That is if xn xn1 and lim xn x n then exn1 exn

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Unformatted text preview: X (that is, Xn increases to X ). 18–1 Proof. Let X ≥ 0 be given and define the sequence (Xn ) by ￿ k +1 , if 2k ≤ X (ω ) < k2n and 0 ≤ k ≤ n2n − 1, n 2n Xn ( ω ) = n, if X (ω ) ≥ n. Then it follows that Xn ≤ Xn+1 for every n = 1, 2, 3, . . . and Xn → X which completes the proof. Proposition 18.2. If X ≥ 0 and (Xn ) is a sequence of simple random variables with Xn ↑ X , then E(Xn ) ↑ E(X ). That is, if Xn ≤ Xn+1 and lim Xn = X, n→∞ then E(Xn+1 ) ≤ E(Xn ) and lim E(Xn ) = E(X ). n→∞ Proof. Suppose that X ≥ 0 is a random variable and let (Xn ) be a sequence of simple random variables with Xn ≥ 0 and Xn ↑ X . Observe that since the Xn are increasing, we have E(Xn ) ≤ E(Xn+1 ). Therefore, E(Xn ) increases to some limit a ∈ [0, ∞]; that is, E(Xn ) ↑ a for some 0 ≤ a ≤ ∞. (If E(Xn ) is an unbounded sequence, then a = ∞. However, if E(Xn ) is a bounded sequence, then a < ∞ follows from the fact that increasing, bounded sequences have unique limits....
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This document was uploaded on 04/07/2014.

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