2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$\mathbb E\left[\sum_{i=1}^\infty Z_i\right]=\sum_{i=1}^\infty\mathbb E[Z_i] \le \mathbb E[Y]<\infty.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.