The number of accidents X that a person has in a given year is a Poisson random variable with mean Y . However, Y ∼ Uniform ([2, 4]). Calculate:
(a) E[X]
(b) Var(X)
Extra
My understanding of the problem is that E[X] = Y, where Y is our λ. So do we find the CDF of Y in order to calculate a and b? I think I'm just overcomplicating the problem.
$\operatorname{E}(X\mid Y)=Y$, so by the law of total expectation, $\operatorname{E}(X) = \operatorname{E}(\operatorname{E}(X\mid Y)) = \operatorname{E}(Y) = 3$.
For the Poisson distribution, the variance is the same as the expected value, so $\operatorname{var}(X\mid Y) = Y$. Then the law of total variance tells us that $$ \operatorname{var}(X) = \operatorname{E}(\operatorname{var}(X\mid Y)) + \operatorname{var}(\operatorname{E}(X\mid Y)) = \operatorname{E}(Y) + \operatorname{var}(Y) = 3 + \frac 1 3. $$