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No4(2003)solutions

# No4(2003)solutions - MATH 587 Solutions to Assignment 4...

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MATH 587 Solutions to Assignment 4 December 2, 2003 1. (a) Consider a subsequence f ( X n i , Y n i , i 1. Since X n i X in P , there is a further subsequence X n i j , j 1 such that X n i j X a.s. Since Y n i j Y in P , there is a further subsequence Y n i j k , j 1 such that Y n i j k Y a.s. Since also X n i j k X a.s. and f is continuous, then f ( X n i j k , Y n i j k ) f ( X, Y ) a.s. Hence f ( X n , Y n ) f ( X, Y ) in P . (b) Apply part (a) with f ( x, y ) = x + y and f ( x, y ) = xy . 2. (a) For a measurable rectangle C = A × B , we have µ X,Y ( A × B ) = P { X A, Y B } = P { X A } P { Y B } = µ X ( A ) µ Y ( B ) = µ X µ Y ( A × B ). Hence µ X,Y and µ X µ Y coincide on the semialgebra S of measurable rectangles, and therefore on the σ -algebra B ( 2 ). (b) Let C = { ( x, y ) | x + y z } and note that C y = { x | x z y } . Then F Z ( z ) = I C ( x, y ) d µ X µ Y = I C y ( x ) X Y = µ X ( C y ) Y = F X ( z y ) dF Y ( y ). 3. By Fubini, ( X, Y ) = φ ( x, y ) dP Y dP X = g ( x ) dP X = Eg ( X ). Next, suppose E | Y | = . Since E | Y | ≤ E | x + Y | + x , then g ( x ) = def E | x + Y | = for all x . But then part (a) implies that E | X + Y | = Eg ( X ) = .

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No4(2003)solutions - MATH 587 Solutions to Assignment 4...

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