# Calculating Bayesian posterior distribution for an exponential distribution

I want to calculate the Bayesian posterior distribution of an exponential distribution where $\lambda$ is distributed according to gamma distribution. I know that I need to calculate $\frac{P(x|\lambda)P(\lambda)}{P(x)}$. I know $P(\lambda)$ because I know the gamma distribution function. I also know that $P(x|\lambda)$ = $\lambda \exp(-\lambda x)$. But I dont know what $P(x)$ is or what it means.

Am I on the right track? What does $P(x)$ mean in this context?

First is a hierarchical modeling problem: you know the distribution of $$x$$ given a parameter $$\lambda$$ ($$P(x|\lambda) \sim$$ exponential), and you have a distribution on that parameter ($$P(\lambda) \sim$$ gamma), but if you want the distribution on $$x$$ by itself ($$P(x) \sim$$ ?), you need to combine those two pieces of information.
This is done by integration; any value of $$x$$ could be achieved by any value of $$\lambda$$, and so $$P(x)=\int_0^\infty P(x|\lambda)P(\lambda)d\lambda$$. (Since the gamma and exponential distributions are closely related, this actually flows pretty smoothly.)
Second is the problem of inferring what the uncertain value $$\lambda$$ is, given we observe a particular value $$x$$ from the resulting exponential distribution. Here the role of $$P(x)$$ becomes clear--we are conditioning on our observation $$x$$, and dividing by $$P(x)$$ normalizes the probability distribution.
As pointed out in @ebb-earl-co's answer, you don't actually need to calculate $$P(x)$$ in this situation, if you're willing to integrate the distribution $$P(x|\lambda)P(\lambda)$$ in order to determine what normalizing constant you need to apply to make it a well-formed probability. The 'magic' of Bayes is that this normalizing constant is necessarily the probability of the observation that you're updating on.
$P(x)$ is the marginal density of $X$, which is called the marginal because you get it by marginalizing (i.e. integrating) out $\lambda$ from the joint density of $X$ and $\lambda$. I am going to rename $P(x)$ as $h(x)$ so that it doesn't get confused with the posterior. Then, I'll write the data model as $f(x|\lambda)$ and the prior as $\pi(\lambda)$. Therefore, your expression is $$p(\lambda|x) = \frac{f(x|\lambda)\pi(\lambda)}{h(x)}$$ Now, the beauty of Bayes' Theorem is that you don't need to know $h(x)$ in order to calculate $P(\lambda|x)$! So, it suffices to know that the posterior is proportional to the product of the data model (a.k.a. likelihood) and the prior on $\lambda$.