Lets say that I have N observations that are poisson and i.i.d. The prior is an exponential with paramater 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 posterior estimate? In must places, they same to just set x=1, but I don't understand why?

This would lead to the the map estimate to be

$ 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?

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?