MATH 28A: Homework #6 Jongha Ryu Due date: November 8, 206 Problem. (Problem 2..2. Soluton. If X,..., X n Bern(p, then T = X s a complete suffcent statstc. Our target s g(p = p, and the nave guess suggested s { f X = X 2 = X = δ(x =. 0 o.w. We Rao-Blackwellze the estmator to get the UMVUE as follows: η(t = E[δ(X T = t] = E[δ(X,..., X n T = t] ( = P X = X 2 = X = X = t = P(X = X 2 = X =, X = t P( X = t = P(X = P(X 2 = P(X = P( n =4 X = t P( X = t ( n = p t p t ( p n t p t ( p n t = = ( n t ( n t Problem 2. (Problem 2.2.. ( n t t(t (t 2 n(n (n 2. Soluton. Assume X,..., X n..d. N (ξ, σ 2 wth σ 2 known. We know that T = X s a complete suffcent statstc. Let X = Y + ξ N (ξ, σ2 σ2 n, so that Y N (0, n.
Soluton of (a. E[X 2 ] = E[(Y + ξ 2 ] = E Y 2 + 2ξ E[Y ] + ξ 2 = σ2 n + ξ2. Thus, s the UMVUE for ξ 2. ˆξ 2 = X 2 σ2 n Soluton of (b. Lkewse, E[X ] = E[(Y + ξ ] = E Y 2 ξ + ξ = σ2 n ξ + ξ. Thus, s the UMVUE for ξ. ˆξ = X σ2 n X Problem. (Problem 2.2.5. Soluton. If X,..., X n Therefore, snce..d. N (ξ, σ 2, then T = (X, S 2 s a complete suffcent statstc. E δ(x = E[X 2 ] E[S2 ] n(n = σ2 n + ξ2 σ2 n = ξ2, so δ s unbased, a functon of T, and thus s the UMVUE for ξ 2. Problem 4. (Problem 2.2.7. Soluton. Assume X N(ξ, σ 2. If an unbased estmator δ of σ 2 exsts when ξ s unknown, E ξ,σ 2[δ(X] = σ 2 ξ, σ 2. As the hnt suggests, for fxed σ = a, X s a complete suffcent statstc for ξ, and thus E ξ [δ(x] = a 2 for all ξ mples δ(x = a 2 almost surely. However, from the unqueness of UMVUEs, ths s a contradcton. Hence, such an unbased estmator of σ 2 does not exst. Problem 5. (Problem 2.2.25. Soluton. Let X,..., X m and Y,..., Y n be..d. as Unf(0, θ and Unf(0, θ, respectvely. We know that (X (m, Y (n s a complete suffcent statstc of the data. (Lehmann and Casella, Example 6.2. Snce 2 θ 2 = E θ [ ] [ ] X( +... + X (m X = Eθ, m m m = X ( s an unbased estmator of θ. Usng Rao-Blackwell Theorem, [ ] 2 ˆθ = E θ m (X ( +... + X (m X (m = 2 ( (m X (m + X m 2 (m = m + m X (m s the UMVUE of θ. Now we wll derve the UMVUE of /θ. Snce Y (n = max{y,..., Y n }, the cdf of Y (n s P(Y (n t = P(Y t P(Y n t = tn θ 2
for t [0, θ ]. Hence, Y (n has pdf f Y(n (t = ntn (θ n t [0,θ ]. Therefore, E θ Hence, the UMVUE of /θ s [ Y (n ] = θ 0 ny n y (θ n dy = n n θ = n. n Y (n Therefore, snce X m and Y n are ndependent, the UMVUE of θ/θ s Problem 6. (Problem 2.2.27. θ θ = ˆθ θ = (m + (n X (m. mn Y (n Soluton of (a. The bas of the ML estmator Φ(u X s bas(ξ = E[Φ(u X] Φ(u ξ. Note that u X N (u ξ, n. Therefore, f ξ = u, then u X N (0, n. Also, snce Φ(z Φ(0 s an odd functon, we get E[Φ(u X Φ(0] = 0, whch mples bas(u = 0. θ. Soluton of (b. R ML (ξ = E ξ [ (Φ(u X Φ(u ξ 2 ], [ ( ( 2 ] n R δ (ξ = E ξ Φ (u X Φ(u ξ. n Then at ξ = u, the dfference of the expectd square error s, [ ( ( n R δ (u R ML (u = E ξ=u Φ (u X n ( n = E ξ=u [Φ (u X n 2 ] [ Φ(0 E ξ=u (Φ(u X Φ(0 2 ] 2 Φ(u X 2 ]. ( 2 However, snce n n (u X > (u X 2 always and Φ( s strctly ncreasng, the ntegrand s always postve. Hence, R δ (u R ML (u > 0. Problem 7. (Problem 2..8.
Soluton. Let X Posson(θ. Suppose there exsts an unbased estmator δ(x of /θ. Then for all θ, θ = E θ θx θ δ(x = e x! δ(x, and thus x=0 x=0 θ x x! = θ x δ(x (x!. x= However, ths does not hold n general, snce the rght hand sde does not have a constant term. Hence, there s no such an unbased estmator of /θ. Problem 8. (Problem 2..20. Soluton of (c. One can easly see that Note that P λ (x = E λ X = On the other hand, we would get Therefore, e λ λ x e λ x! e λ e λ x= for x =, 2,.... λ x (x! = λ e λ. log P λ (x = λ log( e λ + x log λ log x!, λ log P λ(x = e λ + x λ, 2 λ 2 log P e λ λ(x = ( e λ 2 x λ 2. (θ = E X λ 2 e λ ( e λ 2 = λ( e λ e λ ( e λ 2 = e λ λe λ λ( e λ 2. Thus, the CRLB of ths problem s Problem 9. (Problem 2..2. Var λ λ( e λ 2 n( e λ λe λ. Soluton. Let Y Posson(λ and Z = Y {Y a}. Then P λ (Z = z = P(Y = y, Y a P(Y a = λy A(λy! y [0,a], 4
where A(λ = a x=0 λx x! After some algebra, we have. Suppose there exsts an unbased estmator δ(z of λ. Then for all λ > 0, a z=0 δ(z P λ (Z = z = A(λ a z=0 δ(z λ z = λ. z! a ( δ(z λ z + δ(0 = λa+ z! (z! a! z= λ > 0. Ths cannot be true, however, for any choce of δ(, because degrees of LHS and RHS dffer. Hence, there exsts no unbased estmator of λ. Problem 0. (Problem 2..2. Suppose X,..., X n..d. Posson(λ, and consder estmaton of e bλ, where b s known. Soluton of (a. T = X s a complete suffcent statstc. We are gven δ (X = η(t = ( b n T. The expectaton s E η(t = x=0 ( nλ (nλx e b x = e bλ (n bλ ((n bλx e x! n x! x=0 so t s unbased. By Lehmann-Scheffe Theorem, δ s the UMVUE. = e bλ, Soluton of (b. If b > n, δ s postve f T s even, and negatve f T s odd. Therefore, ts behavor s not desrable as an estmator of a postve quantty e bλ. Problem. (Problem.6.29. If a mnmal suffcent statstc exsts, a necessary condton for a suffcent statstc to be complete s for t to be mnmal. Soluton. Suppose that T = h(u s mnmal suffcent and U s complete. If U s not equvalent to T, there exsts a functon ψ such that ψ(u E[ψ(U T ] wth postve probablty. However, by law of terated expectaton, we have E[E[ψ(U T ] ψ(u] = 0, and thus E[ψ(U T ] ψ(u s an unbased estmator of 0. Now, t follows that E[ψ(U T ] ψ(u = E[ψ(U h(u] ψ(u = 0 almost surely from completeness of U, whch s a contradcton. Hence, U s equvalent to the mnmal suffcent statstc T. Problem 2. (Problem.6.2. Soluton of (a. P 0, P are two famles of dstrbutons such that P 0 P and every null set of P 0 s also a null set of P. Assume T s complete for P 0. Then, E F [δ(t ] = 0 F P 0 = δ 0 (a.e. P 0. 5
We have E G [δ(t ] = 0 G P = δ 0 (a.e. P 0 (0. = δ 0 (a.e. P, (0.2 so ths mples T s also complete for P. Note that eq. (0. follows from P 0 P, and eq. (0.2 follows because every null set of P 0 s also a null set of P. Soluton of (b. P 0 = {Bnom(n, p: 0 < p < } where n s fxed, and P = P 0 {Posson(}. E p δ(x = n k=0 ( n p k ( p n k δ(k = 0 p (0, = k Hence, P 0 s complete. However, consderng P, we assume and E p δ(x = n k=0 n k=0 ( n ρ k δ(k = 0 ρ > 0 k = δ(x 0 (a.e. ( n p k ( p n k δ(k = 0 p (0,, k E Posson( δ(x = k=0 e δ(k = 0. k! From the frst restrcton, t s requred that δ(0 =... = δ(n = 0 as we derved. However, the second restrcton δ(k = 0 k! k=n+ can be satsfed wth a smple choce of δ, for example, δ(n + = (n +!, δ(n + 2 = (n + 2!, and δ(x = 0 for x n +. Hence, P s not complete. Problem. (Addtonal problem. Show that any fnte famly of denstes on R wth common support s an exponental famly. If the famly has more than one densty, the parameter space s not natural. Soluton. Let F = {f (x,..., f N (x}. Then we can express ths famly as the followng form. { ( N } F = g η (x: g η (x = exp η log f (x, η {e,..., e N }, = where we denote e as a standard unt vector. Clearly, f N 2, then the parameter space s not natural. Problem 4. (Addtonal problem 2. Defne E (λ X = (E X λ /λ. (a Show lm λ 0 E (λ = e E log X. 6
(b Extendng the defnton through λ = 0, show that E (λ X s monotoncally ncreasng n λ. Soluton of (a. We want to prove Usng L Hoptal s Law, t follows that lm λ 0 λ log E Xλ = E log X. lm λ 0 λ log E Xλ = lm λ 0 E[X λ log X] E[X λ ] = E[X0 log X] E[X 0 ] = E log X. Soluton of (b. Consder η > λ > 0. Then x x η λ s (strctly convex, and thus usng Jensen s Inequalty, we would get (E X λ η λ > E(X λ η λ = E X η, whch mples E (η X > E (λ X. Lkewse, one can prove that t also holds when 0 > η > λ. Snce E (η X > E (0 X > E (λ X for η > 0 > λ, E (λ X s monotoncally ncreasng n λ. Problem 5. (Addtonal problem. For a dstrbuton symmetrc wth respect to ts mean, show that a statstc ( n n n T = X, X 2, s not complete. = = Soluton. Let us denotde T = (T, T 2, T. Let θ := E X. Then by symmetry, we know that E(X θ = 0. Expandng the terms, we would get E(X θ = E X θ E X 2 + θ 2 E X θ = E X θ E X 2 + 2θ = 0. Then we consder a functon of data δ(x,..., X n = (X X. Then = (X X = (X X 2 X + X X 2 X = X X 2 X + nx nx X = T nt T 2 + 2 n 2 T, δ s a functon of T. Also, we observe E(X X = E X E XX 2 + E X X 2 E X = E X ( E X n + (n E X 2 E X + ( E X n 2 + (n E X 2 E X + (n (n 2(E X 7
n ( n E X + n(n E X 2 E X + n(n (n 2(E X = E X n 2 (n (n 2 E X2 E X n 2 (n (n 2 + (E X n 2 2(n (n 2 (n (n 2 ( = E X n 2 E X 2 E X + 2(E X (n (n 2 = n 2 E(X E X = 0. Thus, δ s a nontrval unbased estmator of 0. Hence, T s not a complete statstc. 8