sol7S09

# sol7S09 - PSTAT 120C: Solutions to Assignment #7 June 3,...

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Unformatted text preview: PSTAT 120C: Solutions to Assignment #7 June 3, 2009 1. For Y N ( , 2 ), Z N ( , 2 ), and X = Y + Z (a) The covariance of W = Y- Z. and X is Cov( X,W ) = Cov Y + Z, Y- Z = Cov Y, Y- Cov Y, Z + Cov Z, Y )- Cov Z, Z = ( 2 )- 0 + 0- ( 2 ) = - = 0 (b) We can write Y as a function of X and W 2 2 + 2 X + 2 + 2 W = 2 2 + 2 ( Y + Z ) + 2 + 2 Y- Z = 2 2 + 2 + 2 2 + 2 Y + 2 2 + 2- 2 2 + 2 Z = Y. (c) By the answer to part (a) X and W are independent. Thus, E ( W | X ) = E ( W ) because X has no information about W . Therefore, E ( W | X ) = E Y- E Z = - . (d) By part (b), E ( Y | X ) = E 2 2 + 2 X + 2 + 2 W X = 2 2 + 2 E ( X | X ) + 2 + 2 E ( W | X ) from part (c) = 2 2 + 2 X + 2 + 2 - = 2 2 + 2 X + 2 2 + 2 - 2 2 + 2 = 2 2 + 2 ( X- ) + 2 2 + 2 . 1 2. WMS 16.8 (a) Our distribution is Binomial( n,p ), and our prior on p is a Beta(1 , 1), so our posterior f ( p | x ) can be found as follows: f ( p | x ) n x p n (1- p ) n- x 1 B ( , ) p - 1 (1- p ) - 1 p n + - 1 (1- p ) n- x + - 1 Which we can recognize as the kernel of a Beta distribution with parameters p = + x and p = + n- x ....
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## sol7S09 - PSTAT 120C: Solutions to Assignment #7 June 3,...

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