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No4(2004)solutions

# No4(2004)solutions - MATH 587 Solutions to Assignment 4(1...

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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 = j . It follows that E ( S n ) 4 = + 3 n ( n 1) σ 4 , where ρ = E ( X i µ ) 4 < . Then n =1 EY n < , so Y n 0 a.s., so S n n µ a.s. (c) It is equivalent to show that 1 n n i =1 log X i → − 1 a.s. This follows from part (b) once we observe that E log X = 1 0 (log x ) 4 dx = 0 w 4 e w dw = Γ(5) = 24 (where w = log x ), and E log X =
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