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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.)2011-11-10
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    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?2011-11-10
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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.2011-11-10
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    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.2011-11-10
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    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.2011-11-10
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    @Bullmoose You should be talking about characteristic function, not the moment generating function, this does not exists if $\mathbb{E}(\vert X \vert^k ) = \infty$ for all $k>n$.2011-11-10

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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.2011-11-10
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    @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. :)2011-11-10
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    @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.2011-11-10
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    @Bullmoose: I was referring to the following statement of yours in [this question](http://math.stackexchange.com/questions/80674/physical-meaning-of-higher-moments-their-values-and-their-existence): *I've heard somewhere that all finite moments of A with support R means that the tails of A decay exponentially. Is that true?*2011-11-10
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    Ahh, right. And that statement was shown to not be true. :)2011-11-11