Let N be the number of accidents on the highway on any given day $\lambda$ is uniformly distributed on [0,3]
\begin{equation} Then,\ N \sim Poisson(\lambda)\ and \ \lambda \ is \ also \ RV. \end{equation}
What is the mean and the variance of the number of accidents?
My answer:
\begin{equation} Mean\ of\ N\ =\ \lambda\ since\ this\ is\ a\ property\ of\ Poisson \end{equation} I'm a bit unsure of my answer above. In my head it makes sense that the mean of N is $\lambda$ because it is the property of the poisson distribution - however, does the fact that $\lambda$ is uniformly distributed change anything?
\begin{equation} Var(N) = E(Var[N|\lambda]) + Var(E[N|\lambda]) \end{equation} I think with the above, $Var[N|\lambda]$ should equal to $\lambda$ since N is poisson distributed given $\lambda$. Same with the expectance on the side as well.
Am I approaching the problem correctly? Any corrections or insights would be really helpful!
Yes. What you have is that the conditional expectation of $N$ for a given $\lambda$ is $\lambda$, as is its conditionl variance, since the distribution is Poisson for a given $\lambda$. Since $\lambda$ is itself continuous uniformly distributed over $[0;3]$, thence we know its expectation and variance too.
$$\begin{split}\mathsf E(N\mid \lambda)&=\lambda\\\mathsf{Var}(N\mid \lambda)&=\lambda\\\mathsf E(\lambda)&=\frac 32\\\mathsf {Var}(\lambda)&=\frac{9}{12}\end{split}$$
What you are required to use is the Tower Property, also known as the Law of Iterated Expectation (sometimes "Law of Total Expectation").
$$\mathsf E(N) = \mathsf E(\mathsf E(N\mid \lambda))$$
The Law of Iterated Variance follows from the same principle.
$$\mathsf {Var}(N) = \mathsf E(\mathsf {Var}(N\mid \lambda))+\mathsf {Var}(\mathsf E(N\mid \lambda))$$
The rest is just substitution .