What is V12ar (Y )?

d.

What is E Y |X ≥ ?

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ANSWERS

Problem 1 a.

The distribution of Y |X is the distribution of + βX + o, where o is the only random variable and + β X is fixed, call it C. Then, the distribution of C +e is that of e , shifted by the value C, or U[C — , C + ].

Thus Y |X ~ U + βX — 2 , + 1βX + 2 . 1 b.

c.

1 2

1

2

For a random variable Z with a uniform distribution U[a, b], ƒ 7 (z) = 1/(b — a), h i b ∫b x b2 —a2 a E [Z ] = = 2(b—a) =b+ ; x2 a b—a dz = 2 2(b—a) h i a b ∫ 2 3 3 2 2 b x E [Z 2] = = b —a = b +a 3+2ab ; x3 a b—a dz = 3(b—a) 2a 2 3(b—a) 2 (b—a) 2 2 E [Z ] = 2 b +a +2ab V ar[Z ] = E [Z ] — — (b+a) .

3 4 12 = Therefore, for U[ + βX — 0.5, + βX + 0.5], +βE—0.5+ +β E [Y |X ] = b+a = = + 2βX ; 2 2 E+0.5 (b—a) 2 [ +βE+0.5—( +βE—0.5)] 1.

V ar[Y |X ] = = = 12 12 12 We can calcualte the expected value directly: E [Y ] = E [ + βX + e] = + βE [X ] + E [e] = +β·0+0 =

Or, we could use the law of iterated expectations: E [Y ] = = = =

E E [E Y [Y |X ]]

E E [ + βX ] + βE [X ]

For the variance, we can again calculate either directly or using our con- ditional variance identity.

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V ar[Y ] = =

V ar[ + βX + e] β2 V ar[X ] + V ar[e] + 2βCov(X , e) = β σ 2 2+ = β2σ2 +

[0 + 0.5 — (0 12 — 0.5)] 1

2

+ 2β · 0

12 or

V ar[Y ] = = = =

E (V ar(Y |X )) + V ar(E (Y |X )) 1 E ( ) + V ar( + βX ) 12 1 12 + β2 V ar[X ] 1 12 + β2 σ2

Problem 2 a.

The moment generating function is defined as

tE M E (t) = E e Since X is distributed uniformly over [0, 4], we have M E (t) = E∫ 4e t E 1 e txdx = 0 4 1 = e 4t — 1 4t To find the mean, we need to take the first derivative of the MGF and evaluate it at t = 0, using l'Hopital's rule in the fourth line.

6 6 1 e 4t — 1 | 6t M E(t) | t=0 = 6t 4t t=0 4t — 1 e 4t — = |

e

t=0

4t + 2 1 4te t 4t — e4t = | 2 4t t=0 16te4t + 4e4t — 4e4t = | 8t t=0 = 2 = E (X ) Then, to find the variance, we will take the second derivative of the MGF at t = 0 (again making use of l'Hopital's rule), and then subtract the square

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of the expected value.

6t2

16t 2e 4t— 8t e4t— 1 16t4 t2 4t 4t 4t 4t 4te — e 2te — e + 1 | — 2t3 t=0 2 t 8t2 e4t — 4te4t + e4t — 1 | 2t3 t=0 32t2 e4t + 16te4t — 16te4t — 4e4t + 4e4t 6t2 4t 16e | 3 t=0 16 = E X 2 16 4 3 —4= 4te4t — e4t

62

M E (t) |

=

t=0

= = = = =

V (X ) =

3

—

c.

Pr (X ∈/ (0.5, 3.5)) =

|

t=0

3

b.

The Chebyshev inequality st ates that Pr (| for t > 0. Here: E [X ] = 2, V ar [X ] = , t 43= (4/3) 3 16 Pr |X — 2| ≥ ≤2 = 27.

2 (3/2)

|

t=0

∫

0.5 0

1 dx 4

+

∫

41 3.5 dx 4

V ar[E ] X — E [X ]| ≥ t) ≤ 3 2 t= so: Pr (X ∈ / (0.5, 3.5)) 2

=

1

4<

16

27so

we get a lower

result than in part a. d. The Chebyshev inequality is very useful for evaluating distributions for which you only know the mean and the variance, but not the actual distribution. If you know the actual distribution you can get a more precise answer. But this is only because you are using additional information.

Problem 3 a. As instructed, we will find the pdf of Y using both the 1-step and 2-step methods. Note that Y will take on values in [0, 1]. We begin with the 2-step method by calculating the CDF: F Y (y) = = = ƒ Y (y) = =

Pr (Y ≤ y) = Pr X ≤ y √ √ Pr (— y ≤ X ≤ y) √ Pr (0 ≤ X ≤ y) √ √ F E ( 1 y) = y 1 —2 y for y ∈ [0, 1] 2 = 0 elsewhere

2

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Then we use the 1-step method. Because the transformation is already monotonic on the relevant range, we do not have to worry about dividing the range into monotonic pieces. ƒ Y (y) = ƒ Y (y) =

ƒE

r

—1

(y)

—1 .. 6r (y) .. for y ∈ [0, 1] 6y 1— 2 . for y ∈ [0, 1] .

√ 1 ƒ E ( y) ..y 2 1 —1 2 y for y ∈ [0, 1] = 2 = 0 elsewhere

b.

Now we have Z = —Z ln X . So the range of Z is [0, ∞). We again have a monotonic transformation, so we will use the 1-step method. Note that r—1 (z) = e— Z . x ƒ 7 (z) =

ƒE

r

x

e— Z

—1 .. 6r (z) .. for z ∈ [0, ∞) 6z —x . e Z . — for z ∈ [0, ∞)

(z) x

= ƒE =

—1

e— Z

.

Z . for z ∈ [0, ∞)

Z = 0 elsewhere Note that this is the exponential distribution. It turns out that most dist ributions can be const ructed as a transformation of a U [0, 1] random variable.

Problem 4 Because Y =E +E1 is a one-to-one transformation, we know that there will be only one x value that corresponds to each valid y value.

X =y ƒ Y (y) = Pr (Y = y) = Pr X +1 y y = Pr X = =ƒ E 1 — y 1—y y 1—y 1 2 = 3 3

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for valid values of y. What is the support of Y ? We can substitute in for the first few potential values of X to see the pattern: 0 = 0 0+1 1 = 1 1+1 2 2 = 2+1 2 3 = 3 3+1 3 x for whole numbers x so Y ∈ ( 0, , 12 , 23 , 34... } , or Y ∈ ( y| y = x+1 } 4 Problem 5 a.

We will use the 1-step method, and so we must first divide the range of X into segments for which the transformation function is monotone, (—1, 0) and (0, 1) (we can ignore the endpoints of these intervals because the probability that X equals a particular point is zero). We will use the 1-step method on each segment and then sum our results.

For the segment (—1, 0), our transformation function is r (x) = x2 , and √ √ 1 —Y is (0, 1). So we have the inverse function is r—1 y. The range .. 1 y of ..2 y ∈ (0, 1) ƒ Y (y) = (y) =ƒ — for E (— y) — 2 1 1 (1 — √ 2 y ∈ (0, 1) = y) y —for 4 1 = 1 y — 2 — 1 for y ∈ (0, 1) = 0 4elsewhere

Then, for the segment (0, 1), our transformation function is r (x) = x2 , and the inverse function is r—1 (y) =

√

y. The range of Y is again (0, 1). So we have

ƒ Y (y)

=

√ 1 1 ƒE ( y) y .. for—y..2 ∈ (0, 1) 2 1 √ 1 2 y ∈ (0, 1) = (1 + y) y —for 4 1 1 y — 2 + 1 for y ∈ (0, 1) = 4 = 0 elsewhere

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So adding these two functions together we get 1 y 12 — 1 + y1 ƒ Y (y) = — 4 4 2 for y ∈ (0, 1) y 1 —1 = = 0 2elsewhere b.

+—12 for y ∈ (0, 1) 1

To get the moment generating function of Y , we find E e .

M Y (t) = E e ∫ =

tY

1 0

1 1 e ty y—2 6y 2

which does not have a closed form. c.

We can st ill use the MGF to get E (Y ): 6 E (Y ) = M Y (t) | t=0 6t ∫ 1 6 ty 1 e y = 0 6t ∫ 2 1 1 = ye tyy — 2 ∫01 1 = y 2 6y 1 =

And similarly, we can get E Y : E

Y

=

2

= = = = So V ar (Y ) = — 15 =

9

1

.

4 45

2 1 3

0

3

1

| = y 2 y=0 3

—1 1 2

2

| 6y

t=0

| 6y

t=0

1

2

62 Y (t) | 6t2 M t=0 ∫ 1 62 ty 1 — 1 e y 2 0 6t2 ∫ 2 3 1 1 e ty 2 | 6y y t=0 2 ∫01 3 y 2 6y 1 2 1 5

0

†

y

2

1

| y=0

=

| 6y

t=0

1 5

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d.

We start by finding the pdf of Y |X ≥ . This will 12be the similar to part a., except that our transformation function is monotonic over the relevant range, and we will use the pdf of X | X ≥ 1 , which is just2 the pdf of X over the relevant range scaled up by one over the probability that X is in this range: ƒ E †E ≥

1

12 (x 1 1 (x 1 2 ∫ 2

(x) =

2

=

x2 2

=

+x

8 (x + 1) 7

Thus we can calculate ƒ Y †E ≥ 1 (y) = ƒ E †E ≥ 2

1 for

+ 1)

+ 1) dx x+1

1 2

|

≤ x ≤ 1 (and 0 elsewhere) 2

1

x=

1

2

√ 1 ( y) ..y 2

— .12

1 ,1 4

for y ∈

.

4 1+y 1 —2 7 It is then straightforward t o calculate E Y |X ≥ : ∫ 1 4y 1 E Y |X ≥ = 1 7 2 4 ∫

(and 0 elsewhere)

=

= = =

4 7 4

1

1 4

y

1 2

1+y

— 12

dy

y + y 2 dy

2

+

7101≈20.601 3 168

1

2

3

1

y2 1 | y= 4

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