Looking over an exam and I have no idea how to finish when proving this:

Prove that for a Poisson r.v. X, if the parameter $\lambda$ is not fixed and is itself an exponential r.v. with parameter 1, then:

$P(X = x) = (\frac{1}{2})^{x+1}$

My attempt:

$P(x) = \frac{\lambda^xe^{-\lambda}}{x!}$

$\lambda = \frac{1}{\beta}exp(\frac{-x}{\beta})$, $\beta = 1$

$\lambda = e^{-x}$

$P(x) = \frac{e^{-x^2}e^{e^{-x}}}{x!}$

And now I'm stuck...

Thanks so much in advance, this website is gonna make me pass stats :)

  • 2
    $\begingroup$ I've answered a similar question here: math.stackexchange.com/questions/466929/… Just replace the Gamma(2,5) with your Exponential and work the calculation. $\endgroup$ – baudolino Dec 17 '13 at 20:53
  • $\begingroup$ Thanks, I think I've the hang of it! $\endgroup$ – FRU5TR8EDD Dec 17 '13 at 22:04

The solution basically boils down to:

$$ P(X=x) = \int_0^\infty \dfrac{e^{-\lambda} \lambda^x}{x!} e^{-\lambda} \, \mathrm{d}\lambda $$ which after a little manipulation evaluates to $$ P(X=x) = \dfrac{2^{-(x+1)}\Gamma(x+1)}{x!}=\left(\dfrac{1}{2}\right)^{x+1} $$ since for discrete r.v. (i.e. Poisson), $\Gamma(x+1) = x!$


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.