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Given a sequence of non-negative random variables $(X_i)_{i\in\mathbb{N}}$, I would like to show that

$ \mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$

implies that

$ \sum_{i=1}^\infty {X_i} < \infty $

almost surely

The approach that I have in mind is to condition the expectation on the value of $\sum_{i=1}^\infty {X_i}$ as follows:

$\begin{align} \mathbb{E}[\sum_{i=1}^\infty {X_i}] =& \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] + \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \end{align}$

and then say that since

$\mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$

I can then argue that the above conditions imply that

$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] < \infty$

$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] < \infty$

Given that $\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] = \infty $, this must mean that $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] = 0$

Although this explanation makes sense intuitively, it doesn't seem formal enough (in particular it relies on the notion that $0 \times \infty = 0$). Is there a more elegant or formal approach?

4 Answers 4

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This argument is correct. In measure theory, when you allow functions with value $\infty$, the convention is indeed that $0 \times \infty = 0$, while any positive number times $\infty$ is taken to be $\infty$.

Thus, if a random variable is infinite on a set of measure zero, this doesn't contribute to its expected value. But if the function takes the value $\infty$ over a set of positive measure, then its expected value is infinite. (This is just a rephrasing of your argument.)

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Let $Y$ be the sum. You want to show that $\Pr(Y<\infty)=1$ if $E(Y)<\infty$. If $\Pr(Y<\infty)\ne1$, then $E(Y) = \infty\cdot\Pr(Y=\infty)+\text{a non-negative-valued integral over some subset of the space}.$ This is equal to $\infty$. So what you're trying to prove is the contrapositive of that.

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It is a truth that if $X$ is a nonnegative random variable such that $X = \infty$ with positive probability then $\int X \ dP = \infty$. Depending on how you have introduced the Lebesgue integral, this is either part of the definition (when I learned measure-theoretic probability this was the case) or an elementary fact about the integral derived at the outset. Apply this to $\sum_{i= 1} ^ \infty X_i$; if $\sum_{i = 1} ^ \infty X_i$ had positive probability of being $\infty$ then we know immediately that $E(\sum_{i = 1} ^ \infty X_i) = \infty$.

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    @Alex I am saying $\sum X_i = \infty$ with positive probability implies $E(\sum X_i) = \infty$; the contrapositive of this is that $E(\sum X_i) \ne \infty$ implies $\sum X_i \ne \infty$ almost surely, which is what OP wanted.2019-05-16
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As it turns out the approach that I outlined above is not entirely correct. The reason why is because it has conditioned the expectation on the event $\sum_{i=1}^\infty {X_i} = \infty$, which only makes sense if $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$.

A different approach which uses the same idea but avoids this annoying technicality is to use a proof by contradiction, where we assume that

$\mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$

but there exists some $\omega \in \Omega$

$\sum_{i=1}^\infty {X_i}(\omega) = \infty$

so that $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$.

Accordingly, we can now condition the expectation on the value of $\sum_{i=1}^\infty {X_i}$ since $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$ by assumption.

As shown above, this argument then shows that:

$\begin{align} \mathbb{E}[\sum_{i=1}^\infty {X_i}] &= \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] &+ \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \\ &= M + \infty \\ &= \infty \end{align}$

which yields a contradiction and proves the required result.

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    We also don't typically assign probability to points, so $P(\omega | \sum X_i(\omega)= \infty)$ wouldn't be well defined. We might remedy this by considering $\{\omega\}$ instead, but $\{\omega\}$ itself might not even be measurable.2012-12-11