How can I show convergence in distribution to the normal? Consider the following model:
$$Y|N \sim \mathcal{X}^2_{2N} 
\quad \quad \quad 
N \sim \text{Pois}(\theta).$$
and define the standardised statistic:
$$Z = \frac{Y-\mathbb{E}(Y)}{\sqrt{\mathbb{V}(Y)}}.$$
I want to show that as $\theta \to \infty$ we obtain the convergence $Z \overset{\text{dist}}{\to} \text{N}(0,1)$.

My attempt: I found $\mathbb{E}(Y) = 2 \theta$ and $\mathbb{V}(Y) = 8 \theta$, so $Z= (Y-2\theta)/\sqrt{8\theta}$.  At this point, I feel like the best way to approach this problem is to find the distribution of $Y$, because if $Y$ is a normal distribution, I can easily standardise it to $N(0,1)$.  I have managed to show that:
$$f_{Y,N}(y,n) = \frac{1}{\Gamma(n)2^n}\frac{y^{n-1}e^{-y/2}e^{-\theta}\theta^n}{n!}
\quad \quad \quad 
\text{for } n =0,1,2,... \text{ and } 0 \leqslant y < \infty.$$
$$f_Y(y)=e^{-y/2}e^{-\theta}\Sigma^\infty_{n=0}\frac{y^{n-1}\theta^n}{\Gamma(n)2^nn!}
\quad \quad \quad 
\text{for } 0 \leqslant y < \infty.
\quad \quad \quad \quad \quad \quad $$
That is as far as I can get.  I do not know how to evaluate correctly to show that this is a normal distribution in the limit.
 A: When you are trying to show convergence in distribution, it is often useful to work with characteristic functions instead of distributions.  If you can show convergence of the characteristic function to the desired form, this is sufficient to give convergence in distribution.  For all $t < \tfrac{1}{2}$ the characteristic function of $Y$ is given by:
$$\begin{equation} \begin{aligned}
\varphi_Y(t) 
&= \mathbb{E}(\varphi_{Y|N}(t)|N) \\[6pt]
&= \mathbb{E}((1-2it)^{-N}) \\[6pt]
&= \sum_{n=0}^\infty (1-2it)^{-n} \cdot \frac{\theta^n \exp(-\theta)}{n!} \\[6pt]
&= \exp(-\theta) \sum_{n=0}^\infty \frac{1}{n!} \cdot \Big( \frac{\theta}{1-2it} \Big)^n \\[6pt]
&= \exp(-\theta) \cdot \exp \Big( \frac{\theta}{1-2it} \Big) \\[6pt]
&= \exp \Big( \frac{2it}{1-2it} \cdot \theta \Big). \\[6pt]
\end{aligned} \end{equation}$$
Since $\mathbb{E}(Y) = 2\theta$ and $\mathbb{V}(Y) = 8\theta$, we can write the standardised random variable as:
$$Z = \frac{Y - 2\theta}{2 \sqrt{2\theta}} = \frac{Y}{2 \sqrt{2\theta}} - \frac{\sqrt{\theta}}{\sqrt{2}}.$$
Since the characteristic function (which is a Fourier transform) is a linear operator, for all $t < \sqrt{2 \theta}$ we then have:
$$\begin{equation} \begin{aligned}
\varphi_Z(t) 
&= \exp \Big( -it \cdot \frac{\sqrt{\theta}}{\sqrt{2}} \Big) \cdot \varphi_Y \Big(\frac{t}{2 \sqrt{2\theta}} \Big) \\[6pt]
&= \exp \Big( -it \cdot \frac{\sqrt{\theta}}{\sqrt{2}} \Big) \cdot \exp \Big( \frac{it}{\sqrt{2\theta}-it} \cdot \theta \Big) \\[6pt]
&= \exp \Big( -it \cdot \frac{\sqrt{\theta}}{\sqrt{2}} + \frac{it}{\sqrt{2\theta}-it} \cdot \theta \Big) \\[6pt]
&= \exp \Big( -it \frac{\sqrt{\theta}}{\sqrt{2}} \cdot \frac{\sqrt{2\theta}-it}{\sqrt{2\theta}-it} + \frac{it}{\sqrt{2\theta}-it} \cdot \theta \Big) \\[6pt]
&= \exp \Big( -it \cdot \frac{\theta-it \sqrt{\theta/2}}{\sqrt{2\theta}-it} + \frac{it}{\sqrt{2\theta}-it} \cdot \theta \Big) \\[6pt]
&= \exp \Big( - \frac{\sqrt{\theta/2}}{\sqrt{2\theta}-it} \cdot t^2 \Big) \\[6pt]
&= \exp \Big( - \frac{1}{2 - it \sqrt{2/\theta}} \cdot t^2 \Big). \\[6pt]
\end{aligned} \end{equation}$$
Taking $\theta \rightarrow \infty$ we obtain:
$$\begin{equation} \begin{aligned}
\lim_{\theta \rightarrow \infty} \varphi_Y(t)
= \lim_{\theta \rightarrow \infty} 
\exp \Big( - \frac{1}{2 - it \sqrt{2/\theta}} \cdot t^2 \Big)
= \exp \Big( - \frac{1}{2} \cdot t^2 \Big),
\end{aligned} \end{equation}$$
which is the characteristic function of the standard normal distribution.  Since there is a one-to-one relationship between the characteristic function and the distribution (and since the limits can be transferred across domains) this is sufficient to establish convergence in distribution to the standard normal distribution.
