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Hi: I am reading a text that has the following proposition and it says that tbe proof is in Billingsley, 1979. I only have a later version of Billingsley and cannot find the proof in there. If anyone knows how to prove it or can tell me where I can find, it's appreciated. The proposition is below.

Let $\{Z_{t}\}$ be a sequence of random variables such that $\sum_{t=1}^{\infty} E|Z_{t}| < \infty $.Then $\sum_{t=1}^\infty Z_{t}$ converges almost surely and

\begin{equation} E\left(\sum_{t=1}^{\infty} Z_{t}\right) = \sum_{t=1}^{\infty} E(Z_{t}) < \infty \end{equation}

Thanks.

mlofton
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1 Answers1

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If $Z_t\ge 0$ are non-negative, then the sum and expectation may be interchanged by the monotone convergence theorem (since the partial sums form a non-decreasing sequence). More generally, let $Y = \sum_{t=1}^\infty |Z_t|$ and $S_t = \sum_{i=1}^t Z_i$. By triangle inequality, $|S_t|\le Y$ and by assumption, $\mathbb E[Y]<\infty$, so by dominated convergence, $\sum_{i=1}^t\mathbb E[Z_i]\to \mathbb E[\sum_{i=1}^\infty Z_i]$ as $t\to\infty$, and thus $$\mathbb E\left[\sum_{i=1}^\infty Z_i\right]=\sum_{i=1}^\infty\mathbb E[Z_i] \le \mathbb E[Y]<\infty.$$

J. Soloff
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