I am self studying mathematical statistics and am struggling with the following problem.
Let $N\sim \operatorname{Poi}(\Lambda)$ where $\Lambda$ is also a random variable. Show that $$E[N]=E[\Lambda]$$ and $$\operatorname{Var}[N]= \operatorname{Var}[\Lambda]+E[\Lambda]$$
Here is what I have tried so far.
$$E[N]=E[E[N\mid\Lambda=\lambda]]=E[\lambda]$$
I am confident so far due to the $E[E[X\mid Y]]$ fact, and that $N$ is Poisson $\lambda$.
However, I am not confident that I assumed $\Lambda=\lambda$ so $E[\Lambda]$ is the answer because for if it is true, then $E[\Lambda]=\lambda$ since $\lambda$ is a constant although we have no information regarding what kind of distribution $\Lambda$ is.
The next part is even more iffy for me. I know that
$$\operatorname{Var}[N]=\operatorname{Var}[E[N\mid\Lambda=\lambda]]+E[\operatorname{Var}[N|\Lambda=\lambda]]$$
so,
$$\therefore = \operatorname{Var}[\lambda]+E[\lambda^2]$$
which does not make sense since I am trying to prove the identity.
I appreciate your help.
Concerning the conditional expectation we have the rule:$$\mathbb EX=\mathbb E[\mathbb E[X\mid Y]]$$Applying that here gives:$$\mathbb EN=\mathbb E[\mathbb E[N\mid \Lambda]]=\mathbb E\Lambda$$
Concerning conditional variance we have the rule:$$\mathsf{Var}X=\mathbb E[\mathsf{Var}[X\mid Y]]+\mathsf{Var}[\mathbb E[X\mid Y]]$$
Applying that here gives:$$\mathsf{Var}N=\mathbb E[\mathsf{Var}[N\mid \Lambda]]+\mathsf{Var}[\mathbb E[N\mid \Lambda]]=\mathbb E\Lambda+\mathsf{Var}\Lambda$$