So this is from Casella and Berger 5.36

the question states ``Given that N=n the conditional distribtution of Y is $\chi_{2n}^{2}$. the unconditional distribution $N\sim Poi(\theta)$

part a) is easy Calculate EY and VY the unconditional moments. well using laws of total probability/variance you get EY=$2\theta$ and VY=$8\theta$.

b) is weird show that $\theta \to \infty$ $\frac{Y-EY}{\sqrt{VY}} \to N(0,1)$

this is weird because the central limit theorem is for sequences (samples) of a distribution say for iid random variables. here we don't have a sequence of random variables. we could theoretical sample from Y creating $\{Y_{i}\}_{i=1}^{n}$ where the CLT will show that $\sqrt{n}\frac{\overline{Y}-EY}{\sqrt{VY}} \to_{n \to \infty} N(0,1)$ which is not the same thing as taking the limit of $\theta$

but not sure if this is right?

  • $\begingroup$ You may have to use Stirling's formula on this. $\endgroup$ Apr 5, 2017 at 1:54

1 Answer 1


We know that $$ M_{Y}(t)=E(e^{Yt})=E(E(e^{Yt}|N)) $$ While we know the moment generating function for $\chi^2_{n}$ distribution is given by $$ M_{Y|N=n}(t)=(1-2t)^{2n/2}=(1-2t)^{n} $$ Therefore we have $$ M_{Y}(t)=E((1-2t)^{-n})=\sum^{\infty}_{n=0}(1-2t)^{-n}\frac{\theta^{n}}{n!}e^{-\theta}=e^{-\theta}e^{\theta/(1-2t)}=e^{\frac{2t}{1-2t}\theta} $$ Now the rest follows by using $$ M_{aY+b}(t)=e^{at}M_{Y}(bt) $$ and substitute $EY=2\theta, \textrm{Var}(Y)=8\theta$, then left $\theta\rightarrow \infty$. I will leave out the computational details to you.

My guess is you can also work using brutal force by fiddling with $Y$'s pdf (there is a problem in Casella and Berger solved by me this way earlier), but it is quite a mess in this case.

  • $\begingroup$ should the sum be n=0 to infinity and not i? $\endgroup$ Apr 6, 2017 at 2:45
  • $\begingroup$ it is so great that you can use total expectation on moment generating functions. that is the main key. thank you so very much. $\endgroup$ Apr 6, 2017 at 3:54
  • $\begingroup$ another approach I was working on is that Gamma if alpha $\to \infty$ will approach normal. so chisquare is a subset of gamma. $\endgroup$ Apr 6, 2017 at 4:00
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    $\begingroup$ I think you should justify everything rigorously. It is intuitive to say that $\Gamma(\alpha,\beta)\rightarrow N(\alpha\beta, \alpha\beta^2)$, but this is not even well defined: Like CLT you have to shift and re-scale it. And it might be easier to treat $\Gamma$ distribution as sum of exponentials then use CLT. Since $E(N)=Var(N)=\theta$, the proof for the case when $\theta\rightarrow \infty$ is intuitive but may be quite messy to write down (you have to invoke normal approximation or at least Chebyshev's theorem, etc). $\endgroup$ Apr 6, 2017 at 4:15

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