Poisson-Gamma mix conditional probability problem: Find the expected value of Λ for a policyholder having x accidents this year

This question is unclear, but a reasonable interpretation would be as follows: Suppose $$X \mid \Lambda \sim \operatorname{Poisson}(\Lambda), \quad \Pr[X = x \mid \Lambda] = e^{-\Lambda} \frac{\Lambda^x}{x!}, \quad x = 0, 1, 2, \ldots,$$ and $$\Lambda \sim \operatorname{Gamma}(2,1), \quad f_\Lambda(\lambda) = \lambda e^{-\lambda}, \quad \lambda > 0.$$ Then we are asked to compute $$\operatorname{E}[\Lambda \mid X = x].$$ There are a number of ways to do this. One way is to compute the conditional distribution $\Lambda \mid X$ via Bayes' theorem, then compute the expectation. Another way is to compute the expectation directly from the joint distribution. The first approach described is illustrative, so this is how we will proceed.

Note $$f_{\Lambda \mid X}(\lambda \mid x) = \frac{\Pr[X = x \mid \Lambda = \lambda]f_\Lambda(\lambda)}{\Pr[X = x]}.$$ We do not know the marginal/unconditional distribution $\Pr[X = x]$ in the denominator on the RHS, but at least we can compute the likelihood in the numerator: $$\Pr[X = x \mid \Lambda = \lambda] f_\Lambda(\lambda) = e^{-\lambda} \frac{\lambda^x}{x!} \lambda e^{-\lambda} = e^{-2\lambda} \frac{\lambda^{x+1}}{x!}.$$ If we were to integrate this expression with respect to $\lambda$ on the interval $(0,\infty)$, we would get $\Pr[X = x]$, where upon dividing the likelihood by this expression, we would obtain the desired conditional density. But the kernel of the likelihood with respect to $\lambda$ is $$\lambda^{x+1} e^{-2\lambda},$$ which we get by removing any multiplicative factors of the likelihood that are independent of $\lambda$. This kernel is the same for a gamma density with shape parameter $a = x+2$ and rate parameter $b = 2$. Thus we can immediately conclude that $$\Lambda \mid X \sim \operatorname{Gamma}(x + 2, 2), \quad f_{\Lambda \mid X}(\lambda \mid x) = \frac{2^{x+2} \lambda^{x+1} e^{-2\lambda}}{(x+1)!}.$$ In fact, you would get this same result by actually performing the calculation explicitly as described above.

Finally, because we know that the conditional distribution $\Lambda \mid X$ is gamma with shape $X+2$ and rate $2$, its expectation is simply $$\operatorname{E}[\Lambda \mid X = x] = \frac{x + 2}{2}.$$ We could set up the integral $$\operatorname{E}[\Lambda \mid X = x] = \int_{\lambda = 0}^\infty \lambda f_{\Lambda \mid X}(\lambda \mid x) \, d\lambda,$$ but as we now know, this would merely repeat the process of computing the mean of a general gamma distribution, which we already know to be the shape divided by the rate.

It is worth mentioning that in the language of Bayesian inference, $\Lambda \mid X$ is a posterior distribution, and $\Lambda$ is the conjugate prior when the observed data $X$ is Poisson. But the inference of interest regards the random variable $\Lambda$, not $X$.