851_lectures17_24

# B if h 0 or if either condition in a holds then ehx

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Unformatted text preview: se two inequalities gives E(X ) = E(U ) ≤ lim inf E(Xn ) ≤ lim sup E(Xn ) ≤ E(V ) = E(X ) n→∞ n→∞ so that E(Xn ) → E(X ) as required. Theorem 23.4. Let Xn be a sequence of random variables. (a) If Xn ≥ 0 for all n, then ￿ E ∞ ￿ Xn n=1 ￿ = ∞ ￿ E( Xn ) n=1 with both sides simultaneously being either ﬁnite or inﬁnite. (b) If ∞ ￿ n=1 then E(|Xn |) < ∞, ∞ ￿ Xn n=1 converges almost surely to some random variable Y ∈ L1 . In other words, ∞ ￿ Xn n=1 is integrable with E ￿ ∞ ￿ n=1 Xn ￿ = ∞ ￿ E( Xn ) . n=1 Thus, (23.3) holds with both sides being ﬁnite. Proof. Suppose that Sn = n ￿ k=1 |Xk | and Tn = n ￿ Xk . k=1 Since Sn contains ﬁnitely many terms, we know by linearity of expectation that ￿n ￿ n ￿ ￿ E ( Sn ) = E |Xk | = E( |Xk |) . k=1 k=1 23–3 (23.3) Moreover, 0 ≤ Sn ≤ Sn+1 so that Sn increases to some limit S= ∞ ￿ k=1 |Xk |. Note that S (ω ) ∈ [0, +∞]. By the Monotone Convergence Theorem, E(S ) = lim E(Sn ) = lim n→∞ n→∞ n ￿ k=1 E(|Xk |) = ∞ ￿ k=1 E(|Xk |). If Xn ≥ 0 for all n, then Sn = Tn and (a) follows. If Xn are general random variables, with ∞ ￿ k=1 E(|Xk |) < ∞, then E(S ) < ∞. Now, for every ￿ > 0, since 1S =∞ ≤ ￿S , we conclude P {S = ∞} = E(1S =∞ ) ≤ ￿E(S ). Since ￿ > 0 is arbitrary and E(S ) < ∞, we conclude that P {S = ∞} = 0. Therefore, ∞ ￿ Xn n=1 is absolutely convergent almost surely and its sum is the limit of the sequence Tn . Moreover, |Tn | ≤ Sn...
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## This document was uploaded on 04/07/2014.

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