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Suppose I have a sequence of positive continuous random variables $\{X_k\}_{k=1}^\infty$ with (unknown) MGF's $M_{X_k}(s)$. Furthermore, it is known that \begin{equation}\frac{X_n-n\mu}{\sqrt{n}\sigma}\rightarrow\mathcal{N}(0,1),\end{equation} for some known $\mu$ and unknown $\sigma$. Given the function \begin{equation}F[z,s]=\sum_{n=0}^\infty z^{-n} M_{X_n}(s),\end{equation} is it possible to extract $\sigma$ without the use of inverse transforms?

For example: \begin{equation}F[z,s]=\frac{zs}{1-e^s+zs}.\end{equation} Answer: $\sigma^2=\frac{1}{12}$.

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If $M_n^*(s)=M_{Z_n}(s)e^{-ns\mu}$ and $G(z,s)=\sum_{n=0}^{\infty}z^{-n}M_n^*(s)=F(ze^{\mu s},s)$ then for a suitable circle $C$ in the complex plane

$$\frac{1}{2i\pi}\int_Cz^{n+1}G(z,s/\sqrt{n})dz=M_n^*(s\sqrt{n})\rightarrow e^{\sigma^2s^2/2}.$$

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