Let $X_1,X_2,...$ be iid with mean $\mu$ and variance $\sigma^2$. Let $N_{\lambda}$ be poisson($\lambda$) independent of the $X_i$'s.
(a) Find the limit in distribution as $\lambda \rightarrow \infty$ for $$\frac{\sum_{i=1}^{N_{\lambda}}X_i-N_{\lambda}\mu}{\sqrt{\lambda}}.$$
(b) Find the limit in distribution as $\lambda \rightarrow \infty$ for $$\frac{\sum_{i=1}^{N_{\lambda}}X_i-\lambda\mu}{\sqrt{\lambda}}.$$
(c) For which random variables $X$ will the two limits be the same?
My solution: Let $Y_{\lambda}$ be the quotient given in the problem. Then $\mathbb{E}e^{it Y_{\lambda}}=\sum_{n=0}^{\infty}\mathbb{E}(e^{itY_{\lambda}}1_{N_{\lambda}=n})=\sum_{n=0}^{\infty}(\phi_{X'_1}(\frac{t}{\lambda} ))^n\frac{e^{-\lambda} \lambda^n}{n!}=\sum_{n=0}^{\infty}(1-\sigma^2\frac{t^2}{2\lambda^2}+o(\frac{t^2}{\lambda^2}))^n \frac{e^{-\lambda}\lambda^n}{n!}$
where $X'=X-\mu$. Can the above sum be simplified as $\lambda$ goes to $\infty$?
Note that $\sum_{n=0}^\infty z^n\mathbb P(N_\lambda =n)=e^{\lambda(z-1)}$ is a probability generating function of $N_\lambda$. You need also to replace $\lambda$ in denominator by $\sqrt{\lambda}$. Then $$\mathbb{E}e^{it Y_{\lambda}}=\sum_{n=0}^{\infty}\mathbb{E}(e^{itY_{\lambda}}1_{N_{\lambda}=n})=\sum_{n=0}^{\infty}\left(\phi_{X'_1}\left(\frac{t}{\sqrt{\lambda}}\right)\right)^n\frac{e^{-\lambda} \lambda^n}{n!} $$ $$ =e^{\lambda\left(\phi_{X'_1}\left(\frac{t}{\sqrt{\lambda}}\right)-1\right)}= e^{\lambda\left(1-\sigma^2\frac{t^2}{2\lambda}+o\left(\frac{t^2}{\lambda}\right)-1\right)} \to e^{-\frac{t^2\sigma^2}{2}}. $$