No4(2004)solutions

No4(2004)solutions - MATH 587 Solutions to Assignment 4...

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MATH 587 Solutions to Assignment 4 November 30, 2004 (1) E ( Y X ) 2 = E [( Y E ( X |G ))+( E ( X |G ) X )] 2 = E [ Y E ( X |G )] 2 + E [( Y E ( X |G ))( E ( X |G ) X )] + E [ E ( X |G ) X )] 2 . But E [( Y E ( X |G ))( E ( X |G ) X )] = E [ E ( Y X |G )( E ( X |G ) X )] = E { E [ E ( Y X |G )( E ( X |G ) X ) |G ] } = E { E ( Y X |G ) E [ E ( X |G ) X |G ] } = 0. Hence E ( Y X ) 2 = E [ Y E ( X |G )] 2 + E [ E ( X |G ) X )] 2 . The LHS will be a minimun when Y = E ( X |G ). (2) (a) By Fubini, E i =1 X i = i =1 EX i < ,so i =1 Y i < a.s. and therefore Y n 0 a.s. (b) Let W i = X i µ . Then E ( S n ) 4 = E ( W 1 + ··· + W n ) 4 . By independence and since EW =0, the only terms in the expansion with non-zero expectation are of the form EW 4 and E ( W 2 i ) E ( W 2 j ) where i 6 = j . It follows that E ( S n ) 4 = +3 n ( n 1) σ 4 , where ρ = E ( X i µ ) 4 < . Then n =1
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This note was uploaded on 01/25/2011 for the course STAT 235a at Stanford.

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