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Let $X_1, X_2, . . .$ be independent, $L(a)$-distributed random variables, and let $N \in Po(m)$ be independent of $X_1, X_2, . . . . $

Determine the limit distribution of $S_N = X_1 + X_2 + · · · + X_N$ (where $S_0 = 0)$ as $m \to \infty$ and $a \to 0$ in such a way that $m · a^2 \to 1$.

I think that using the moment generating might be the key. I did try to look for $\lim_{n \to \infty}M_{S_N}(t)$ and condition on N being a Poisson.

$$\lim_{n \to \infty}M_{S_N}(t)=\lim_{n \to \infty}E(e^{tS_N})=\lim_{n \to \infty}EE(e^{tS_N}|N)= \lim_{n \to \infty}P(N=n)E(e^{tS_n})=\lim_{n \to \infty} \frac{m^ne^{-m}}{n!}(\frac{a^2}{a^2-t^2})^n=\lim_{n \to \infty} \frac{(ma^2)^ne^{-m}}{n!(a^2-t^2)^n}=e^{-m}/n!(a^2-t^2)^n $$

Any hint would be appreciated.

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    $\begingroup$ Maybe helpful to recall that $M_{S_N}(t) = M_N (\ln M_X(t))$. And $M_N(z) = e^{m(e^z-1)}$ since $N$ is $\textsf{Poisson}(m)$. What distribution is $L(a)$ referring to? $\endgroup$ – Minus One-Twelfth Apr 12 at 23:12
  • $\begingroup$ I believe L(a) is a Laplace distribution with param a $\endgroup$ – Mahamad A. Kanouté Apr 12 at 23:17
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    $\begingroup$ Ah OK. So its moment generating function (using the formula at Wikipedia) is $M_X(t) = \frac{1}{1-a^2 t^2}$ for $|t|< 1/a$. So $$M_{S_N}(t) = e^{m(M_X(t)-1)} = e^{m\cdot \frac{a^2t^2}{1-a^2t^2}}.$$ Try taking the limits now. $\endgroup$ – Minus One-Twelfth Apr 12 at 23:22
  • $\begingroup$ The book answers claim that it should be a $N(0,2)$ but the answer I keep getting contain a "a" I see that you term $ma^2 \to to 1$ as $n \to \infty$ you'd still be left with $e^{\frac{t^2}{1-a^2t^2}}$ $\endgroup$ – Mahamad A. Kanouté Apr 12 at 23:26
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    $\begingroup$ The $a^2t^2$ in the denominator would become $0$ because we are taking $a\to 0$ also. $\endgroup$ – Minus One-Twelfth Apr 12 at 23:50

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