London School of Economics
ST 202
ST202 – LT Solutions to problem set 3 1. We have already worked out that X(n) has CDF FX(n) (x) = x θ n for 0 ≤ x ≤ θ. If we set U = X(n)/θ, an increasing function, the CDF becomes FU (u) = FX(n) (θu) = u n for 0 ≤ u ≤ 1. The statistic U is pivotal for θ, as its distribution does not depend on any unknown parameters. [More specifi
...[Show More]
ST202 – LT Solutions to problem set 3 1. We have already worked out that X(n) has CDF FX(n) (x) = x θ n for 0 ≤ x ≤ θ. If we set U = X(n)/θ, an increasing function, the CDF becomes FU (u) = FX(n) (θu) = u n for 0 ≤ u ≤ 1. The statistic U is pivotal for θ, as its distribution does not depend on any unknown parameters. [More specifically, it is a Beta(n, 1) distribution.] 2. (a) Using the fact that Y2 − Y3 ∼ N(0, 2), we have E(U1) = E(Y1) + E[(Y2 − Y3) 2 ] − 2 = µ + 2 − 2 = µ , so U1 is unbiased for µ. It is clearly not consistent, as it does not change as n increases. (b) Recall that Y¯ has mean µ and variance 1/n, so we have E(U2) = E(Y¯ 2 ) − 1 n = 1 n + µ 2 − 1 n = µ 2 , and we deduce that U2 is unbiased for µ 2 . If we let Y¯ n denote the mean of a sample of size n, we know that Y¯ n p −−→ µ as n → ∞, so Y¯ 2 n p −−→ µ 2 . The term 1/n converges to zero, so we conclude that U2 p −−→ µ 2 , that is, U2 is a consistent estimator of µ 2 . 3. Notice that S ∼ Bin(n, p). We have E[S(S − 1)] = E(S 2 ) − E(S) = np(1 − p) + (np) 2 − np = n(n − 1)p 2 , so E( ˆθ) = p 2 , which means that it is unbiased. To establish that it is consistent, we use the fact that Y¯ = S/n p −−→ p. We can see that S − 1 n − 1 = nY¯ − 1 n − 1 → Y¯ as n → ∞ (the terms in n dominate), so this also converges to p in probability. We conclude that ˆθ p −−→ p 2 and the esti
[Show Less]
-by-Gary-Donell-SOLUTIONS-MANUAL-preview.jpeg)
-by-Gary-Donell-INSTRUCTOR’S-SOLUTIONS-MANUAL-preview.jpeg)
