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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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    @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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    Ahh, right. And that statement was shown to not be true. :)2011-11-11