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.

  • 1
    $\begingroup$ This brute-force method might not work unless you're sufficiently familiar with Bessel functions, because $f_Y$ involves $I_0,$ a modified Bessel function. $\endgroup$
    – whuber
    Nov 23 '19 at 23:02
  • $\begingroup$ @whuber I have not reached the Bessel functions yet; how would you proceed in solving this problem? $\endgroup$
    – Jen Snow
    Nov 24 '19 at 1:56

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.

  • $\begingroup$ I really like this approach since it is using characteristic functions, which are new to me. However, can this also be solved with mgfs, since the Wiki article says they accomplish much of the same thing? If so, I am getting $\endgroup$
    – Jen Snow
    Nov 25 '19 at 0:06
  • $\begingroup$ I am getting $M_Z(t)=exp[\frac{-tE(Y)}{\sqrt{Var(Y)}}]M_Y(\frac{t}{\sqrt{Var(Y)}})$, where $Z=\frac{Y-E(Y)}{\sqrt{Var(Y)}}$ $\endgroup$
    – Jen Snow
    Nov 25 '19 at 0:11
  • 2
    $\begingroup$ Yes, frequently you can do the job with mgfs, but characteristic functions are more general. For example, not every distribution for which the CLT holds has an mgf. $\endgroup$
    – Glen_b
    Nov 25 '19 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.