If $N_{\mid\Lambda = \lambda} \sim$ Poisson ($\lambda$) and $\Lambda \sim$ unif$(0,5)$, find the probability of zero occurring. The number of storms in the upcoming rainy season is assumed to be Poisson distributed, but with a parameter $\Lambda$ that is also random and uniformly distributed on $(0,5)$. That is, $\Lambda \sim$ unif$(0,5)$ and given that $\Lambda = \lambda$, the conditional distribution of the number of storms $N$ is Poisson with mean $\lambda$: $N_{\mid\Lambda = \lambda} \sim$ Poisson ($\lambda$).
So far I've gotten (for the first parts of the question; not pertaining to the question below):
E$(N\mid \Lambda) = \lambda$
E$(N) = \frac{5}{2}$
Var$(N\mid\Lambda) = \lambda$
Var$(N) = \frac{55}{12}$
(It would be greatly appreciated if someone checked that)
Now what I need to answer is: 
(i) Find the probability that zero storms occur this season (I think I need to integrate the pmf of Poisson distribution but I'm getting stuck)
(ii) Given that zero storms occur this season, what is the conditional distribution of $\Lambda$?
How do I go about answering these questions?
 A: Yes, all your previous calculations look fine.


*

*Recall that to find the unconditional distribution of $N$ can be computed by
$$P(N = k) = \int_0^5 P(N=k\mid\Lambda = \lambda)f_\Lambda(\lambda)\,d\lambda.$$

*Notice that you are told $N$, and so you are seeking the conditional distribution of $\Lambda$ given $N$, which is
$$f_{\Lambda|N}(\lambda\mid k) = \frac{f_{\Lambda,N}(\lambda,n)}{P(N=k)} = \frac{P(N=k\mid\Lambda=\lambda)f_{\Lambda}(\lambda)}{P(N=k)}.$$
A: Since $N\mid\Lambda=\lambda \;\sim\;\mathcal{Pois}(\lambda)$ you know:
$$\mathsf P(N=0\mid \Lambda=\lambda) = \dfrac{\lambda^0 \mathsf e^{-\lambda}}{0!}$$
Since $\Lambda\sim\mathcal{U}(0;5)$ you also know $f_\Lambda(\lambda) = \frac 1 5 \mathbf 1_{\lambda\in(0;5)}$
And you should know how to find marginal distributions.
$$\mathsf P(N=0) =\int_0^5\mathsf P(N=0\mid\Lambda=\lambda) f_\Lambda(\lambda)\operatorname d \lambda$$
Put it together.
For (ii) use what you know about conditional probability (Bayes' Rule).  The mix of probability densities and masses is not an issue for this question.
$$f_{\Lambda\mid N}(\lambda\mid n) \;=\; \frac{\mathsf P(N=n\mid\Lambda=\lambda)~f_{\Lambda}(\lambda)}{\mathsf P(N=n)}$$
A: A possibility for (i): define the random variable (indicator) $X=\mathbf{1}_{N=0}$, so that
$
\mathbb{P}\{N=0\} = \mathbb{E} X
$.
Now, write 
$$
\mathbb{E} X = \mathbb{E}[\mathbb{E}[ X\mid \Lambda ]]
$$
so we can first deal with the random variable
$$
\mathbb{E}[ X\mid \Lambda ] = e^{-\Lambda}
$$
and then get 
$$
\mathbb{E} X = \mathbb{E}  e^{-\Lambda} = \int_{0}^5 e^{-x}\frac{dx}{5}
= \frac{1}{5}(1-e^{-5}).
$$
