# Non-existence of moment generating function

If moment generating function $E[e^{tX}]$ exists only for $t=0$, does it imply that every moment of $X$ does not exist(is not finite)?

• Is this work for some subject? homework, for example? If not, how do you come to face such a question? – Glen_b Sep 9 '17 at 8:21
• This is homework and if this would be true then I could prove that no moments of Cauchy distribution exist because its mgf doesn't exist. – Ethan Sep 9 '17 at 9:06
• 1. So how does the question arise? Is it work for some subject other than homework? 2. It's unclear why you keep referring to the Cauchy; to answer a question "does having attribute A imply B" you don't examine some examples of B and see if they have attribute A. You need to know there are no examples of not-B which have attribute A. The Cauchy is logically useless for answering the question you asked, it's an example for which the proposition is true, which doesn't help you figure out whether all cases are true; for that you need to show there are no counterexamples. But there are! – Glen_b Sep 9 '17 at 9:38
• This is why my answer discusses the $t$ and the lognormal. Either is sufficient, since they're counterexamples to the proposition. If you want to prove no moments exist for the Cauchy, try it for the first moment (which is easy). That implies all the others. – Glen_b Sep 9 '17 at 9:40

## 2 Answers

No.

Consider a distribution whose first $$k$$ moments are finite (perhaps a $$t_{k+1}$$-distribution) but whose higher order moments aren't. Does its MGF exist in a neighborhood of 0? (See if you can show it)

Indeed, even if every moment exists, it can be that the MGF doesn't exist in a neighborhood of 0. The lognormal is a commonly given example.

• Well, but the OP said "exists only for $t=0$", not an open interval about zero. Are there counterexamples also for that case? There cannot be, because mgf always exist for $t=0$ (and is equal to 1 there). Not very useful, yes. – kjetil b halvorsen Sep 9 '17 at 14:16
• @kjetil Wow, I could have sworn I saw a $(-\epsilon,\epsilon)$ in the question before I answered. Turns out it was another question I was looking at on the same day. – Glen_b Sep 10 '17 at 1:17

Again, let us use the lognormal as example. Let $$X, Y$$ be two iid lognormal variables. Let $$D = X - Y$$. Then all moments of $$D$$ exists (they can be calculated from the lognormal moments), but the mgf of $$D$$ only exists for $$t=0$$.

Some details here: Difference of two i.i.d. lognormal random variables