For what range of $\beta$ is random walk heavy-tailed?

Question

I have $\beta > 0$ and $S_0 = 0$, and $S_n = \varepsilon_1 + \cdots + \varepsilon_n, n \geq 1$, a random walk with i.i.d increments $\lbrace \varepsilon_n \rbrace$ having a common distribution $$ P(\varepsilon_1 = -1) = 1 - C_{\beta} \text{ and } P (\varepsilon_1 > t) = C_{\beta}e^{-t^{\beta}}, t \geq 0,$$ where $C_{\beta} \in (0,1)$ such that $\mathbb{E}\varepsilon_1 = -1/2$.

I want to find the range of values of $\beta$ for which this random walk is heavy-tailed.

So far I have tried using the fact that a heavy-tailed distribution has infinite moment-generating function, as follows:

\begin{align*} \varphi(t) = \mathbb{E}e^{t \varepsilon} = e^{-t}(1-C_\beta) + \int_0^\infty x (1-C_\beta e^{-t^\beta}) \, dx \end{align*} but that route doesn't seem to lead anywhere, given that the integral itself is always infinite. I have the same problem when trying to determine the value of $C_\beta$ by using fact that the expectation of $\varepsilon_1$ is $-1/2$.

Can anybody see what I'm doing wrong? Or have any ideas for me to try?

Answer 1

I used a different definition of heavy-tailedness: If a distribution is heavy-tailed, then $\lim_{t \to \infty} e^{\lambda t} \overline{F}(t) = \infty, \quad \forall \lambda >0$.

Therefore \begin{align*} \lim_{t \to \infty} e^{\lambda t} C_{\beta} e^{-t^{\beta}} = \infty \implies \lambda t - t^{\beta} >0 \implies \lambda > t^{\beta-1} \implies \beta - 1 < \frac{\log \lambda}{\log t} \end{align*}

For large $t$ and fixed $\lambda$, $\frac{\log \lambda}{\log t} \approx 0$, and we therefore have that \begin{align*} \lim_{t \to \infty} e^{\lambda t} C_{\beta} e^{-t^{\beta}} = \infty \implies \beta < 1 \end{align*}

Therefore the distribution is heavy-tailed if $\beta \in (0,1)$.