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I have a question related to this post: Expected value of infinite sum

Is the condition listed necessary/sufficient (or both?)

For instance, I'm thinking of $X_n=\frac{1}{n}Z_n$, where $Z_n \sim$N(0,1) are iid. It "feels like" $E\sum_{n=1}^\infty X_n$=0, but $\sum_{n=1}^\infty E|X_n|$ =$\sum_{n=1}^\infty \frac{1}{n}E|Z_n|$ =$E|Z_1|\sum_{n=1}^\infty \frac{1}{n}=\infty$. Thus, we can't use the result listed above to justify interchanging expectation and summation. But I was wondering if my feeling is right.

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Jason
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1 Answers1

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The condition provided is sufficient. See Fubini Theorem and scroll down to Fubini-Tonelli Theorem. This states that says:

If $\mathbb{E}\Big[\sum\limits_{i=1}^{\infty}\big|X_i\big|\Big] < \infty$ or $\sum\limits_{i=1}^{\infty} \mathbb{E}[|X_i|] < \infty$ then we may apply Fubini Theorem and compute the double integral using iterated integrals.

As a note, we may use Fubini-Tonelli because probability measures are $\sigma-$finite and the counting measure on $\mathbb{N}$ is $\sigma-$finite.

Consider what happens if, $$X_n = \left\{ \begin{array}{ll} (-1)^{n}\frac{1}{n} & \text{with probability} \; \frac{1}{2}\\ 0 & \text{with probability} \; \frac{1}{2} \end{array} \right.$$

Both $\mathbb{E}\Big[\sum\limits_{n=1}^{\infty}\big|X_n\big|\Big] = \infty$ and $\sum\limits_{n=1}^{\infty} \mathbb{E}[|X_n|] =\infty$ the second is an easy series result while the first can be shown using Borel-Cantelli Lemma. However, the Fubini Theorem still holds for this example.

J. Stewart
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  • Thanks for sharing that. If I may, the author also initially asked if its "feeling" was right i.e. if the infinite sum of normal variables (0,1) iid was 0. How shall we show that in such a case? – yeahman269 Mar 11 '22 at 15:12