# Probability Density function of Poisson distribution

This is an assignment I got for my course on Stochastic Processes:

Let us consider a random variable X distributed as a Poisson P (λ) where λ ∼ [0.5, 1].

(a) Which are the unconditional mean and variance for variable X? (DONE)

(b) Which is the probability density function of X? (Not need to solve the integral)

I managed to do the first part (a) but the second part (b) doesn't make sense to me.

How can X have a probability density function if X is a Poisson and the poisson is discrete? Am I missing something? Also, the professor says that there is no need to solve the integral, but how can there be an integral if the Poisson is discrete?

I guess the answer lies in this part:

λ ∼ [0.5, 1]

But I can't find it.

This is what I did to solve (a)

• @Xi'an I don't know, she didn't add anything to that. – Edoardo Busetti Jan 15 at 9:57
• @Xi'an Could you please explain me how to do that? For question (a) I simply demonstrated that the mean and variance of a poisson is given by λ. – Edoardo Busetti Jan 15 at 10:10
• I suspect the vague notation "$\lambda\sim[0.5,1]$" might be intended to stipulate that $\lambda$ is a random variable with a uniform distribution on the interval $[0.5,1].$ – whuber Jan 15 at 13:11

$$\mathbb{E}(X)=\mathbb{E}[\mathbb{E}(X\mid \lambda)])$$ where $$\mathbb{E}(X\mid \lambda)=\lambda$$ and $$\operatorname{Var}(X)=\mathbb{E}[\operatorname{Var}(X\mid \lambda)] + \operatorname{Var}(\mathbb{E}[X\mid \lambda])$$ where $$\operatorname{Var}(X\mid \lambda)=\lambda$$
Question (b): If the integral need not be solved, the marginal density of $$X$$ can be written as the integral (when $$k\in\mathbb{N}$$)
$$p(k)=2\int_{0.5}^1 \frac{\lambda^k}{k!}e^{-\lambda}\,\text{d}\lambda$$
• $\text{var}(X|\lambda)$ is the variance of the Poisson $P(\lambda)$ distribution. – Xi'an Jan 15 at 12:52