The question is
An insurance company supposes that the number of accidents that each of its policyholders will have this year is Poisson distributed, with a mean depending on the policyholder: 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 ($x = 0, 1, 2, \ldots$)
I am very unsure of how to solve this problem. I know that we need to solve for $E(\Lambda\mid X=x).$ I am pretty sure, but not sure what comes after this.
For any fixed $x=1,2,3,\ldots$ $$ \mathbb E(\Lambda\mid X=x) = \int_0^\infty \lambda\cdot f_{\Lambda\mid X}(\lambda\mid x)\, d\lambda $$ is the expected value of a random variable with conditional pdf $$ f_{\Lambda\mid X}(\lambda\mid x) \propto f_{\Lambda}(\lambda)\cdot f_{X\mid \Lambda}(x\mid \lambda). $$ Here the second term is the pmf since $X$ is discrete r.v. $$ f_{X\mid \Lambda}(x\mid \lambda) = \mathbb P(X=x \mid \Lambda=\lambda) = \frac{\lambda^x}{x!}e^{-\lambda} $$ and therefore $$ f_{\Lambda\mid X}(\lambda\mid x) \propto f_{\Lambda}(\lambda)\cdot f_{X\mid \Lambda}(x\mid \lambda)=\lambda e^{-\lambda} \cdot \frac{\lambda^x}{x!}e^{-\lambda}\propto \color{red}{\lambda^{x+1} e^{-2\lambda}}. $$ So conditional distribution of $\Lambda$ given $X=x$ is $\Gamma(2,x+2)$. Indeed, the pdf of $\Gamma(2,x+2)$ is $$ \gamma_{2,x+2}(\lambda) = c\cdot \color{red}{\lambda^{x+2-1}e^{-2\lambda}} $$ This is exactly the pdf that we got.
The expected value this distributon is $\frac{x+2}{2}$. So the answer is $$ \mathbb E(\Lambda\mid X=x) = \frac{x+2}{2}. $$