hw4sol - Math 280A, Fall 2011 Homework 4 Solutions 2....

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Unformatted text preview: Math 280A, Fall 2011 Homework 4 Solutions 2. Problem 17 (page 116). Lets begin with part (c), confirming that Y n is integrable. This will quell any doubts one might have about the finiteness of E [ Y n ] in part (a). (c) Observe that Y + n X + 1 + X + 2 + + X + n , so E [ Y + n ] n summationdisplay k =1 E [ X + k ] < , because each X k is integrable. On the other side, Y n X 1 , so Y n X 1 , and therefore E [ Y n ] E [ X 1 ] < , since X 1 is integrable. It follows that E | Y n | = E [ Y + n ] + E [ Y n ] < . (a) E [ Y n ] = E [ Y + n ]- E [ Y n ] E [ Y + n ] < , in view of part (c). (b) Clearly X k Y n for each k , so E [ X k ] E [ Y n ] for each k . (d) Here is a simple example with n = 2 and E | X 1 | > E | Y 2 | . In this example, the sample space has just two points, say = { , 1 } . Each point of is given weight 1 / 2 by the probability measure P . (For F we take { , , { } , { 1 }} .) Our random variables....
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This note was uploaded on 12/11/2011 for the course MATH 280a taught by Professor Driver,b during the Fall '08 term at UCSD.

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hw4sol - Math 280A, Fall 2011 Homework 4 Solutions 2....

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