Conditional Expectation Problem? An insurance company supposes that the number of accidents that each of its customers will have this year is Poisson distributed, with a mean depending on the customer: the Poisson mean $\Lambda$ of a randomly chosen person has a Gamma distribution with the $\Gamma(2, 1)$-density function $f_\Lambda(\lambda) = \lambda e^{−\lambda}$, $(\lambda > 0)$. Find the expected value
of $\Lambda$ for a policyholder having $x$ accidents this year $(x = 0, 1, 2 \ldots)$?
Not quite sure how to start this, I was thinking it's $\mathbb E[\Lambda|X=x]$, but I can't find any documentation on how conditional expectation is solved when the distributions aren't both discrete or both continuous.

The Wikipedia article on Bayes's theorem has forms for all three cases of continuous variables being involved. In this case, we use the first form:
$$f_{\Lambda}(\lambda \mid X = x) = \frac{\Pr[X = x \mid \Lambda = \lambda] f_{\Lambda}(\lambda)}{\Pr[X = x]}.$$
In this expression,


*

*$\Pr[X=x \mid \Lambda=\lambda]$ is just the probability that a $\operatorname{Poisson}(\lambda)$ customer will have $x$ accidents, 

*$f_{\Lambda}(\lambda)$ is the density function you're given, and

*$\Pr[X=x]$ is the constant factor we need to divide by to make $f_{\Lambda}(\lambda \mid X = x)$ integrate to $1$.


Once you know $f_{\Lambda}(\lambda \mid X = x)$, you can find the expected value of $(\Lambda \mid X=x)$ just as you would with any other density function.
