Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A sort of continuation of 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 $c$?

share|improve this question
1  
@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 '12 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 '12 at 5:31
add comment

1 Answer

up vote 1 down vote accepted

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 $\exp(-x+x^{1/2})$.

share|improve this answer
    
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 '12 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 '12 at 19:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.