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This is a follow-up to this previous question.

Suppose I have a mean-zero symmetrically-distributed random variable $X$ over the support $\mathbb{R}$. If $X$ has a moment-generating function $M_X(t)$ that is smooth around 0, $X$ has an exponentially decaying tail probability, by Chernoff bound (Lemma 11.9.1 in Cover and Thomas's "Elements of Information Theory" 2nd edition).

Now, suppose that $X$ has an $M_X(t)$ that is not smooth around 0. Suppose that $\mathbf{E}[X^k]=\infty$ for all even $k>n$, where $n$ is a positive integer. Is there $X$ that has exponentially-decaying tail probability in that case? Or would the tail probability always be a power-law?

Also, what happens to the tail if $M_X(t)$ is not defined, i.e. the integral in the transform diverges?

EDITS: Clarified the question based on helpful comments from @cardinal.

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A somewhat pedantic response to your question is that no such random variable $X$ can exist in the first place based on the set of conditions you've placed on it. Do you see why? (Hint: Consider $k > n$ where $k$ is odd.) –  cardinal Nov 10 '11 at 16:42
    
Hmmm... I see what you are saying. But does this mean that any symmetric zero-mean $X$ has to have all finite moments? Or is there a symmetric zero-mean $X$ that has all the moments but for which $M_X(t)$ is not smooth around 0? Which condition should I weaken? –  M.B.M. Nov 10 '11 at 17:07
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No, quite the opposite. A symmetric distribution about zero need not have any (raw) odd moments at all, but all (raw) even moments will exist, even if they are not finite. –  cardinal Nov 10 '11 at 17:16
    
I think I am confused about what it means by "not having moment $\mu_n$". I interpret that only as "$\mu_n=\infty$". That is, my interpretation of "having moments" means "having finite momemts, possibly equal to zero". Would removing "Suppose that $\mathbf{E}[X_k]=\infty$ for all $k>n$, where $n$ is a positive integer" put more sense into my question? I'm mainly interested in what happens for $X$ with $M_X(t)$ that is not smooth around zero. –  M.B.M. Nov 10 '11 at 17:26
    
And I was confused by your first comment because Student's t has finite moments up to its degree of freedom and further even moments are infinite or the integral in the mgf diverges for odd ones. Anyway, my bad -- I've edited my question. –  M.B.M. Nov 10 '11 at 17:33

1 Answer 1

up vote 5 down vote accepted

I assume that "exponentially decaying tail probability" means that $P(|X| > t) \le C e^{-\epsilon t}$ for some $C, \epsilon$. Any such random variable has finite moments of all orders. This follows from the formula $$E[|X|^p] = \int_0^\infty p t^{p-1} P(|X| > t) dt$$ which you can prove with Fubini's theorem and the fundamental theorem of calculus.

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Thanks! That's the result I've been looking for. –  M.B.M. Nov 10 '11 at 18:54
    
@Bullmoose: Note that this is the converse of the statement you originally quoted in another question. So, just be sure it actually is the result you were looking for. :) –  cardinal Nov 10 '11 at 22:04
    
@cardinal Basically, I wanted to know whether not having finite moments of all orders implies that the tail probabilities do not decay exponentially. I think this answer answers it (as a contrapositive.) The statement I made in the comment to the other question states that if MGF is smooth around 0 then its tails decay exponentially, and it has all finite moments. I understand that having finite moments of all orders doesn't necessarily result in having exponential tails (the lognormal example), but for my problem it suffices to state that having a smooth MGF implies moments and exp-tails. –  M.B.M. Nov 10 '11 at 22:49
    
@Bullmoose: I was referring to the following statement of yours in this question: I've heard somewhere that all finite moments of A with support R means that the tails of A decay exponentially. Is that true? –  cardinal Nov 10 '11 at 23:20
    
Ahh, right. And that statement was shown to not be true. :) –  M.B.M. Nov 11 '11 at 0:03

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