**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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