Convergence in Distribution\CLT 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 hints appreciated.
 A: 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...
A: This can be shown via the relationship to the noncentral chisquared distribution. There is a good wikipedia article on that which I will reference freely!  https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution
You have given that $Y|N=n$  is distributed chisquare with $2 n $ degrees of freedom, for 
$n=0,1,\dots, \infty$. Here $N$ has the Poisson distribution with expectation $\theta$.
Then we have that the density function of $Y$ (unconditionally) can be written, using the law of total probability, as
$$
  f_Y(y; 0, \theta) = \sum_{i=0}^\infty \frac{e^{-\theta} \theta^i}{i!} f_{{\chi^2}_{2i}}(y)
$$
Which is almost  the density of a non-central chisquared  variable, except the degree of freedom parameter is $k=0$, which is really undefined.  (this is given in the definition section of wikipedia article).
So to get something well-defined, we replace the above formula with
$$
  f_Y(y; k,\theta) = \sum_{i=0}^\infty \frac{e^{-\theta} \theta^i}{i!} f_{{\chi^2}_{2i+k}}(y)
$$
which is the density of a noncentral chisquared variable with $k$ degrees of freedom and non-centrality parameter $2\theta$.  So, in our analysis, we must remember to take the limit when $k \rightarrow 0$ after taking the limit $\theta \rightarrow \infty$.  This is unproblematic, because in the limit of $\theta \rightarrow \infty$ the probability of $N=0$ goes to zero, so the point mass at zero disappears (chisquared variable with zero degrees of freedom must be interpreted as a pointmass at zero, so, have no density function).  
Now, for each fixed $k$, use the result in wiki , section related distributions, normal approximations, which gives the sought-for standard normal limit for each $k$.  Then, take the limit when $k$ goes to zero, which gives the result.  
