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.

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I guess, the following reference may be useful for you, it's also pretty recent: An Introduction to Heavy-Tailed and Subexponential Distributions, 2011. Perhaps, available in Russian as well. Furthermore, this paper provides untight rates of convergence which hold however even in case you only have first two moments –  Ilya Jan 19 at 11:44
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 $\sum_j e^{xe^{(2 \pi i j)/k}}/k$
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).