Let $(X_n)_{n\in\mathbb N}$ be a sequence of non-negative iid random variables with $\mathbb E[X] < \infty$.

How could one go about showing that $\sum^{\infty}_{k=0} e^{X_k} c^k < \infty$ almost surely for some $c \in (0,1)$? I've tried using the Borel-Cantelli lemma but I just can't make it work. Any suggestions?

  • $\begingroup$ Note that if the sum converges for $c_0\in(0,1)$ then it converges on $(0,c_0]$ due to monotonicity. Interesting question though; I'm not sure how to bound $e^{X_k}$ given that $X_k\in L^1$... $\endgroup$
    – Math1000
    Jul 9, 2016 at 23:49

2 Answers 2


The series

$$\sum_{k=0}^{\infty}e^{X_k} c^k = \sum_{k=0}^{\infty} \left( e^{X_k/k} c \right)^k$$

converges almost surely for some $c \in (0,1)$, if there exists a constant $C>0$ such that

$$\limsup_{k \to \infty} e^{X_k/k} \leq C \quad \text{almost surely}. \tag{1}$$ Since, by the non-negativity of the random variables,

$$\frac{X_k}{k} \leq \frac{X_1+\ldots+X_k}{k} =: \frac{S_k}{k}$$

it follows from the strong law of large numbers that

$$\limsup_{k \to \infty} e^{X_k/k} \leq \limsup_{k \to \infty} e^{S_k/k} = e^{\mathbb{E}(X_1)}$$

almost surely, i.e. $(1)$ holds for $C:=e^{\mathbb{E}(X_1)}$.


By the Borel-Cantelli lemma, $X_k/k\to 0$ almost surely, hence for almost every $\omega\in \Omega$, $X_k(\omega)/k\lt 1$ for each $k\geqslant K(\omega)$. This implies $\exp\left(X_k\left(\omega\right)\right)\lt \exp\left(k\right)$ and we and choose $c$ such that $ec\lt 1$.


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