CS 361 Computational Probability - University of Illinois_Homework 8.

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CS361 Daman Mulye Homework 8 1. We can write out the likelihood function for the exponential distribution as L(θ) = P (x|θ) = θe−θx If we assume that all the values in the set are found independently we can write the likelihood function as follows: N Y θe−θxi X∈i

Next we can take the log of both sides to yield: ln(L(θ)) = ln(

N Y

X∈i N X

N X

θe−θxi ) =

ln(θexp(−θxi )) =

X∈xi

ln(θ) + ln(exp(−θxi ) =

X∈xi

N X

ln(θ) − θxi

X∈xi

We can take the log of theta term and factor it out of the summation along with the theta constant multiplied by x to get: N X Ln(L(θ)) = N ln(θ) − θ xi X∈xi

We can differentiate both sides to give: N d d X d xi = ln(L(θ)) = N ln(θ) − θ dθ dθ dθ X∈x i

N X N − xi θ X∈x i

Setting this equal to 0 and solving for theta gives: N θ̂ = PN

X∈xi xi

1

P 2. The maximum likelihood function for a Poisson distribution is: L(λ) = N1 N i=1 Xi Since the max likelihood function is the average of the number of observations in a given period of time we can calculate he average number of popup windows for each day separately. (a) i. For day 1: (3 + 1 + 4 + 2) ∗ 14 = 2.5 counts/hr ii. For day 2: (2 + 1 + 2) ∗ 13 ≈ 1.667 counts/hr iii. For day 3: (3 + 2 + 2 + 1 + 4) ∗ 15 = 2.4 counts/hr (b) Since for day 4, the observer already summed the number of observations and we have the given time we can just divide the sum by the period of time they were recording observations for to get: 13 ≈ 2.167 counts/hr 6 (c) The MLE for the intensity for all the days taken together is just the sum of all the occurrences 1 ≈ divided by the total time spend browsing: (3+1+4+2+2+1+2+3+2+2+1+4+13)∗ 4+3+5+6 2.22

2

3. (a) The MLE for the probability of success in a geometric distribution is estimated by θ̂ = 1r , where r is the number of failures until the first success. Since r, is recorded as the number of observations 1 of failures. We can get an estimate of the probability of the ball landing in the zero slot as r+1 .

1 Which means that the number of zero slots should be proportional to r . Thus, to solve for the 1 x = r+1 → x = 36 , so the MLE number of slots we can solve for x in the following equation: 36+x r 36 estimate for the total number of slots on the wheel is: 36 + r (b) The estimate is not very reliable, since we’re essentially estimating the probability with just one sample. To improve our confidence in our estimate we should repeat this experiment multiple times. (c) To find the MLE for number of slots with this repeated experiment we follow these steps: k Y L(θ) = (1 − θ)r θ i=1

Then we take the log of both sides: k k k Y X X ri ri ln(L(θ)) = ln( (1 − θ) θ) = ln((1 − θ) θ) = kln(θ) + ri ln(1 − θ) i=1

i=1

i=1

Now that we have a simplified form for our likelihood function we can differentiate and set that function equal to 0 and solve for theta. k k X d d k X ri d ln(L(θ)) = kln(θ) + (ri ln(1 − θ)) = − dθ dθ dθ θ 1−θ i=1 i=1

Solving this equation for 0 gives: θ̂ =

k+

k Pk

i=1 ri

Following the same steps as in part a. If x represents the number of zero slots we can then solve the following equation for the number of slots on the wheel: 36k k x k →x= ÷ (1 − ) = Pk Pk Pk 36 + x k + i=1 ri k + i=1 ri k + i=1 ri This expression simplifies down to: 36k Pk

i=1 ri

Since we solved for the maximum likelihood of the number of zero slots, we can add 36 to this estimate to calculate the total number of slots: 36k Pk

i=1 ri

3

+ 36

z 4. (a) P (n|z) can be calculated by 36+z : 2 3 1 , P (n|3) = 39 P (n|0) = 0, P (n|1) = 37 , P (n|2) = 38

(b) P(z = 0 | observations) 6= 0, when the observations don’t contain the event when the ball lands in the 0 slot. (z) (c) P(z | observations) = P (X=2|z)P . Let’s calculate each component of the formula individually.

P (X=2)  0.1    0.2 P (z) =  0.4    0.3

z=0 z=1 z=2 z=3

To calculate P (X = 2|z) we can use the likelihood function for binomial distribution: L(z) = 36 z 36 34 ( 36+z )2 ( 36+z ) . Plugging in the different values of z’s yield: 2  0 z=0    0.181 z = 1 P (X = 2|z) =  0.2776 z = 2    0.245 z = 3 Next, we can multiply the respective probabilities together to get:  0 z=0    0.03626 z = 1 P (X = 2|z)P (z) =  0.11104 z = 2    0.07356 z = 3 (z) The bayesian formula is as follows: P (X=2|z)P . However, we can recognize that this probability P (X=2) is proportional to P (X = 2|z)P (z) and since using bayesian inference is like a pdf in that the values should all add up to one we can find some scalar k to multiply all the probabilities by such that their sum equals 1. This factor turns out to be 4.527755 Thus, leading to a final answer of  0 z=0    0.16418 z = 1 P (z|observations) =  0.50276 z = 2    0.33306 z = 3

4

5. We can model all of the experiments as a binomial distribution, and hence their respective likelihood functions are: 35 (a) L(θ) = 50 θ (1 − θ)15 35

(b) L(θ) = 99 θ54 (1 − θ)45 In this experiment we can forget about the last misplaced coin and instead 54 of 100 trials being conducted only 99 were.

6 (c) L(θ) = 13 θ (1 − θ)7 The total number of trials is 13.

6

5