Existence of a random variable $X$ such that the moment generating function of $X$ is given by $\exp(t^3c)$ for some number $c$?

Question

We know that for a normal random variable $N(0,a^{2})$ the moment generating function (MGF) $E \exp(tX)$ is given by $\exp(t^2a^2/2)$. I am trying to find a random variable $X$ such that the MGF of $X$ is given by $\exp(t^3 c)$ for some number $c$, and more generally $\exp(t^{n} c)$ for an integer $n \geq 3$ and a number $c$.

Answer 1

J. Marcinkiewicz derived conditions under which functions of a certain form can be characteristic functions. Among them was :

If the moment generating function of a random variable $X$ is the exponential of a polynomial $P$ i.e. $E[e^{tX}] = e^{P(t)}$, then $P$ has degree at most two and $X$ is a normal random variable.

Therefore, there are no random variables with MG function $e^{t^3c}$ or $e^{t^n c}$ for $n > 2$, in fact far more cases get excluded thanks to the above theorem. (Note that ditto conditions carry over for the MGF)

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As it turns out, thanks to Bochner's theorem the condition for a function to be a characteristic function (generalization of MGF) is only positive definiteness, continuity at the origin, and being $1$ at the origin. These conditions hold for $e^{P(t)}$ if $P(0) = 0$, so only positive definiteness has to be checked, and it turns out that this is the condition violated if the power of $t$ is higher than $2$.

Answer 2

There is a simple argument to show that you cannot have powers higher than $2$ in the exponent. Suppose $Ee^{tX}=e^{ct^{n}}$ with $ n >2$. A simple calculation shows that the second derivative of $e^{ct^{n}}$ at $0$ is $0$. But this means $EX^{2}=0$ so $X=0$ a.s. and $Ee^{tX}\equiv 1 $.