Convergence in distribution for $\frac{Y}{\sqrt{\lambda}}$

Given a sequence of independent r.v's $\{X_n\}_{n\geq 1}$ such that

$P(X_n=x)=\frac{1}{2}$ if $x=-1$ and/or $x=1$

Let $N\in Po(\lambda)$ be independent of $\{X_n\}_{n\geq 1}$ and we set that $Y=X_1+X_2+\dots+X_N$

One have to show that $\frac{Y}{\sqrt{\lambda}}$ converges in distribution to $N(0,1)$ as $\lambda$ goes to infinity.

How can this be shown? I have been thinking of some kind of use of central limit theorem but I don't get any idea for how to use it to show the convergence...

• You might want to compute the characteristic function, the result follows. – Did Dec 29 '13 at 0:08
• With that done: $\varphi_X(t)=\frac{1}{2}e^{-it}+\frac{1}{2}e^{it}$ and $N$ we have, $\varphi_N(t)=e^{^{(\lambda e^{it}}-1)}$ But I don't see any pattern... – Elekko Dec 29 '13 at 0:15
• Thus $\varphi_X(t)=\cos t$. And you want $\varphi_Y$... – Did Dec 29 '13 at 0:17
• $\varphi_Y(t)=(\varphi_X(t))^N$ where $N$ is Poisson distributed.. – Elekko Dec 29 '13 at 0:20
• $\varphi_Y(t)\ne\varphi_N(\varphi_X(t))$. You need to be more careful (are you actually writing down these computations in full? you should...). – Did Dec 29 '13 at 0:33

The characteristic function of $Y$ is $\varphi_Y(t)=\mathbb{E}[\varphi_X(t)^N]=\mathbb{E}[\cos(t)^N]=e^{\lambda(\cos(t)-1)},$ and so the characteristic function of $Y/\sqrt{\lambda}$ converges, as $\lambda\to\infty$, $$\varphi_{Y/\sqrt{\lambda}}(t)=e^{\lambda(\cos(t/\sqrt{\lambda})-1)}\to e^{-t^2/2}=\varphi_{N(0,1)}(t).$$
First show the independence hypotheses imply that $\Bbb E(e^{itY}\mid N=k)=\varphi_X(t)^k$. This means that $\Bbb E(e^{itY}\mid N)=\varphi_X(t)^N$. Then $\Bbb E(e^{itY})=\Bbb E(\Bbb E(e^{itY}\mid N))=\Bbb E(\varphi_X(t)^N)$, as noted by Didier in the comments. Finally $$\Bbb E(s^N)=\sum_{k\geqslant 0} s^k P(N=k)=\sum_{k\geqslant 0} s^k e^{-\lambda} \frac{\lambda^k}{k!}=e^{\lambda(s-1)}$$
gives $\varphi_Y(t)=e^{\lambda(\cos t-1)}$. The rest follows by Byron's answer, since $\cos t =1-\dfrac{t^2}2+o(t^2)$.