# Determine the limit distribution of $S_N = X_1 + X_2 + · · · + X_N$

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.

• 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? – Minus One-Twelfth Apr 12 at 23:12
• I believe L(a) is a Laplace distribution with param a – Mahamad A. Kanouté Apr 12 at 23:17
• 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. – Minus One-Twelfth Apr 12 at 23:22
• 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 $e^{\frac{t^2}{1-a^2t^2}}$ – Mahamad A. Kanouté Apr 12 at 23:26
• The $a^2t^2$ in the denominator would become $0$ because we are taking $a\to 0$ also. – Minus One-Twelfth Apr 12 at 23:50