I'm struggling with this one:

If $\theta $ is a Gamma$(p,\lambda)$ random variable with $p>1$ and $\lambda>0$.

We give the density of the gamma distribution: $ f(x) = \frac {\lambda^p}{\phi(p)} x^{p-1} \exp(-\lambda x) 1_{t>0}$

$\phi (p)$ is referring to the function $\phi$, for which we give: $\phi(z+1)=z\phi (z) $ for $z>0$, and $\phi (n+1)=n!$.

If $N$ is a random variable such that, given $\theta=t $, $N$ is a Poisson distribution of parameter $t$.

  • What is the distribution of $N$, $P(N=n)$?.

  • What is $E(N)$ ? What is $\operatorname{Var}(N)$?

If I apply the formula of conditional probability, I will have something like this:

$$P(N=n\mid\theta =t)=\frac {\frac {e^{-t} t^n}{n!} \times \frac{\lambda^p}{\phi (p)} t^{p-1} \exp(-\lambda t) 1_{t>0}}{\frac{\lambda^p}{\phi(p)} t^{p-1} \exp(-\lambda t) 1_{t>0}} =\frac {e^{-t} t^n}{n!} $$

It looks rather strange to me. It is as if, $\theta $ will cancels itself out on the numerator and denominator and therefore has no influence on the Poisson distribution.

Thanks in advance.


  • 1
    $\begingroup$ What you write as $\phi$ is usually called the Gamma function $\Gamma(\cdot)$. Also, note that your calculation of the conditional probability gives the correct result: $P(N=n\mid \theta = t)$ does equal what you find it to be; in any case, it does need to be computed since you are given that $N$ is conditionally Poisson-distributed with parameter $t$. What you are asked to find is the unconditional probability $P(N=n)$ which is given by $$P(N=n) = \int_0^\infty P(N=n\mid \Theta = t)f_\Theta(t)\,\mathrm dt.$$ $\endgroup$ – Dilip Sarwate Jul 13 '14 at 13:37
  • $\begingroup$ So i need to compute this integral: $P(N=n)=\int _{ 0 }^{ +\infty }{ \frac { { e }^{ -t }{ t }^{ n } }{ n! } \times \frac { { \lambda }^{ p } }{ \phi (p) } { t }^{ p-1 }exp(-\lambda t){ 1 }_{ t>0 } } dt$ ? $\endgroup$ – XCoder Jul 13 '14 at 14:04
  • $\begingroup$ Sorry for the Gamma function, I could not figure out the latex code $\endgroup$ – XCoder Jul 13 '14 at 14:04
  • $\begingroup$ Yes, you need to compute the value of this integral. The integral itself is a version of what is called the law of total probability. You can use \Gamma to get the Gamma function symbol. $\endgroup$ – Dilip Sarwate Jul 13 '14 at 14:13
  • $\begingroup$ Thanks Dilip I will try this out $\endgroup$ – XCoder Jul 13 '14 at 14:32

The result is a Negative Binomial distribution.


The function that you call $\phi$ is usually called $\Gamma,$ the gamma function, and that is what I will call it here. \begin{align} & \Pr(N=n) \\[8pt] = {} & \operatorname E\big( \Pr(N=n\mid \theta) \big) \\[8pt] = {} & \operatorname E\left( \frac{\theta^n e^{-\theta}}{n!} \right) \\[8pt] = {} & \int_0^\infty \frac{t^n e^{-t}}{n!} \cdot \frac{(\lambda t)^{p-1} e^{-\lambda t} (\lambda\, dt)}{\Gamma(p)} \\[8pt] = {} & \frac {\lambda^p} {n!\Gamma(p)} \int_0^\infty t^{n+p-1} e^{-(\lambda+1)t} \, dt \\[8pt] = {} & \frac {\lambda^p} {n!\Gamma(p)}\cdot\frac 1 {(\lambda+1)^{n+p}} \int_0^\infty \big((\lambda+1)t\big)^{n+p-1} e^{-(\lambda+1)t} \big((\lambda+1)\, dt\big) \\[8pt] = {} & \frac 1 {n!\Gamma(p)} \left( \frac\lambda {\lambda+1} \right)^p \left( \frac 1 {\lambda+1} \right)^n \int_0^\infty u^{n+p-1} e^{-u} \, du \\[8pt] = {} & \frac {\Gamma(n+p)} {n!\Gamma(p)} \cdot r^p (1-r)^n \\[8pt] = {} & \frac{p(p+1)(p+2) \cdots (p+n-1)}{n!} r^p (1-r)^n \\[8pt] = {} & \binom {p+n-1} n r^p (1-r)^n \end{align} with $r:=\frac{\lambda}{\lambda+1}$. So this is a negative binomial distribution.


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