What distribution has MGF=0?

Question

I found a MGF with 2 variables, n and t, to do with a transformation to an average of a sum of independent random variables following the same distribution. The limit of this MGF as n approaches infinity is equal to 0 and I am wondering what distribution this follows? Edit: I'm wondering about the limiting distribution of Zn

$f(x)=0.75*0.25^{x-1} , x=1, 2, 3, ...$

$Y_n=\sum_{i=1}^{n}{X_i}$

$Z_n=\frac{3}{2}\sqrt{n}\bar{Y_n}-2\sqrt{n}$

$M_\bar{Y_n}(t)=[M_{Y_n}(\frac{t}{n})]^n=[M_{X_n}(\frac{t}{n})]^{n^2}=[\frac{0.75e^\frac{t}{n}}{1-0.25e^\frac{t}{n}}]^{n^2}$

$M_{Z_n}(t)=E(e^{tz})=E(e^{t(\frac{3}{2}\sqrt{n}\bar{Y_n}-2\sqrt{n})})=E(e^{\frac{3t}{2}\sqrt{n}\bar{Y_n}}e^{-2t\sqrt{n}})=e^{-2t\sqrt{n}}E(e^{\frac{3t}{2}\sqrt{n}\bar{Y_n}})=e^{-2t\sqrt{n}}M_{\bar{Y_n}}(\frac{3t\sqrt{n}}{2})$

$\lim_{n->\infty}(M_{Z_n}(t))=\frac{1}{\infty}=0$

Answer 1

Your limit is wrong. $X_i$ has geometric distribution with mean $1/(1-p)=4/3$, variance $(1-p)/p^2$ which by CLT implies that $\sqrt{n}\bar{Y}_n-\sqrt{n}/(1-p)$ approaches a normal random variable with mean 0 and variance $(1-p)/p^2$

$Z_i$ has mean $(3/2)\sqrt{n}(4/3)-2\sqrt{n}=0$, and should also correspondingly also approach a normal random variable via CLT.

Answer 2

There does not exist an MGF that is identically equal to zero:

Let $f:D\to\mathbb{R}$, then for $f$ to be a valid pdf then $\int_Df(x)\mathrm{d}x=1$ and $f(x)\geq0~\forall~x\in D$.

For the MGF to be identically zero then $\int_{D}e^{tx}f(x)\mathrm{d}x$ would have to be 0 for all values of $t$. However, for $t=0$ we get $$\int_De^{0 x}f(x)\mathrm{d}x=1\neq0$$.

This also holds in the discrete case:

$$\sum_{x\in D}e^{tx}p(x)\leadsto\sum_{x\in D}e^{0 x}p(x)=1$$

Edit: I now realize that I'm answering your title and not your actual question. Sorry about that, heh.