# Let $\{X_i\}_{i=1}^{\infty}$ be a sequence of iid random variables. If $P(X_i\geq0)=1$ and $P(X_i\neq0)>0$ for all $i$, then $\sum X_i=\infty$ a.s.

I came across a proposition, which is stated without proof. It appears in a section on the 0-1 Laws.

Proposition. Let $$\{X_i\}_{i=1}^{\infty}$$ be a sequence of iid random variables on some probability space. If $$P(X_i \geq 0) = 1$$ and $$P(X_i \neq 0) > 0$$ for all $$i$$, then the sum of the $$X_i$$'s is $$\infty$$ a.s. (almost surely).

This doesn't quite seem trivial to me. So, I try to prove it:

Proof (so far): Let $$S = \sum_{i=1}^{\infty} X_i$$. If $$P(S = \infty) = 1$$, it follows that $$S = \infty$$ a.s. So, the problem boils down to showing that:

1. $$S \in \mathcal{T}$$, where $$\mathcal{T}$$ is the tail $$\sigma$$-field, and $$S$$ is considered a tail event.
2. Any $$A \in \mathcal{T}$$ has probability $$P(A) \in \{0, 1\}$$. We need to show that specifically $$P(S = \infty) = 1$$.

Since $$\{X_i\}_{i=1}^{\infty}$$ iid, by Kolmogorov's 0-1 Law, there exists a $$\mathcal{T}$$ defined as in (1). Not sure how to show that $$S$$ is a tail event, and that $$P(S = \infty) = 1$$? Can we use the fact that $$P(X_i \neq 0) > 0$$ and $$P(X_i \neq 0) > 0$$ in some way to conclude this? I don't see how.

The fact that $$(S=\infty)$$ is tail event follows from the fact that for any $$N$$, $$S=\infty$$ iff $$\sum\limits_{k=N}^{\infty} X_i=\infty$$ and the event $$\sum\limits_{k=N}^{\infty} X_i=\infty$$ belongs to $$\sigma (X_N,X_{N+1},...)$$.

One way of showing that $$S=\infty$$ a.s is to show that $$Ee^{-S}=0$$.
Note that $$Ee^{-S}=\prod Ee^{-X_i}=\lim_N (Ee^{-X_1})^{N}$$. Can you use the hypothesis $$P(X_i \neq 0 ) >0$$ to check that $$Ee^{-X_1}<1$$ (which shows that $$\lim_N (Ee^{-X_1})^{N}=0$$)?

Maybe it's worth discussing what it means intuitively for $$S = \infty$$ to be a tail event. $$S = \infty$$ is the event that the series $$\sum_{i = 1}^\infty X_i$$ diverges to positive infinity.

• If I tell you the values of $$X_1, X_2, X_3, X_4 \dots$$, it's possible for you to determine whether $$\sum_{i = 1}^\infty X_i$$ diverges to positive infinity. (i.e. $$S = \infty$$ is in the sigma algebra $$\sigma(X_1, X_2, X_3, X_4, \dots)$$).
• If I only tell you the values of $$X_2, X_3, X_4, \dots$$, it's still possible for you to determine whether $$\sum_{i = 1}^\infty X_i$$ diverges to positive infinity. (i.e. $$S = \infty$$ is in the sigma algebra $$\sigma(X_2, X_3, X_4, \dots)$$).
• If I only tell you the values of $$X_3, X_4, \dots$$, it's still possible for you to determine whether $$\sum_{i = 1}^\infty X_i$$ diverges to positive infinity. (i.e. $$S = \infty$$ is in the sigma algebra $$\sigma(X_3, X_4, \dots)$$).

And so on. Since the event $$S = \infty$$ is in all of the sigma algebras of the form $$\sigma(X_n , X_{n+ 1}, \dots)$$, for all $$n \in \mathbb N$$, the event $$S = \infty$$ is in the tail sigma algebra.

As for showing that $$P(S = \infty)$$ cannot be zero, the method that sprang to my mind is more pedestrian than the one suggested by Kavi Rama Murthy.

First, observe that since $$P(X_i > 0) > 0$$, there exists a $$c > 0$$ and a $$p > 0$$ such that $$P(X_i \geq c ) = p$$, for each $$i$$. [If this wasn't true, then $$P(X_i \geq \tfrac 1 n ) = 0$$ for all $$n \in \mathbb N$$, hence $$P(X_i > 0) = P(\cup_{n \in \mathbb N} \{X_i \geq \tfrac 1 n\}) = 0$$, a contradiction.]

Therefore, the probability that all but finitely many of the $$X_i$$'s are less than $$c$$ is $$0$$, since there are countable many ways of choosing a finite subset of the $$X_i$$'s, and for each choice, the probability that all of the $$X_i$$'s outside of our finitely chosen subset are less than $$c$$ is $$\lim_{k\to\infty} (1-p)^k = 0$$. This implies that $$S=\infty$$ with probability $$1$$.

Although now that I have written this answer, I realise that I've shown that $$P(S = \infty) = 1$$ rather than that $$P(S = \infty) > 0$$, i.e. I've proven the statement directly without appealing to the Kolmogorov zero-one law. Never mind.

• No worries on directly appealing, I nonetheless found this answer very helpful, and I am sure others who come across this will appreciate it as well! I find it unfortunate I am not be able to accept two solutions, otherwise I would. – EzioBosso Sep 27 at 10:26