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...
View
Full
Document
This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

Click to edit the document details