Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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
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.

share|cite|improve this answer
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

Your Answer


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.