Math 445 – Assignment 3 Solutions

CH7.15 (WT; 5 points)

1. We have a random sample from a Rayleigh distribution, and we’re told that . Hence . Then as an estimator of , consider (since the first sample moment is a reasonable estimate of the first distribution moment). Now consider the expectation of our estimator: , since the Xs are identically distributed. This then reduces to . Hence, is an unbiased estimator of .

2. For the sample of 10 measurements, , so our estimate is pmol/L-squared.

CH7.20 (WA; 5 points)

1. First note that . So the bias of the estimator is . Also, the variance of the estimator is . Then the mean square error of the estimator is .

Note that the MSE does not depend on the unknown parameter, p. It on depends on the sample size, n.

2. Recall the MSE of the usual estimator is . Included in the table below are the two MSEs for different values of p and n (in each case, the smaller MSE is bolded). The usual estimator has lower MSE for smaller values of p, and, because of the symmetry, for larger values of p.

p	n	MSE—Usual Estimator	MSE—Bayes Estimator
0.1	20	0.0045	0.00835
	50	0.0018	0.00384
	100	0.0009	0.00207
0.2	20	0.008	0.00835
	50	0.0032	0.00384
	100	0.0016	0.00207
0.3	20	0.0105	0.00835
	50	0.0042	0.00384
	100	0.0021	0.00207
0.4	20	0.012	0.00835
	50	0.0048	0.00384
	100	0.0024	0.00207
0.5	20	0.0125	0.00835
	50	0.005	0.00384
	100	0.0025	0.00207

Furthermore, consider the ratio of MSEs: (Usual Estimator MSE) divided by (Bayes Estimator MSE). If this ratio is greater than 1, then the Bayes estimator has lower MSE; if the ratio is smaller than 1, then the usual estimator has smaller MSE. Shown below is a graph of MSE ratio for different values of n and p. Again, it’s clear that the MSE for the usual estimator is lower for smaller (and larger) values of p (especially at the small samples sizes). Otherwise, the MSE of the Bayes estimator is lower.

Another interesting note is that as n gets very large, the MSE of the Usual Estimator is smaller for more values of p:

CH7.23 (WA; 5 points)

1. For this distribution, . Then we find the method-of-moments estimator of by solving the equation . For the given sample of data, , so the method-of-moments estimate of is .

2. In this case, the likelihood function is and the log-likelihood is . To maximize this log-likelihood, we differentiate with respect to , set the derivative equal to 0, and then solve for : . Hence, the maximum-likelihood estimator of is . For the given sample of data, , so the maximum-likelihood estimate of is .

CH7.29 (WT; 5 points)

1. In this case, the likelihood function is and the log-likelihood is . To maximize this log-likelihood, we differentiate with respect to , set the derivative equal to 0, and then solve for : . Hence, the maximum-likelihood estimator of is . This MLE is the same as the unbiased estimator suggested in Problem 15.

2. If we can express the median in terms of , then we can use the invariance property of MLEs to answer this question. We know the median is value with area under the density curve to the left equal to 0.5. Hence, . MLEs have an invariance property (as discussed in class). That is, if is the MLE of , then is the mle of , for any function g.

Hence, the MLE of the median is , where .

CH7.30 (WA; 5 points)

1. Note that the pdf can written as , where . Then the likelihood function is , where is the smallest x value (the smallest order statistic). Also, as a function of , the exponent in the likelihood function is positive, so as increases, the likelihood function increases. Then to maximize the likelihood, we must choose as large as possible. But we know . Hence, the maximum-likelihood estimator of is the first order statistic—the minimum of all the Xs): .

Now reconsider the likelihood as only a function of ( is now a fixed value, once it’s MLE, , is substituted):

. Then the log-likelihood is . To maximize this log-likelihood, we differentiate with respect to , set the derivative equal to 0, and then solve for : . So the maximum-likelihood estimator of is.

2. Based on the given sample of 10 headway times (in seconds), the MLE estimates are and

CH8.10 (WA; 5 points)

See me if you have any questions.

CH8.11 (WA; 5 points)

For each random confidence interval, there is a 0.95 chance it contains the true value of . Also, the confidence intervals are independent. For a sample of 1000 intervals, the number of intervals, X, that contain follows a binomial distribution with n = 1000 and p = 0.95. Hence, we’d expect np = 950 of the intervals to capture the value of (because np is the mean for a binomial distribution).

Since the sample size is “large” (both and ), we know by the Central Limit Theorem that X has an approximate normal distribution with mean = 950 and standard deviation = .

Then, applying the continuity correction and “standardizing”:

Additional Problem (WT; 10 points)

Sketch of the solutions (see me if you need more detail):

a. For the given distribution, . The method of moments sets the first moment, E(X), equal to the first sample moment, , and solves for : .

b. In this case the likelihood function reduces to , and the log-likelihood is . Finally, the derivative of the log-likelihood with respect to is . Setting this derivative equal to zero and solving for (note that ) gives the maximum-likelihood estimator: .

c. The moment-generating function of the MLE is , where the last equality holds because the Xs are independent and identically distributed. For the given distribution, . Then the moment-generating function of our MLE simplifies to . Using the Devore-Berk textbook’s parameterization, this is the moment-generating function of a gamma random variable with and . By the uniqueness of moment-generating functions, we know our MLE has this particular gamma distribution.

d. Using known information about the gamma distribution (again, based on the Devore-Berk textbook’s parameterization), we know the expected value of our MLE is . Hence, the MLE is an unbiased estimator of . Furthermore, the variance of our MLE is . Since is a constant, and n is in the denominator of the variance, the variance clearly converges to 0 as n goes to infinity. So our MLE is an unbiased, consistent estimator of . Yeah!