If $N_{\mid\Lambda = \lambda} \sim$ Poisson ($\lambda$) and $\Lambda \sim$ unif$(0,5)$, find the probability of zero occurring.

Question

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?

Answer 1

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.

$$f_{\Lambda\mid N}(\lambda\mid n) \;=\; \frac{\mathsf P(N=n\mid\Lambda=\lambda)~f_{\Lambda}(\lambda)}{\mathsf P(N=n)}$$

Answer 2

Yes, all your previous calculations look fine.

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

2. Notice that you are told $N$, and so you are seeking the conditional distribution of $\Lambda$ given $N$, which is $$f_{\Lambda|N}(\lambda\mid k) = \frac{f_{\Lambda,N}(\lambda,n)}{P(N=k)} = \frac{P(N=k\mid\Lambda=\lambda)f_{\Lambda}(\lambda)}{P(N=k)}.$$

Answer 3

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 $$ \mathbb{E} X = \mathbb{E} e^{-\Lambda} = \int_{0}^5 e^{-x}\frac{dx}{5} = \frac{1}{5}(1-e^{-5}). $$