Given that $N = n$, the conditional distr. of $Y$ is $\chi ^2(2n)$. $N$ has marginal distr. of Poisson($\theta$), $\theta$ is a positive constant.

Show that, as $\theta \rightarrow \infty$, $\space \space (Y – E(Y))/ \sqrt{\operatorname{Var}(Y)} \rightarrow N(0,1)$ in distribution.

Could anyone suggest strategies to solve this. It seems like we need to use CLT (Central Limit Theorem) but it looks tough to get any information on $Y$ on it’s own. Is there a rv that can be introduced to take a sample of, to generate $Y$?

This is homework so

hintsappreciated.

**Answer**

I provide a solution based on properties of characteristic functions, which are defined as follows $$\psi_X(t)=E\exp{(itX)}.$$

We know that distribution is uniquelly defined by characteristic function, so I will prove that

$$\psi_{(Y-EY)/\sqrt{Var(Y)}}\rightarrow \psi_{N(0,1)}(t), \text{ when } \theta \rightarrow \infty,$$

and from that follows the desired convergence.

For that I will need to calculate mean and variance of $Y$, for which I use law of total expectations/variance – http://en.wikipedia.org/wiki/Law_of_total_expectation.

$$EY=E\{E(Y|N)\}=E\{2N\}=2\theta$$

$$Var(Y)=E\{Var(Y|N)\}+Var\{E(Y|N)\}=E\{4N\}+Var(2N)=4\theta+4Var(N)=8\theta$$

I used that the mean and variance of Poisson distribution are $EN=Var(N)=\theta$ and mean and variance of $\chi^2_{2n}$ are $E(Y|N=n)=2n$ and $Var(Y|N=n)=4n$.

Now comes the calculus with characteristic functions. At first I rewrite the definition of $Y$ as $$Y=\sum_{n=1}^{\infty}Z_{2n}I_{[N=n]}, \text{ where } Z_{2n}\sim \chi^2_{2n}$$

Now I use theorem which states

$$\psi_Y(t)=\sum_{n=1}^{\infty}\psi_{Z_{2n}(t)}P(N=n)$$

The characteristic function of $\chi^2_{2n}$ is $\psi_{Z_{2n}(t)}=(1-2it)^{-n}$, which is taken from here: http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)

So now we calculate the characteristic function for $Y$ using Taylor expansion for $\exp(x)$

$$\psi_Y(t)=\sum_{n=1}^{\infty}\psi_{Z_{2n}(t)}P(N=n)=\sum_{n=1}^{\infty}(1-2it)^{-n}\frac{\theta^n}{n!}\exp{(-\theta)}=\sum_{n=1}^{\infty}\left(\frac{\theta}{(1-2it)}\right)^n\frac{1}{n!}\exp{(-\theta)}=\exp(\frac{\theta}{1-2it})\exp(-\theta)=\exp(\frac{2it\theta}{1-2it})$$

At the end we use the properties of characteristic functions

$$\psi_{(Y-EY)/\sqrt{Var(Y)}}(t)=\exp(-i\frac{EY}{\sqrt{VarY}})\psi_Y(t/\sqrt{VarY})= \\\exp(-\frac{t^2}{2})\exp{(-1+2i\frac{t}{\sqrt{8\theta}})}\rightarrow \exp(-\frac{t^2}{2})=\psi_{N(0,1)}(t), \text{ when } \theta \rightarrow \infty$$

I jumped over the calculus because it is too lengthy by now…

**Attribution***Source : Link , Question Author : user42102 , Answer Author : Fimba*