# Tails of sums of Weibull random variables

Suppose that $X_1, X_2, \ldots, X_n$ are i.i.d random variables distributed according to Weibull distribution with shape $0 < \epsilon < 1$ (it means that $\mathbf{Pr}[X_i \geq t] = e^{-\Theta(t^{\epsilon})}$).

Now consider the random variable $S_n = X_1 + X_2 + \ldots + X_n$, when $n$ tends to infinity. Clearly, $\mathbf{E}[S_n] = O_{\epsilon}(n)$. Is it true that for some $C = C(\epsilon)$ we have $\mathbf{Pr}[S_n \geq C n] \leq e^{-\Omega_{\epsilon}(n^{\alpha})}$ for some $\alpha = \alpha(\epsilon) > 0$? If so, what is the largest $\alpha$ one can get?

The standard MGF-based methods that work nicely in similar situations are not applicable here due to the fact that $X_i$'s are heavy-tailed. My feeling is that this question must be studied somewhere.

The affirmative answer can be found in this paper by A.V.Nagaev: essentially "the conjecture" is true for $\alpha = \varepsilon$ (which is clearly the best possible).
1. $f(x) = cosh(x^{\alpha})$ is convex for $\alpha > \frac 12$ so 2. $\mathbb E f(\frac {S_n} n) \le \mathbb E f(X)$ which, with chebyshev, implies your claim for any $\frac 12 \le \alpha \le \epsilon$. 3. For general case, replace $cosh$ with $\sum \frac {x^{nk}}{(nk)!}$ where K satisfies $\alpha k > 1$. You have to know that this thing is about $e^x$, which can be gotten from the fact that it is $\sum_j e^{xe^{(2 \pi i j)/k}}/k$