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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 :)

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    $\begingroup$ 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
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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), $\Gamma(x+1) = x!$

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