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MTH 525 Introduction to Mathematical Statistics Spring 2017 Problem Set # 1 This problem set is due on Thursday, February 2, 2017 at the beginning
MTH 525 Introduction to Mathematical Statistics Spring 2017 Problem Set # 1 This problem set is due on Thursday, February 2, 2017 at the beginning of class. 1. Let X1 , X2 , . . . be a sequence of independent random variables with E(Xi ) = and P p n as n . Var(Xi ) = i2 . Show that if n2 ni=1 i2 0, then X 2. (a) Show that if T tn (Student's t distribution with n degrees of freedom), then T 2 F1,n (F distribution with 1 and n degrees of freedom). (b) Show that if X Fn,m , then X 1 Fm,n . (c) Show that if X and Y are independent Exp(1) random variables, then X/Y follows an F distribution. Also, identify the degrees of freedom. 3. If X Gamma(n, 1), how large should n be so that \u0012 \u0013 X P 1 > 0.01 < 0.01 n (a) according to Chebyshev's inequality? (b) according to the central limit theorem? (Hint: Let Xi Gamma(1, 1) for i = 1, , n. Then X n n = X 1 n Pn i=1 Xi .) 4. The number of floods that occur in a certain region over a given year is a random variable having a Poisson distribution with parameter = 2, independently from one year to the other. Moreover, the time period (in days) during which the ground is flooded, at the time of an arbitrary flood, is an exponential random variable with parameter = 1/5. We assume that the durations of the floods are independent. Use the central limit theorem to calculate (approximately) the probability that (a) over the course of the next 50 years, there will be at least 80 floods, (b) the total time during which the ground will be flooded over the course of the next 50 floods will be smaller than 200 days. 5. Suppose (X1 , X2 ) is a random sample of size 2 from a population of size N , and consider the following as an estimator of the population mean : = 1 X1 + 2 X2 , X where 1 and 2 are fixed numbers. (a) Find a condition on 1 and 2 so that the estimator is unbiased. (b) Show that the choice of 1 and 2 that minimizes the variances of the estimate subject to this condition is 1 = 2 = 1/2. 1
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