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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?

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up vote 3 down vote accepted

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. – guy Dec 10 '12 at 4:15

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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There is no reason to bring conditioning into this, and the existence of an $\omega$ such that $\sum X_i(\omega) = \infty$ doesn't imply that $P(\sum X_i = \infty) > 0$ anyways. We are also allowed to condition on probability $0$ events, but the theory behind such things is tricky. For nonnegative $X$, the fact that $P(X = \infty) > 0$ implies that $\int X \ dP = \infty$ is one of the first properties of the Lebesgue integral that one proves (e.g. it is immediate if you define the integral by approximations using simple functions), if it isn't part of the definition to begin with. – guy Dec 11 '12 at 2:52
Thanks for this! One question: the complement of $ \sum X_i(\omega) < \infty$ almost surely should be, there exists $\omega \in \Omega$ such that $\mathbb{P}(\omega | \sum X_i(\omega) = \infty) > 0$ - right? I realize that this is not what I wrote, but I'd like to get this straightened out, if possible. – Elements Dec 11 '12 at 19:56
No, that shouldn't be the complement. The probability assigned to $\omega$ might be $0$. Let $\Omega_0$ be the set of $\omega$ such that $\sum X_i(\omega) < \infty$. The statement $\sum X_i < \infty$ almost surely is equivalent to the statement that $\Omega_0 ^ c$ is contained in some measurable set $B$ such that $P(B) = 0$. The negation of this is that, for all measurable sets $B$, if $P(B) > 0$ then $\Omega_0 ^ c$ is not contained in $B$. This is a little involved because we have to deal with the fact that $\Omega_0 ^ c$ might not be measurable. (continued) – guy Dec 11 '12 at 20:42
If $\Omega^c_0$ is measurable, then $\sum X_i < \infty$ almost surely is the statement that $P(\Omega^c_0) = 0$ while the negation is that $P(\Omega^c_0) > 0$. But $\Omega^c_0$ certainty doesn't have to be empty, in either case. In most examples one runs into, $\Omega^c_0$ is probably not empty. For example, let $X_i$ be equal to either $0$ or $1$ and the $X_i$ are independent with $P(X_i = 1) > 0$ but $\sum P(X_i = 1) < \infty$. Then there is an $\omega$ corresponding to $X_i = 1$ for all $i$, but the series still converges almost surely. – guy Dec 11 '12 at 20:46
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. – guy Dec 11 '12 at 20:49

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