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### hw9_stat210a_solutions

Course: STAT 210a, Fall 2006
School: Berkeley
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Word Count: 1013

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Berkeley UC Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 9 Fall 2006 Issued: Thursday, November 2, 2006 Due: Thursday, November 9, 2006 Graded exercises Problem 9.1 (a) First, notice that ga () = P(X1 a) = P(X1 - a - ) = (a - ). Additionally, by the CLT, we have that: n( - Xn ) = n (a - Xn ) - (a - ) N (0, 1) d Letting h(.) = (.) and using the delta...

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Berkeley UC Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 9 Fall 2006 Issued: Thursday, November 2, 2006 Due: Thursday, November 9, 2006 Graded exercises Problem 9.1 (a) First, notice that ga () = P(X1 a) = P(X1 - a - ) = (a - ). Additionally, by the CLT, we have that: n( - Xn ) = n (a - Xn ) - (a - ) N (0, 1) d Letting h(.) = (.) and using the delta method yields: So: n (a - Xn ) - ga () N 0, [(a - )]2 p d n (a - Xn ) - (a - ) N 0, (a - ) d 2 To get to the result, it is enough to prove that n (Xn ) - (a - Xn ) 0. This follows from continuity of and the fact that: n (a - Xn ) - (a - Xn ) = n-1 n p - 1 (a - Xn ) 0 n-1 (b) We know that I(Xi a) is a Bernoulli variable with mean P(Xi a) = FX (a) = ga () and variance ga ()(1 - ga ()). Using the central limit theorem: n( (X) - ga ()) = n Pn i=1 I(Xi a)-ga n N (0, ga ()(1 - ga ())) d Under normality, ga () = (a - ) and the result follows. (c) The asymptotic relative efficiency between n and n is: ARE( , ) = = (a - ) [1 - (a - )] 2 (a - ) (a - )( - a) 2 (a - ) From the plots below, we can see that the non-parametric estimator is less efficient that the parametric one. Furthermore, the efficiency of the non-parametric estimate degrades exponentially fast as we move towards the tails of the distribution. 1 (a) ARE( , )vs.a - Problem 9.2 (b) [ARE( , )]-1 vs.a - (a) We have that Y0 Poisson() and given Yj-1 , Yj Poisson(yj-1 ). It follows that the log-likelihood of the sequence of observations Y0 , Y1 , . . . , Yn is given by: n L(y; ) = log p(y0 ) j=1 p(yj |yj-1 , . . . , y0 ) n j=1 n = log p(y0 ) p(yj |yj-1 ) log (p(yj |yj-1 ) j=1 n = log (p(y0 )) + = - y0 log() + (yj-1 - yj log(yj-1 )) j=1 = 1 + n j=1 n j=0 yj + C(Y ) yj-1 - log() Noting that this is a concave differentiable function on it is enough to set: L(Y ; ) ^ = 0 (Y ) = ^ n j=0 Yj n-1 j=0 Yj 1+ (b) We know that we can compute the information about from: 2 L(Y ; ) In () = E - 2 n 1 = E Yj = -2 2 j=0 n E (Yj ) j=0 2 Finally, notice that: E Y0 = E (Yj ) = E (E (Yj |Yj-1 )) = E (Yj-1 ) =j+1 where the last equality follows from induction. We conclude that: n n+1 In () = -2 j=0 j+1 = -2 j=1 j When > 1, we have that limn In () = , so the variance of the MLE estimate shrinks to zero asymptotically and consistency follows (L2 -convergence convergence in probability). (c) If < 1, we have that: n+1 k=1 k = k=1 - k=n+2 k k = - n+2 1- Hence, we have: n lim In () = < 1- ^ As a result, - has asymptotic distribution N (0, 1- ). It follows that the variance does not decay at a fast enough to rate yield consistency for the MLE. Problem 9.3 Defining the conditional density on Rn , we know that: + p(X|) = and so the posterior of given X is such that: (|X) = I( - Mn 0) n () (t) Mn tn dt I(Mn ) Now, define Yn = n( - Mn ), whose density at y can be computed as a linear transform of as: Y (y|X) = = 1+ y I(y 0) (Mn + n ) y n (t) n Mn tn dt y I(y 0) (Mn + n ) y nMn n 0 I(y Mn n (t)dt t Mn + Mn ) 3 For n , we know: Mn y n y Mn + n y (Mn + ) n n y 1+ nMn Now, it is enough to prove that 0 ( ), exp 0 I(y p p p p by continuity of y , Mn n (t)dt t Mn ) min{1, ( ). We know that: I(y Mn ) 0 Mn t n (t) dt 0 Mn t n }(t) dt 0 (t)dt = 1 and I(y Mn ) Mn t n ( ), poitwise as n where (x) denotes the Dirac delta at x. Using the dominated convergence theorem: n 0 lim I(y Mn ) Mn t n (t)dt = 0 n lim I(y Mn ) Mn t n (t)dt = ( ) and the result follows. Problem 9.4 We want to prove that, for all , P ( S(X)) 1 - . To prove that, we first notice that, according to the definition of S(X), S(X) X A(). It follows that: P ( S(X)) = P (X A()) 1 - , for all where the inequality follows from the definition of a acceptance region for a level test. Problem 9.5 n 2 (a) Set Yn N (0 , ). It follows that n(Y -0 ) N (0, 1). Let q = -1 (1 - ) be the n 1 - quantile of the standard normal distribution. It follows that: n(Yn - 0 ) P 0 q = and hence: q P0 Yn 0 + n and we can set t = 0 + q . n q = E0 I Yn 0 + n = 4 (b) From the definition of power: (1 ) = E1 ((Y )) n(Yn - 0 ) = P 1 q n(Yn - 1 + 1 - 0 ) = P 1 q n(Yn - 1 ) n(1 - 0 ) = P 1 q - From part a: q = n(Yn -1 ) n(t - 0 ) so, noticing that Z = N (0, 1) under H1 : = 1 - (-z - ) = (z + ) (1 ) = P (Z -z - ) 1 (c) From the definition of Xi , we have that E0 (Xn ) = 2 and var0 (Xn ) = binomial approximation to the normal, that yields that: 1 4n . Using the 1 n Xn - 2 1 approx. N (0, ) 4 So, following the notation of items a and b above, a significance level threshold against H1 is given by: ~ t = 1 1 + z 2 2 n Now, notice that under H1 , E1 Xn = P1 (Yi 0 ) = P1 ( Yi -1 0 -1 ) = ( 0 -1 ) = n n ) = 1 ( n ) 1 - ( n ) . Using the normal approxi( ). Furthermore, var (X n 1 n n n mation to the binomial once again: ~ X (1 ) = P1 Xn t ~ n Xn - n n t - n = P 1 1 2 n 1 - n n 1- n n n n 1 - 1 - 2 n z 2 n n n 1 n 2 n 1 - n n n n n - n - 2n z 2 n = 1 2 n n n n 1 - n 5 1 2 n n n n 2 2 (d) Since < 1, we have that z + n < z + n and, as is strictly increasing, the censoring will result in a loss of power for the test. 6
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