I will just recapitulate the complete problem first, then show my solution.
Problem
$\{X_{n}\}_{n\geq1}$ is a sequence of independent r.v.'s, $X_{n}\in\text{Po}(\mu)$ for each $n$. $N$ is independent of $\{X_{n}\}_{n\geq1}$, and $N\in\text{Po}(\lambda)$. Set
\begin{align}
S_{N} = X_{1}+X_{2}+\dots+X_{N},\quad S_{0}=0.
\end{align}
Let now $\lambda\rightarrow \infty$, while at the same time $\mu\rightarrow 0$, so that $\mu\lambda\rightarrow \upsilon>0$. Show that
\begin{align}
S_{N}\stackrel{d}{\rightarrow} \text{Po}(\upsilon).
\end{align}
My solution is this, to show convergence in distribution of the sum to $\text{Po}(\upsilon)$ we show that the characteristic function associated with $S_{N}$, i.e. $\varphi_{S_{N}}(t)$, converges to
\begin{align}
\exp{\left\{\upsilon(e^{it}-1)\right\}},\quad \text{as}\; \lambda\rightarrow\infty\;\text{and}\;\mu\rightarrow0.
\end{align}
The characteristic function of $X_{n}$ for each $n$ is
\begin{align}
\varphi_{X_{n}}(t) = \exp{\left\{\mu(e^{it}-1)\right\}},
\end{align}
and the characteristic function of $S_{N}$ then becomes (for $N>0$)
\begin{align}
\varphi_{S_{N}}(t) &= \mathbb{E}\left[e^{itS_{N}}\right] = \{\text{Law of total expectation}\} = \mathbb{E}\left[\mathbb{E}\left[e^{itS_{N}}\vert N\right]\right] = \mathbb{E}\left[\varphi_{X_{n}}^{N}(t)\right] = \{\text{Law of total expectation}\}\\
&= \sum_{k=0}^{\infty}\mathbb{E}\left[\varphi_{X_{n}}^{N}(t)\big\vert N=k\right]p_{N}(k) = \sum_{k=0}^{\infty}\varphi_{X_{n}}^{k}(t)e^{-\lambda}\frac{\lambda^{k}}{k!} = e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda\varphi_{X_{n}}(t))^{k}}{k!} = e^{-\lambda}e^{\lambda\varphi_{X_{n}}(t)} = e^{\lambda(\varphi_{X_{n}}(t)-1)}.
\end{align}
We may expand this expression a bit and obtain
\begin{align}
\varphi_{S_{N}}(t) = \exp{\left\{\lambda\left(\exp{\left\{\mu\right(e^{it}-1\left)\right\}}-1\right)\right\}}.
\end{align}
But how do I find the limit of this expression? Maybe this is wrong, but I don't see the product $\mu\lambda$ appearing anywhere. However, if $\mu\rightarrow 0$, then $\exp{\left\{\mu\right(e^{it}-1\left)\right\}}-1\rightarrow 0$, while on the other hand when $\lambda\rightarrow \infty$ the expression $\lambda\left(\exp{\left\{\mu\right(e^{it}-1\left)\right\}}-1\right)$ grows. But what is the limit?
Best Answer
Start from \begin{align} \varphi_{S_{N}}(t) = \exp{\left\{\lambda\left(\exp{\left\{\mu\right(e^{it}-1\left)\right\}}-1\right)\right\}}= \exp{\left\{\mu\lambda\frac{1}{\mu}\left(\exp{\left\{\mu\right(e^{it}-1\left)\right\}}-1\right)\right\}}. \end{align} As $\mu$ goes to $0$, $\frac{1}{\mu}\left(\exp{\left\{\mu\right(e^{it}-1\left)\right\}}-1\right)\to e^{it}-1$ and $\mu \lambda \to\nu$.