Probability distribution from moment-generating function?

Question

The moment-generating function of $X\sim\chi^2(\nu)$ is given by $$M_X(t)=(1-2t)^{-\nu/2}$$ for all $t<1/2$. For $\chi^2$ distribution, the degree of freedom $\nu$ is positive. Is it possible that the moment-generating function of some distribution looks like $M_X(t)$ but with a negative $\nu$? My guess is no, but how can I prove it?

Answer 1

Let's try. Say we have a random variable $Y$ such that $$M_Y(t) = 1-2t.$$ What does this say about $Y$? Well, we know that $$M'_Y(0) = \operatorname{E}[Y],$$ so $$\operatorname{E}[Y] = -2.$$ We also know that $$M''_Y(0) = \operatorname{E}[Y^2].$$ But that's $0$, and in fact, all higher moments are $0$. How is this possible? It would also imply $$\operatorname{Var}[Y] = \operatorname{E}[Y^2] - \operatorname{E}[Y]^2 = 0 - (-2)^2 = -4.$$ So no such variable exists.

Can we extend this reasoning further to the general case? Try it: let $$M_Y(t) = (1-2t)^\nu, \quad \nu > 0.$$ (We can ignore the factor of $1/2$ in the exponent.) What happens when you try to calculate $\operatorname{Var}[Y]$?