MATH 255 - Probability and Statistics Final Exam Solutions

MATH 255 - Probability and Statistics Final Exam Solutions Problem 1. [8pt] Let X1, X2, . . . be independent random variables that are uniformly distributed over [0, 1]. Show that the sequence of Y1, Y2, . . . converges with probability 1 to some limit and identify the limit, for the case where Yn is the sampled geometric mean, given by Yn = Yn i=1 Xi !1/n Solution: limn→∞ Yn i=1 Xi !1/n = limn→∞ exp  log Yn i=1 Xi !1/n  = limn→∞ exp 1 n log Yn i=1 Xi !! = limn→∞ exp 1 n Xn i=1 log (Xi) ! = exp limn→∞ 1 n Xn i=1 log (Xi) ! = exp (E[log(Xi)]) and E[log(Xi)] = Z 1 0 log(x)dx = −1. Hence, we have Yn → 1/e with probability 1. 1 This study source was downloaded by from CourseH on :34:25 GMT -05:00 Problem 2. [5pt] The probability of heads of a possibly biased coin is modeled as a random variable Θ whose prior distribution is uniform over the interval [0, 1]. The coin is tossed n times, independently, and we observed x heads out of n tosses. What is the linear LMS estimate of Θ given x? Express it as a function of the parameters n and x. Solution: Note that X is a binomial random variable with parameters θ and n conditioned on Θ = θ. The LLMS estimate is given by (+1pt) θb LLMS = E[Θ] + cov(Θ, X) var(X) (X − E[X]). Since Θ is uniformly distributed over [0, 1], we have E[Θ] = 1/2 and var(Θ) = 1/12. The total expectation theorem yields that (+1pt) E[X] = E[E[X | Θ]] = E[nθ] = n 2 , The total variance theorem yields that (+1pt) var(X) = E[var(X | Θ)] + var(E[X | Θ]) = E[nΘ(1 − Θ)] +