# 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?

This is an exercise from Achim Klenke's probability book, and it appear previous to the chapter about the laws of large numbers, so we need to handle it without this.

As $$X_1$$ is non-negative then from the previous chapter of the book we knows that $$\mathrm{E}X_1= \int_{[0,\infty )}\Pr [X_1\geqslant t] \mathop{}\!d t\tag{*}$$ Then for any chosen $$\epsilon >0$$ we have that \begin{align*} \epsilon \Pr [X_1\geqslant (n+1) \epsilon ]\leqslant \int_{n \epsilon }^{(n+1) \epsilon }\Pr [X_1\geqslant t] \mathop{}\!dt \leqslant \epsilon \Pr [X_1\geqslant n \epsilon ]\\ \therefore\quad \epsilon\sum_{n\geqslant 0}\Pr [X_1\geqslant (n+1)\epsilon ]\leqslant \mathrm{E}X_1\leqslant \epsilon \sum_{n\geqslant 0}\Pr [X_1\geqslant n \epsilon ] \end{align*}\tag1

Now we can compare the value of $$e^{X_k}$$ with $$c^k$$, that is, if $$\Pr [e^{X_k}\geqslant c^{-k} \text{ i.o. }]=\Pr [X_k\geqslant k\log(c^{-1})\text{ i.o. }]=1\tag2$$ for any chosen $$c\in(0,1)$$ then this would imply that $$\sum_{k\geqslant 1}e^{X_k}c^k=\infty$$ almost sure. Therefore if $$\mathrm{E}X_1=\infty$$ then from the Borel-Cantelli lemma and $$(1)$$ the conclusion follows.

Now, to prove the other assertion it would be enough to show that $$\mathrm{E}X_1<\infty \implies \Pr [e^{X_k}< c^{-k/2}\text{ eventually }]=1\tag3$$ However the last condition is equivalent to $$\Pr [X_k\geqslant k\log(c^{-1/2})\text{ i.o. }]=0$$, and this follows immediately again from $$(1)$$ and the Borel-Cantelli lemma. $$\Box$$

• Why does your proof for $E[X_1]<\infty$ have $e^{X_k}<c^{-k/2}$? Dec 4 at 20:25

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. Nov 18, 2018 at 21:00
• @SangchulLee Yes of course; thanks for your comment.
– saz
Nov 18, 2018 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? Jan 21, 2020 at 13:20