For an independent trial for the random variable X with density $f_X(x)=\frac{e^{|x|}}{2}$. If $S_n = X_1 + ... X_n$, $A_n = S_n/n$, and $S_n^*=\frac{S_n-n\mu}{\sqrt{n\sigma^2}}$, I found the following:
MGF for $X$:
$$g_X(t)= \int_{-\infty}^01/2e^{x+tx}dx+\int_{0}^{\infty}1/2e^{-x-tx}dx$$ $$g_X(t)= 1/2[\frac{1}{1+t}+\frac{1}{1-t}]=\frac{1}{(1+t)(1-t)}$$
MGF for $S_n$:
$$g_{S_n}(t)=(g_X(t))^n=\frac{1}{(1+t)^n(1-t)^n}$$
MGF for $A_n$:
$$g_{A_n}(t)=(g_X(t/n))^n=\frac{1}{(1+t/n)^n(1-t/n)^n}$$
MGF for $S_n^*$
$$g_{S_n^*}(t)=e^{\frac{-n\mu t}{\sqrt{n\sigma^2}}}(g_X(t/\sqrt{n\sigma^2}))^n=\frac{e^{\frac{-n\mu t}{\sqrt{n\sigma^2}}}}{(1+t/\sqrt{n\sigma^2})^n(1-t/\sqrt{n\sigma^2})^n}$$
Are the computations above correct? If so, is each MGF only valid for $|t|<1$ based on the convergence of the integral for $g_X(t)$?
Given $\mu_X=0$ and $\sigma_X^2=2$, I think $S_n^*=\frac{1}{(1+t/\sqrt{2n})^n(1-t/\sqrt{2n})^n}$. If so, does this mean that as $n\to\infty$, both $g_{A_n}(t)$ and $g_{S_n^*}(t)\to 1$? Is there any significance to this?
This is the Laplace distribution with $\mu = 0$ and b = 1, you can check it on wikipedia (especially the "Related Distributions" section of the article): https://en.wikipedia.org/wiki/Laplace_distribution