# Find the Posterior distribution- prior: $exp(1)$, likelihood: $poisson(\lambda)($

I have a prior $\lambda \sim exp(1)$ and a likelihood $X \sim poisson(\lambda)$, and I observed in a sample of $n=5$ a mean of $3$. What is the posterior distribution of $\lambda$?

Here is my asnwer:

$f(x|\lambda) = \frac{\lambda^{x}e^{-\lambda}}{x!}$ $f(\lambda) = \theta e^{-\theta \lambda}, \theta = 1 => f(\lambda) = e^{-\lambda }$

So, the posterior:

$f(\lambda|x) \propto e^{-\lambda} \lambda^{x} e^{-\lambda +(-\lambda)} = \lambda^{x}e^{-2 \lambda}$

This posterior is some known distribution (e.g. exponential)?

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Since the standard (and standardized) procedure fully applies, you might want to explain what is causing you trouble here. (Furthermore, note that stricto sensu you asked no question.) –  Did May 28 '13 at 5:55
I have done the calculations but I don't know it my results are right. I would like to know if this posterior would be some know probability distribution (conjugate or not). –  Filipe Ferminiano May 31 '13 at 20:07
I have done the calculations but I don't know it my results are right... Then show them instead of leaving everybody in the blue. –  Did May 31 '13 at 23:20
Ok, I edited my post. Take a look please. –  Filipe Ferminiano Jun 1 '13 at 13:04
Which part of my answer is not the standardized procedure I alluded to in my first comment and, retrospectively, was causing you problems? –  Did Jun 1 '13 at 16:34

Assume one observed $x=(x_k)_{k\leqslant n}$, then $f(x\mid\lambda)=f(x_1\mid\lambda)\cdots f(x_n\mid\lambda)\propto\lambda^{|x|}\mathrm e^{-n\lambda}$ where $|x|=x_1+\cdots+x_n$ and $\propto$ refers to the fact that one omits multiplicative constants independent of $\lambda$. Thus, $f(\lambda\mid x)\propto f(\lambda)f(x\mid\lambda)\propto\lambda^{|x|}\mathrm e^{-(n+1)\lambda}$. To find the normalizing constant, recall that, for every positive $u$ and $v$, $$\int_0^\infty\lambda^{u-1}\mathrm e^{-v\lambda}\mathrm d\lambda=v^{-u}\Gamma(u),$$ hence, finally, $$f(\lambda\mid x)=(n+1)^{-|x|-1}\,(|x|)!\,\lambda^{|x|}\mathrm e^{-(n+1)\lambda}.$$ This is the gamma distribution with parameters $(n+1,|x|+1)$.

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