Poisson distribution with exponentially distributed parameter Let $X$ be a Poisson distributed random variable with parameter $\Lambda$, where $\Lambda \sim \text{exp}(\mu)$. I don't know how to get $$ \mathbb{P}[X=k|\Lambda]=\frac{\Lambda^k e^\Lambda}{k!}.$$ More precisely, 
i. It is clear that the random variable $\Lambda^k e^\Lambda/k!$ is $\sigma(\Lambda)$-measurable. But how can I show $$\mathbb{E}\left[\frac{\Lambda^k e^\Lambda}{k!} \mathbb{1}_{A} \right]=\mathbb{E}[\mathbb{1}_{\{X=k\}}\mathbb{1}_{A}]$$ for every $A \in \sigma(\Lambda)$?
ii. I just don't know how to calculate the expected probability $\mathbb{P}[X=k|\Lambda]$. By the factorisation Lemma one has $\mathbb{P}[X=k|\Lambda]=g(\Lambda)$ for some measurable function $g \colon \mathbb{R}  \to \mathbb{R}$. So it seems that I should finally have $$g(z)=\frac{z^ke^z}{k!}$$ by some theorem. What could be this theorem?
Thanks for your help.  
PS: Does the random vector $(X, \Lambda)$ have a density?

You can calculate $\mathbb{P}(X=k \mid \Lambda \leq j)$ for each $j\in \mathbb{R}$ and get the function $g$ that you say. This is the "substitution principle" (ie calculate $\mathbb{P}(X=k \mid \Lambda \leq j)$ and replace $j\to \Lambda$ to obtain $\mathbb{P}(X=k \mid \Lambda)$.
