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)

  • $\begingroup$ @Xi'an I don't know, she didn't add anything to that. $\endgroup$
    – user210809
    Jan 15, 2019 at 9:57
  • $\begingroup$ @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. $\endgroup$
    – user210809
    Jan 15, 2019 at 10:10
  • 1
    $\begingroup$ 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 $[0.5,1].$ $\endgroup$
    – whuber
    Jan 15, 2019 at 13:11

1 Answer 1


Question (a): One need call the Law of Total Expectation and Law of Total Variance.

$$\mathbb{E}(X)=\mathbb{E}[\mathbb{E}(X\mid \lambda)])$$ 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 $k\in\mathbb{N}$)

$$p(k)=2\int_{0.5}^1 \frac{\lambda^k}{k!}e^{-\lambda}\,\text{d}\lambda$$

  • $\begingroup$ I'm having trouble understanding how to calculate Var(X)=𝔼[Var(X∣λ)]+Var(𝔼[X∣λ]). I think it is: Var(X)=𝔼[Var(X∣λ)] + Var(λ). Where the Var(λ) is the variance of an uniform distribution, but I don't understand how to calculate 𝔼[Var(X∣λ)]. $\endgroup$
    – user210809
    Jan 15, 2019 at 11:19
  • $\begingroup$ $\text{var}(X|\lambda)$ is the variance of the Poisson $P(\lambda)$ distribution. $\endgroup$
    – Xi'an
    Jan 15, 2019 at 12:52

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