851_lectures17_24

# The next theorem tells us that continuous functions

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: j 1Aj . j =1 Observe that if pj = 0 for inﬁnitely many j , then X is simple and can be written as X (ω ) = j ￿ 1Aj j =1 for some n &lt; ∞ which implies that E(X ) = j ￿ j =1 P {Aj } = n ￿ j =1 j P {X = j } . On the other hand, suppose that pj ￿= 0 for inﬁnitely many j . In this case, X is not simple. We can approximate X by simple functions as follows. Let Aj,n = {ω : X (ω ) = j, j ≤ n} so that Aj,n ⊆ Aj,n+1 and ∞ ￿ Aj,n = Aj . n=1 22–1 That is, Aj,n ↑ Aj and so by continuity of probability we conclude lim P {Aj,n } = P {Aj } . n→∞ If we now set Xn (ω ) = n ￿ X (ω )1Aj,n = j =1 so that E(Xn ) = n ￿ j 1Aj,n j =1 n ￿ j =1 j P {Aj,n } , then (Xn ) is a sequence of positive simple random variable with Xn ↑ X . Proposition 18.2 then implies E(X ) = lim E(Xn ) = lim n→∞ n→∞ n ￿ j =1 j P {Aj,n } = ∞ ￿ j =1 P {Aj } = ∞ ￿ j =1 j P {X = j } as required. Computing Expectations of Continuous Random Variables We will now turn to that other formula for computing expectatio...
View Full Document

## This document was uploaded on 04/07/2014.

Ask a homework question - tutors are online