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 ,wehave 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 ( R 2 ). (b) Let C = { ( x, y ) | x + y z } and note that C y = { x | x z y } . Then F Z ( z )= RR I C ( x, y ) X µ Y = R £ R I C y ( x ) X ¤ Y = R µ X ( C y ) Y = R F X ( z y ) dF Y ( y ). 3. By Fubini, ( X, Y )= R £ R φ ( x, y ) dP Y ¤ dP X = R g ( x ) dP X = Eg ( X ). Next, suppose E | Y | = . Since
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This note was uploaded on 01/25/2011 for the course STAT 235a at Stanford.

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

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