Maximum a posteriori estimate with exponential prior Lets say that I have N observations that are poisson and i.i.d. The prior is an exponential with parameter 2. I know that the exponential distribution is given by
$
\lambda e^{(-\lambda x)}
$
But how does it work with x when you multiply the prior times the likelihood to get the maximum a posteriori (MAP) estimate? In most places, they seem to just set $x=1$, but I don't understand why.
This would lead to the the MAP estimate to be
$
\operatorname{argmax} \quad \lambda^{\sum x_n}e^{-\lambda N}e^{-2\lambda}
$
So my question is, how do you handle $x$ in the exponential distribution when it is a prior?
 A: I'm not clear on the question, so I'm going to write some exposition and then finish my answer when you've clarified.
The Maximum A Posteriori estimate is, as you've said, is
$$ \hat{\lambda} =  \underset{\lambda \in \mathbb{R}_+}{\mbox{argmax}} \left\{ \ell(\lambda; x) + \log(p(\lambda)) \right\} $$
Here, $\ell(\lambda;x)$ is the log likelihood and $p(\lambda)$ is the prior density. The prior for $\lambda$ depends on a different parameter which we will call $k$.  The likelihood (poisson) is
$$ \dfrac{\lambda ^x e^{\lambda}}{\Gamma(x+1)}$$
The prior is (remember, $\lambda$ is the variable, $k$ is the parameter)
$$ k \exp(-k \lambda) $$
The log posterior is then (if my algebra is correct)
$$ N \bar{x}\log(\lambda) - N \lambda - \sum_i\log(\Gamma(x_i+1)) - \log(k) - k \lambda $$
From here, you can differentuate with respect to lambda and solve.
$$ 0 =  \dfrac{N \bar{x}}{\lambda} - N -k \implies \lambda = \dfrac{N}{N+k} \bar{x} $$
Does your question concern $k$, the parameter in the prior?
