# IID random variables $(X_n)$ have $\sum e^{X_n} c^n < \infty$ a.s.

I'm working on the following exercise:

Let $$X_1, X_2, \ldots$$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $$c \in (0,1)$$, $$\sum_{n=1}^\infty e^{X_n} c^n \begin{cases} < \infty \textrm{ a.s.} & \textrm{if } \mathbb E[X_1] < \infty; \\ = \infty \textrm{ a.s.} & \textrm{if } \mathbb E[X_1] = \infty \end{cases}$$

I'm trying to show $$\sum_{n=1}^\infty \mathbb P\left[\sum_{k=1}^n e^{X_k} c^k \geq M\right] < \infty$$ for some large $$M > 0$$. For then, Borel-Cantelli gives us that $$\mathbb P\left[\limsup \left\{ \sum_{k=1}^n e^{X_k} c^k \geq M\right\}\right] = \mathbb P\left[\sum_{k=1}^\infty e^{X_k} c^k \geq M\right] = 0$$ and we're done. But I don't know how to show $$\sum_{n=1}^\infty \mathbb P\left[\sum_{k=1}^n e^{X_k} c^k \geq M\right] < \infty$$. Any suggestions?

If $$\mathbb{E}(X_1)< \infty$$ then it follows from the strong law of large numbers that $$S_n := \sum_{j=1}^n X_j$$ satisfies

$$\lim_{n \to \infty} \frac{S_n}{n} = \mathbb{E}(X_1) \quad \text{a.s.};$$

hence

$$\lim_{n \to \infty} \frac{X_n}{n} = \lim_{n \to \infty} \left( \frac{S_n}{n} - \frac{S_{n-1}}{n} \right)=0 \quad \text{a.s.}$$

Consequently, there exists for almost all $$\omega \in \Omega$$ some $$N \in \mathbb{N}$$ such that $$\left| \frac{X_n(\omega)}{n} \right| \leq -\log(\sqrt{c}) \quad \text{for all n \geq N}$$ for fixed $$c\in (0,1)$$, and so $$\sum_{n \geq N} e^{X_n(\omega)} c^n \leq \sum_{n \geq N} \sqrt{c}^n < \infty.$$

If $$\mathbb{E}(X_1)=\infty$$ then

$$\sum_{n \geq 1} \mathbb{P}(X_n \geq n)=\sum_{n \geq 1} \mathbb{P}(X_1 \geq n) =\infty,$$

and therefore it follows from the Borel Cantelli lemma that $$\mathbb{P}(X_n \geq n \, \, \text{infinitely often})=1,$$ i.e. $$e^{X_n} \geq e^n \quad \text{for infinitely many n with probability 1.}$$ This implies $$\sum_{n \geq 1} e^{X_n} c^n = \infty$$ almost surely for $$c:= 1/e$$.

• For the fact that $\mathbb{E}[X_1] < \infty$ implies $X_n/n \to 0$ a.s., it may as well be proved by Borel-Cantelli: $$\forall \epsilon > 0 \ : \quad \sum_{n=1}^{\infty} \mathbb{P}[X_n > \epsilon n] \leq \frac{1}{\epsilon}\mathbb{E}[X_1] < \infty.$$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach. – Sangchul Lee Nov 18 '18 at 21:00
• @SangchulLee Yes of course; thanks for your comment. – saz Nov 18 '18 at 21:02
• The case $\mathbb{E}[X_1]=\infty$ does not solve the question, he needs it for all $c\in(0,1)$. Couldn't you do an analogous proof to the first part using that $\limsup_{n\rightarrow\infty} \frac{X_n}{n} =\infty$ (which follows for example by Borel-Cantelli and $\mathbb{E}[X_1]=\sum_{n\ge 1}\mathbb{P}[X_1\ge n]$) in this case. Using that you can find an almost surely diverging subsequence $X_{m_n}$ of $X_n$ such that $\exp(X_{m_n}/n)>c$ for all $m_n$ and hence the part of the sum with the indices in the subsequence would diverge thus giving the claim, right? – Yannick Jan 21 at 13:20