I have a problem with a random walk I'm trying to work with.
Suppose I have a random walk
$$S_n = \xi_1 + ...+\xi_n, n \geq 1, S_0= 0$$
with i.i.d. increments $\{\xi_n\}$ with common distribution
$$\mathbf{P}(\xi = -1) = 1 - C_{\beta}$$
$$\mathbf{P}(\xi_1 \gt t) = C_\beta e^{-t^{\beta}}, t \geq 0.$$
Where $\beta \gt 0$ and $C_\beta \in(0,1)$ and $\mathbf{E}\xi_1 = -\frac{1}{2}.$
I'm trying to find the range of values of $\beta$ where the distribution $\xi_1$ is heavy-tailed, i.e. the moment generating function
$$\phi(t) = \mathbf{E}e^{t\xi_1} = \infty.$$
How could I do this?
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*}
$\implies$ heavy tailed in $\beta \in (0,1)$. Do you agree?