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

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 $2$.
A: 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 $.
