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What are the general results for when it is allowed to take the expectation and variance operator inside an infinite sum involving random variables?

So, when can I say $\operatorname E \sum_{i = 1}^\infty X_i = \sum_{i = 1}^\infty \operatorname E X_i$, and likewise for the variance? What conditions must be satisfied?

For this question, assume $(X_i)$ is always iid, and all infinite sums exist almost surely.

Michael Hardy
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eofwjof
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  • In the i.i.d the case the question is entirely trivial. (See my answer). If ${X_n}$ is independent but not identically distributed then the answer by Michael Hardy is useful. – Kavi Rama Murthy Oct 04 '18 at 00:37

2 Answers2

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$$ \operatorname E\left( \sum_i X_i \right) = \sum_i \operatorname E X_i \text{ ?} \tag 1 $$

  • Line $(1)$ holds if there are only finitely many indices $i.$
  • Line $(1)$ holds if $\Pr(X_i\ge0)=1$ for every value of $i.$
  • Line $(1)$ holds if $\displaystyle\operatorname E \left( \sum_i \left|X_i\right| \right)<+\infty,$ or equivalently if $\displaystyle \sum_i \operatorname E\left| X_i \right|<+\infty.$

The above is true regardless of whether the random variables are i.i.d., but the third one assumes convergence of the infinite sum. But if all of the terms have equal expectation, then of course the sum of expectations diverges to $+\infty$ or $-\infty$ unless the expectation of each term is $0.$

With variances it's more complicated. With finitely many terms in the sum, the sum of the variances equals the variance of the sum if the random variables are uncorrelated. That is a weaker assumption than independence. And it is weaker than pairwise independence. So it holds if they're independent. Since you assumed independence, that's all you need. Maybe I'll add some more on this later$\,\ldots$

Michael Hardy
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If $\{X_n\}$ is i.i.d then $\sum_n X_n$ exists almost surely iff $X_n=0$ almost surely for each $n$.

Kavi Rama Murthy
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