Maximum a posteriori estimate with exponential prior

Question

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?

Answer 1

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?