Dr. Hackney STA Solutions pg 70

Dr. Hackney STA Solutions pg 70 - 5-4 Solutions Manual for...

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Unformatted text preview: 5-4 Solutions Manual for Statistical Inference 2 3 4 = E(Xi - )2 = 2 = E(Xi -)3 = E(Xi - )2 (Xi - ) (Stein's lemma: Eg(X)(X - ) = 2 Eg (X)) = 2 2 E(Xi - ) = 0 = E(Xi - )4 = E(Xi - )3 (Xi - ) = 3 2 E(Xi - )2 = 3 4 . 4 1 1 2 2 b. VarS 2 = n (4 - n-3 2 ) = n (3 4 - n-3 4 ) = n-1 . n-1 n-1 2 c. Use the fact that (n - 1)S 2 / 2 2 n-1 and Varn-1 = 2(n - 1) to get Var 2 (n - 1)S 2 2 = 2(n - 1) which implies ( (n-1) )VarS 2 = 2(n - 1) and hence 4 VarS 2 = 2(n - 1) (n - 1) / 4 2 = 2 4 . n-1 Remark: Another approach to b), not using the 2 distribution, is to use linear model theory. For any matrix A Var(X AX) = 22 trA2 + 42 A, where 2 is 2 , = EX = 1. Write 2 n 1 1 S 2 = n-1 i=1 (Xi - X) = n-1 X (I - Jn )X.Where 1 1 1 1- n - n - n . . - 1 1- 1 . n I - Jn = . n . . .. . . . . . 1 1 -n 1- n Notice that trA2 = trA = n - 1, A = 0. So VarS 2 = 1 (n - 1) 2 Var(X AX) = 1 (n - 1) 2 2 4 (n - 1) + 0 = 2 4 . n-1 5.11 Let g(s) = s2 . Since g() is a convex function, we know from Jensen's inequality that Eg(S) g(ES), which implies 2 = ES 2 (ES)2 . Taking square roots, ES. From the proof of Jensen's Inequality, it is clear that, in fact, the inequality will be strict unless there is an interval I such that g is linear on I and P (X I) = 1. Since s2 is "linear" only on single points, we have ET 2 > (ET )2 for any random variable T , unless P (T = ET ) = 1. 5.13 E c S2 = c = c 2 E n-1 2 n-1 0 S 2 (n - 1) 2 q n-1 2 n-1 1 q ( 2 )-1 e-q/2 dq, (n-1)/2 2 Since S 2 (n - 1)/ 2 is the square root of a 2 random variable. Now adjust the integrand to be another 2 pdf and get E c S2 = c 2 (n/2)2n/2 n - 1 ((n - 1)/2)2((n-1)/2 0 1 1 q (n-1)/2 - e-q/2 dq . 2 (n/2)2n/2 =1 since 2 n pdf So c = ( n-1 2 ) n-1 2( n ) 2 gives E(cS) = . ...
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