Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I don't know how to solve Exercise 8, Section 5.2 from Geoffrey G. Grimmett, David R. Stirzaker, Probability and Random Processes, Oxford University Press 2001. For those who don't have this book:

Let $X$ have a Poisson distribution with parameter $\Lambda$, where $\Lambda$ is exponential with parameter $\mu$. Show that $X$ has a geometric distribution.


$X \sim Poiss(\Lambda),\ \ \Lambda \sim Exp(\mu)$.

So we know that generating function of $X$ is $G_x(s) = \sum_{i=0} s^i \frac{\Lambda^i}{i!} e^{-\Lambda}= e^{\Lambda(s-1)}$.

Probability density function of $\Lambda$ is $f_{\Lambda} = \mu e^{-\mu x}$.

And I don't know what I should do next. How to decompose $\Lambda$ in $G_x$ (or maybe this is not a good idea?).

Thanks in advance for your help.

share|improve this question
    
Why use generating functions? One can compute directly $\mathbb P(X=n)$ for each $n$. –  Did Jan 17 '13 at 21:56
    
And how to do this? –  user52354534 Jan 18 '13 at 1:05
    
See my answer... –  Did Jan 18 '13 at 7:13

2 Answers 2

For every nonnegative integer $n$, $\mathbb P(X=n\mid\Lambda)=\mathrm e^{-\Lambda}\Lambda^n/n!$ hence $$ \mathbb P(X=n)=\int_0^{+\infty}\left(\mathrm e^{-\lambda}\frac{\lambda^n}{n!}\right)\,f_\Lambda(\lambda)\,\mathrm d\lambda=\int_0^{+\infty}\left(\mathrm e^{-\lambda}\frac{\lambda^n}{n!}\right)\,\mu\mathrm e^{-\mu\lambda}\,\mathrm d\lambda. $$ The change of variable $x=(1+\mu)\lambda$ yields $$ \mathbb P(X=n)=\frac{\mu}{(1+\mu)^{n+1}}\int_0^{+\infty}\mathrm e^{-x}\frac{x^n}{n!}\mathrm dx=\frac{\mu}{(1+\mu)^{n+1}}. $$ One gets $$ \mathbb P(X=n)=(1-p)p^n,\qquad p=\frac1{1+\mu}. $$ Thus, the distribution of $X$ is geometric with parameter $p$.

share|improve this answer
    
After changing of variable I get different result, please correct me: $$x = \mu \lambda \\ dx = \mu d\lambda \\ \int_0^{\infty} e^{-\frac{x}{\mu}} \frac{x^n}{\mu^n n!} \mu e^{-x} \frac{1}{\mu} dx = \frac{1}{\mu^n} \int_0^{\infty} e^{-x} \frac{x^n}{n!} e^{-\frac{x}{\mu}} dx$$ –  user52354534 Jan 18 '13 at 17:26
    
And why does $\int_0^{\infty} e^{-x} \frac{x^n}{n!} dx$ equal $1$ ? –  user52354534 Jan 18 '13 at 17:45
    
Oh, I see - this is the Gamma Function, so I guess there is no simple explanation? :) Could you also explain me the first two equality: $$P(X=n|\Lambda) = e^{-\Lambda} \Lambda^n / n! \\ P(X=n) = \int_0^{\infty} e^{-\lambda} \frac{\lambda^n}{n!} f_{\Lambda} (\lambda) d\lambda$$ What formula or theorem it is? –  user52354534 Jan 18 '13 at 18:08
    
The correct change of variable is $x=(1+\mu)\lambda$, sorry about that. –  Did Jan 18 '13 at 20:45
    
Regarding the formula for $P(X=n\mid\Lambda)$, I wonder what is to explain. Here is a question: how would you translate the hypothesis that the distribution of $X$ is Poisson with parameter $\Lambda$? –  Did Jan 18 '13 at 20:47

Be careful, $\Lambda$ is a random variable! So your computation only shows that $$ E[s^X \mid \Lambda] = \sum_{n=0}^\infty \frac{(s\Lambda)^n}{n!}e^{-\Lambda} = e^{\Lambda (s-1)}. $$

Now you should be able to compute $$G_X(s) = E[s^X] = E\left[E[s^X\mid \Lambda]\right].$$

share|improve this answer
    
I don't quite understand second equality, could you explain it? Why $s^X = E(s^X | \Lambda)$ ? Provided that, $G_X(s) = E(e^{\Lambda (s-1)}) = \int e^{x(s-1)} \mu e^{-\mu x} dx = \frac {\mu e^{x(s-1-\mu)}}{s-1-\mu}$. Is it okay? If yes, how to deduce that $X$ has a geometric distribution? –  user52354534 Jan 17 '13 at 22:26
1  
I never wote such a thing as $s^X = E(s^X\mid\Lambda)$. I am only using the very basic property of conditional expectation $E[Z]=E\left[E[Z\mid \mathcal{F}]\right]$. Besides, the value of the integral cannot depend on the variable of integration $x$... –  Siméon Jan 18 '13 at 11:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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