MATH 281A: Homework #6

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1 MATH 28A: Homework #6 Jongha Ryu Due date: November 8, 206 Problem. (Problem 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 ( 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.

2 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 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 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 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

3 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 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.

4 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 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 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

5 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 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 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

6 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

7 (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 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

8 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