# Prove long-tailed distribution is heavy-tailed

Consider a sequence of iids $X_is$.

I know a distribution is heavy-tailed if the $E(e^{tX_i}) = \infty$ for all $t>0.$. Additionally a distribution is long-tailed if $\frac{\bar{F}(x+1)}{\bar{F}(x)} \rightarrow 1$ as $x \rightarrow \infty$. But how do I use these two definitions to prove that any long-tailed distribution is heavy-tailed?

Note $F(x)$ is the CDF of my $X_is$ and $\bar{F}(x) = 1- F(x)$. This question is from a chapter regarding the applications of Random Walks.

• What is $\bar F$? The CDF? Why the bar?
– MPW
Feb 22, 2014 at 11:32
• I guess $\bar{F} = 1-F$, the tail probability function. I have never seen those definitions before. Any reference for them? Feb 22, 2014 at 12:35

First of all, note that by Fubini's theorem

$$\mathbb{E}e^{t X} \geq \mathbb{E}(e^{tX} \cdot 1_{\{X \geq 0\}}) = t \int_{[0,\infty)} e^{tx} \cdot \mathbb{P}(X \geq x) \, dx. \tag{1}$$

Fix $\varepsilon>0$. Now, as $\frac{\bar{F}(x+1)}{\bar{F}(x)} \to 1$, we can choose $n \in \mathbb{N}$ sufficiently large such that

$$\frac{\bar{F}(x+1)}{\bar{F}(x)} = \frac{\mathbb{P}(X \geq x+1)}{\mathbb{P}(X \geq x)} \geq 1-\varepsilon$$

for all $x \geq n$. In particular, this entails

$$\mathbb{P}(X \geq n+k) \geq (1-\varepsilon)^k \cdot \mathbb{P}(X \geq n).\tag{2}$$

Combining $(1)$ and $(2)$ yields

\begin{align*} \mathbb{E}e^{tX} &\geq t \int_{[n,\infty)} e^{tx} \mathbb{P}(X \geq x) \, dx \\ &\geq t \, \mathbb{P}(X \geq n) \sum_{k \geq 1} e^{t(n+k)} (1-\varepsilon)^k. \end{align*}

Since $t>0$, we can choose $\varepsilon>0$ sufficiently small such that $e^t \cdot (1-\varepsilon)>1$. Hence,

$$\mathbb{E}e^{tX} = \infty.$$

• I was wondering if you could provide a bit more detail on the second equality in Eq. (1). I don't understand why Fubini's theorem is invoked. Instead, it looks as if one integrates by parts… but that can't be either, since the boundary terms are missing. Aug 4, 2021 at 15:25