The next theorem tells us that continuous functions

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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 < ∞ 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...
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This document was uploaded on 04/07/2014.

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