Bounds on tails with moments

A sort of continuation of http://mathoverflow.net/questions/96165/comparing-distributions-with-moments

Suppose I have some estimates of the moments of a non-negative random variable $X$: $$\log \mathbb{E}(X^n) = n \log n + (\beta-1)n + O(\log n),$$ with $\beta > 0$. Can I conclude that the tail $$\mathbb{P}(X>x) \sim A e^{-c x}$$ for some $A$ and $c$?

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@MichaelRenardy: a thin peak would effect the pdf but not the cdf? Or is there something obvious that I'm just not seeing? The way that I've now set up the problem, it seems like even if $X$ was discrete something like this should still hold? – genneth May 9 2012 at 3:54
I don't know the answer to your question, but there are indeed circumstances where a distribution is asymptotically determined by its high moments. See for example the case of normal distributions in this paper of Gao and Wormald: springerlink.com/content/hd9v0fehfjrknc3a . Their situation isn't quite the same as yours but it may give you ideas. – Brendan McKay May 9 2012 at 5:31

You need a weaker conjecture, since the cdf $(1+x)e^{-x}$ for $x\ge 0$ has moments of the desired form yet it isn't like $1-Ae^{-cx}$. More generally, the distribution with cdf $(1+x^k)e^{-x}$ has moments with logarithm $$n \ln n - n + (1/2+k)\ln n + O(1) .$$ Try cdf $\exp(-x+x^{1/2})$.
 In jstor.org/stable/10.2307/171802 (section 8) it is noted that if $\mathbb{P}(X>x) \approx A e^{-\eta x}$ then $\mu_n = \mathbb{E}(X^n) \approx A\eta^{-n}n!$. In particular, there exists limits of $A_n = \eta^n_n \mu_n / n!$ and $\eta_n = n\mu_{n-1}/\mu_n$, which gives the constants $A$ and $\eta$. They then say "we will assume that if the limits $A$ and $\eta$ exist that also implies that the associated distributions have exponential tails." but I'm not entirely sure what their definition of exponential tails actually is. (cont.) – genneth May 9 2012 at 19:31 (cont.) Looking up the references, it seems to be that they believe (but again, without proof?) that \$\lim_{x \rightarrow \infty} e^{-\eta x} \mathbb{P}(X>x) = a}. I can't quite tell if their conditions are stronger or if their conclusion is weaker. In any case, I think this problem is sufficiently solved/conjectured for me. – genneth May 9 2012 at 19:33