SOLUTION FOR HOMEWORK 7, STAT p(x σ) = (1/[2πσ 2 ] 1/2 )e (x µ)2 /2σ 2.

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1 SOLUTION FOR HOMEWORK 7, STAT We have (for a general case) Denote p (x) p(x σ)/ σ. Then p(x σ) (1/[2πσ 2 ] 1/2 )e (x µ)2 /2σ 2. p (x σ) p(x σ) 1 (x µ)2 +. σ σ 3 Then E{ p (x σ) p(x σ) } σ 2 2σ 4 E(X µ) 2 + σ 6 E(X µ) 4 σ 2 2σ 2 + 3σ 4 /σ 6 2σ 2 : I(σ). Note that for an optimal ˆσ this implies that n 1/2 (ˆσ σ) L N(0, σ 2 /2). Now let θ σ 2 be the estimand. Then, using the Delta method, n 1/2 (ˆσ 2 σ 2 ) L N(0, (2σ) 2 (σ 2 /2)) N(0, 2σ 4 ). But we know (or you can calculate it) that I(θ) 1/[2σ 4 ], that is, we got the efficient estimator. 2. Let p(x θ) p(x θ), x (, ). Then [change of variable u x θ] I(θ) [p (x θ)] 2 p 1 (x θ)dx [p (u)] 2 p 1 (u)du C. In other words, the Fisher information for location family is constant. Thus, for an unbiased estimate θ n based on n observations we have E θ {( θ n θ) 2 } [nc] Consider p(x θ) θ 1 f(x/θ), θ > 0. Then p (x θ) (1/θ 2 )f(x/θ) θ 3 xf (x/θ). We conclude that I(θ) θ 2 θ 2 [f(x/θ) + xθ 1 f (x/θ)] dx θ 1 f(x/θ) 1

2 [make change of variable u x/θ so du dx/θ] θ 2 [f(u) + uf (u)] 2 du f(u) θ 2 [uf (u)/f(u) + 1] 2 f(u)du. 4. Here the idea is that if Y θz > 0 then ln(y ) ln(θ) + ln(z) becomes the location model for which, as we know from Problem 1, the Fisher information is constant. Now let us prove the wished assertion. The density is e ξ f(xe ξ ). Then I(ξ) [change of variable u xe ξ, so du e ξ dx] [ e xi f(xe ξ ) xe 2ξ f (xe ξ )] 2 dx e ξ f(xe ξ ) [f(u) + uf (u)] 2 du. f(u) As you see, the last expression does not depend on ξ. 5. Let us recall the idea of the Information inequality, 0 [ p θ (x) θ]dx p θ (x)dx. Then θ E θ{δ(x)} p θ (x)δ(x)dx p θ (x)(δ(x) g(θ)dx. Then we use Cauchy-Schwarz for p θ x) p θ (x) [ p θ (x)(δ(x) g(θ))]dx, and get the Information inequality. Now remember that we have equality in Cauchy-Schwarz iff p θ (x) p θ (x) C p θ (x)(δ(x) g(θ)) which yields δ(x) g(θ) + C p θ(x) p θ (x). 2

3 Finally E(δ(X) g(θ)) 2 (g(θ))2 I(θ), and this implies that C 2 [g (θ)] 2 /I 2 (θ). We conclude that δ(x) g(θ) + g (θ) I(θ) p θ(x)/ θ 6. Here X Norm(θ, 1), and then Further, g(θ) + g (θ) I(θ) ln(p θ(x))/ θ. p X X (a,b) θ (x) (2π)1/2 e (x θ)2 /2 b a (2π)1/2 e (u θ)2 /2 du. P θ (X (b ǫ, b) X (a, b)) F θ(b) F θ (b ǫ) F θ (b) F θ (a) F(b ǫ θ). F(a θ) Now, if θ then F(x θ) becomes small. It is known that F(x) e x2 2π x (1 + o x (1)), x Thus as θ. Also, from and (1) we get as θ. F(a θ) O(e2(b a)θ ) (1) P θ (X (b ǫ, b) X (a, b)) 1 F(b ǫ θ) O(e 2ǫθ ) 0 F(b ǫ θ) 7. Here P θ (X i x) θ x e θ /[x!(1 e θ )], x 1, 2,... We are seeking the maximum of L(x n, θ) ln(p θ (X x i )) x i ln(θ) nθ ln(x i!) n ln(1 e θ). i1 i1 i1 3

4 Set the derivative to zero, 0 L (X n, θ) X i /θ n ne θ i1 1 e θ ni1 x i n[1 + e θ θ 1 e n x θ i /θ n/(1 e θ ) 0. i1 We get x θ/(1 e θ ). Now consider a function We write f(θ) : θ 1 e θ. f (θ) (1 e θ ) θe θ (1 e θ ) 2 1 (1 + θ)e θ (1 e θ ) 2. As we see, f (θ) is positive for both small and large θ. Let us check the possibility of f (θ) 0. In this case e θ 1 + θ, and this occurs only if θ 0, and this case is not of interest to us. We conclude that f(θ) is increasing in θ and the solution x f(θ) is unique. 8. Here L(θ) : L(X n, θ) ln(p(x n θ)). Then [using Taylor s formula] L(θ 0 + 1/ n) L(θ 0 ) + (1/2)I(θ 0 ) n 1/2 L (θ 0 ) + [2n] 1 L (θ 0 ) + [6n 3/2 ] 1 L ( θ) + (1/2)I(θ 0 ). By the CLT By the LLN n 1/2 L (θ 0 ) n 1/2 n i1 p (X i θ) p(x i θ) L N(0, I(θ)). (2n) 1 L (θ 0 ) P (1/2)E θ0 { p (X θ 0 )p(x θ 0 ) (p (X θ 0 )) 2 } (1/2)I(θ p 2 0 ). (X θ 0 ) Also note that n 1 L ( θ) < C. These relations imply that [L(θ 0 + 1/n 1/2 ) L(θ 0 ) + (1/2)I(θ 0 )]/I 1/2 (θ 0 ) L N(0, 1). 9. Assume that f(x) is differentiable. Then ln(f(x)) ψ(x) and we get for the considered function that ψ (x) < 0. Further, f (x)/f(x) ψ (x). We need to ask about f(x) 0 as x. Then f (x) has the sign of ψ (x). Further, ψ (x) 4

5 cannot strictly positive or negative, so f(x) increases and then decreases. This yields one mode if we assume that ψ (x) < Let us consider just the case of one observation. Then L (x, θ) 0 implies f (x)/f(x) 0 because strictly monotone in θ implies unique root. If f (x)/f(x) is decreasing then L(x, θ) is convex (or the send derivative is positive). For n observations, L(X n, θ) n i1 ln(p(x l θ), and L (X n, θ) n i1 f (X i θ)/f(x i θ). Then you just repeat the above-provided arguments. 5