# 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 '14 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 '14 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 at 15:25