Limit of $S_N = \sum_{i=0}^N X_i$ where $X_i$ is Laplace distributed and $N$ is Poisson distributed. Let $S_N = \sum_{i=0}^N X_i$ where $X_i$ is Laplace distributed with parameter $a$ and independent. Let $N \sim \text{Poi}(m)$ independent of the $X_i$. I want to find the limit distribution of $S_N$, given that when $m \to\infty$ and $a \to 0$ then $m  a^2 \to 1$. 
My attempt is $\text{E}[S_n]=0$ since $X_i$ independent and $\text{E}[X_i]=0$. Furthermore, $\text{Var}[S_n]=2ma^2$ given $N=n$, since $X_i$'s and $N$ independent and $\text{Var}[X_i]=2a^2$. Then I believe Central limit theorem and Slutsky's theorem would fit nicely, however I'm not sure how to proceed due to how the $N,m$ tends to infinity. I also tried with using characteristic functions, but that didnt seem make any more sense.
All help is greatly appreciated. 
 A: The moment generating function of $X_1$ is given by
$$
M_X(t) := \mathbb E\left[ e^{tX}\right] = \int_{-\infty}^\infty e^{tx}\frac1{2a}e^{-\frac{|x|}a}\ \mathsf dx = \frac1{1-a^2t^2},\ |t|<\frac1a
$$
and the moment generating function of $N$ by
$$
M_N(t) := \mathbb E\left[ e^{tN}\right] = \sum_{n=0}^\infty e^{tn} e^{-m}\frac{m^n}{n!} = e^{me^t-1}.
$$
It follows from the law of total expectation that
\begin{align}
M_{S_N}(t) &= \mathbb E\left[e^{tS_N} \right]\\
&= \mathbb E\left[e^{t\sum_{i=0}^N X_i} \right]\\\\
&= \mathbb E\left[\prod_{i=0}^N e^{tX_i} \right]\\
&= \mathbb E\left[\left(e^{tX_1}\right)^N \right]\\
&= \sum_{n=0}^\infty\mathbb E\left[\left(e^{tX_1}\right)^n\mid N = n \right]\mathbb P(N=n)\\
&\sum_{n=0}^\infty \left(\frac1{1-a^2t^2}\right)^ne^{-m}\frac{m^n}{n!}\\
&= e^{-m} \sum_{n=0}^\infty \left(\frac m{1-a^2t^2} \right)^m/n!\\\\
&= \exp\left(\frac{ma^2t^2}{1-a^2t^2}\right).\tag1
\end{align}
As $m\to\infty$ and $a\to0$ we have $ma^2\to1$ and hence $(1)$ has limit $e^{t^2}$. This is the moment-generating function of a normal distribution with mean zero and variance $2$, and so we conclude that $S_N$ converges in distribution to $\mathcal N(0,2)$.
