hw3soln - CME308: Assignment 3 Due: Thursday, May 6, 2010...

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Unformatted text preview: CME308: Assignment 3 Due: Thursday, May 6, 2010 Due Date: This assignment is due on Thursday, May 6, 2010, by 5pm under the door of 380-383V. See the course website for the policy on incentives for L A T E X solutions. Problem 1 (10 pts): Quantile Estimation Solution: 1. For a strictly positive and continuous density f , the cdf F is invertible. Then F- 1 ( U ) is distributed with cdf F when U Unif[0 , 1]. From this and the Delta Method, we will have n 1 / 2 ( Q b pn c- q p ) D = n 1 / 2 ( F- 1 ( U b pn c )- F- 1 ( p )) D N (0 , ( F- 1 ( p ) ( p ) 2 ) , where ( p ) 2 is the asymptotic variance in the CLT for U b pn c (and will naturally be p (1- p )) while ( F- 1 ( p )) = 1 f ( q p ) . We will show later on that ( p ) 2 = p (1- p ), but for now we have just reduced the problem to a base case concerned only with uniform random variables. 2. For the k-th order statistic U ( k ) , we have that P ( U ( k ) u ) = P (At least k of { U 1 ,...,U n } is less than u ) Note that P ( U i u ) = u for all i since each U i is uniform on [0 , 1]. If we let Y = n i =1 1 { U i u } , so that Y is the total number of the U i that are less than u , then Y is distributed Binom( n,u ). Therefore, P ( U ( k ) u ) = P (At least k of { U 1 ,...,U n } is less than u ) = P ( Y k ) = n X i = k n k u k (1- u ) n- k . We differentiate this to get the pdf of U ( k ) : d du n X i = k n i u k (1- u ) n- k ! = n X i = k i n i u i- 1 (1- u ) n- i- n X i = k ( n- i ) n i u i (1- u ) n- i- 1 = n n X i = k n- 1 i- 1 u i- 1 (1- u ) n- i- n X i = k n- 1 i u i (1- u ) n- i- 1 ! = n n- 1 X i = k- 1 n- 1 i u i (1- u ) n- i- 1- n X i = k n- 1 i u i (1- u ) n- i- 1 ! = n n- 1 k- 1 u k- 1 (1- u ) n- k- n- 1 n u n (1- u )- 1 = n n- 1 k- 1 u k- 1 (1- u ) n- k = n ! ( k- 1)!( n- k )! u k- 1 (1- u ) n- k The last expression is the pdf for a Beta( k,n- k + 1) random variable. For k = b pn c , we then have U ( b pn c ) Beta( b pn c ,n- b pn c + 1) . 3. From the hint, we have that U ( b pn c ) D = b pn c i =1 Z i n +1 i =1 Z i , Z i Exp(1) , i.i.d. We can equivalently write, U ( b pn c ) D = 1 n +1 b pn c i =1 Z i 1 n +1 n +1 i =1 Z i , 1 CME308: Assignment 3 Due: Thursday, May 6, 2010 which means by the Delta Method for f ( x,y ) = x y , we have n ( U b pn c- p ) = N (0 , f ( p, 1) T f ( p, 1)) , where f ( p, 1) = (1 ,- p ) T and thus, f ( p, 1) T f ( p, 1) = p (1- p ) . Extra Credit: To show n + 1 " 1 n +1 b pn c i =1 Z i 1 n +1 n +1 i =1 Z i !- p 1 # D N (0 , ) , (1) we use the fact that convergence in distribution is equivalent to convergence of characteristic functions. For characteristic function n ( t 1 ,t 2 ) of the vector on the LHS of ( 1 ), n ( t 1 ,t 2 ) = E exp i n + 1 t 1 n + 1 b pn c X i =1 Z i + t 2 n + 1 n +1 X i =1 Z i- t 1 p- t 2 By independence of the Z i s, we can rewrite this as n ( t 1 ,t 2 ) = E...
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hw3soln - CME308: Assignment 3 Due: Thursday, May 6, 2010...

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