X1 1 p p 1 x p p 1 1 p p x p x p 1 1 p p x example

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Unformatted text preview: always assume that all of the indicated expected values exist. In addition, as in ordinary integration one can take constants outside of the integral. Lemma 5.1. If X is a constant c on A, then E (XY |A) = cE (Y |A). Proof. Since X = c on A, XY 1A = cY 1A . Taking expected values and pulling the constant out front, E (XY 1A ) = E (cY 1A ) = cE (Y 1A ). Dividing by P (A) now gives the result. Being an expected value E (·|A) it has all of the usual properties, in particular: Lemma 5.2. Jensen’s inequality. If E ( (X )|A) is convex then (E (X |A)) Our next two properties concern the behavior of E (Y ; A) and E (Y |A) as a function of the set A. Lemma 5.3. If B is the disjoint union of A1 , . . . , Ak , then E (Y ; B ) = k X E (Y ; Aj ) j =1 Pk Proof. Our assumption implies Y 1B = j =1 Y 1Aj , so taking expected values, we have 1 0 k k k X X X Y 1A j A = E (Y 1Aj ) = E (Y ; Aj ) E (Y ; B ) = E (Y 1B ) = E @ j =1 j =1 j =1 Lemma 5.4. If B is the disjoint union of A1 , . . . , Ak , then E (Y |B ) = k X j =1 E (Y |Aj ) · In...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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