Finding conditional covariance and expectation for Poisson process Let $\{N_t:t \ge 0 \}$ be a Poisson process with rate $\lambda$. Find the followings:
(a) $\mathrm{Cov}(N_{3t}, N_{5t}|N_t)$
(b) $\mathbb E[\mathrm{Cov}(N_t, N_{3t}|N_{5t})]$
To find (a), using that $N_{3t} $ and $N_{5t}-N_{3t}$ are independent when given $N_t$, $\mathrm{Cov}(N_{3t}, N_{5t}|N_t)=\mathrm{Cov}(N_{3t}, N_{5t}-N_{3t}+N_{3t}|N_t)=\mathrm{Cov}(N_{3t}, N_{5t}-N_{3t}|N_t)+\mathrm{Cov}(N_{3t}, N_{3t}|N_t)=\mathrm{Var}(N_{3t}|N_t)=\mathrm{Var}(N_{3t}-N_t+N_t|N_t)=\mathrm{Var}(N_{3t}-N_t|N_t)+\mathrm{Var}(N_{t}|N_t)$
Where the last equality is from the fact that $N_{3t}-N_t$ and $N_t$ are independent.
Also, Since $N_t$ is already given, $\mathrm{Var}(N_{t}|N_t)=0$, and $\mathrm{Var}(N_{3t}-N_t|N_t)=\mathrm{Var}(N_{3t}-N_t)=2\lambda t$. Am I right?
For (b), since $N_{5t}$ is given, we cannot argue that $N_t$ and $N_{3t}-N_t$ are independent because they cannot exceed $N_{5t}$. What should I do for (b)?
 A: First, let's figure out the conditional probability:
$$\begin{align} 
P(N_t=n_t,N_{3t}=n_{3t}|N_{5t}=n_{5t})
&=\frac{P(N_t=n_{t},N_{3t}=n_{3t},N_{5t}=n_{5t})}{P(N_{5t}=n_{5t})}\\
&=\frac{P(N_t=n_{t},N_{3t}-N_t=n_{3t}-n_t,N_{5t}-N_{3t}=n_{5t}-n_{3t})}{P(N_{5t}=n_{5t})}\\
&=\frac{P(N_t=n_{t})P(N_{3t}-N_t=n_{3t}-n_t)P(N_{5t}-N_{3t}=n_{5t}-n_{3t})}{P(N_{5t}=n_{5t})}\\
&=\frac{\frac{e^{-\lambda t} (\lambda t)^{n_t}}{n_t!} \frac{e^{-(3\lambda -\lambda) t} ((3 \lambda-\lambda) t)^{n_{3t}-n_t}}{(n_{3t}-n_t)!} \frac{e^{-(5\lambda-3\lambda) t} ((5\lambda-3\lambda )t)^{n_{5t}-n_{3t}}}{(n_{5t}-n_{3t})!}}{\frac{e^{-5\lambda t} (5\lambda t)^{n_{5t}}}{(n_{5t})!}}\\
&=\frac{n_{5t}!}{n_t!(n_{3t}-n_t)!(n_{5t}-n_{3t})!} \left(\frac{1}{5}\right)^{n_t}\left(\frac{2}{5}\right)^{n_{3t}-n_{t}}\left(\frac{2}{5}\right)^{n_{5t}-n_{3t}} \end{align}$$
$N_t$, $N_{3t}-N_t$ has the trinomial distribution when $N_{5t}=n_{5t}$ and we know that $Cov(N_t, N_{3t}-N_t|n_{5t})=-\frac{1}{5}\frac{2}{5}n_{5t}$.
$N_t$ has the binomial distribution $binomial(n_{5t},1/5)$.
$$Var(N_t|n_{5t})=n_{5t}\frac{1}{5}\frac{4}{5}$$
Then, we can rewrite the covariance as
$$Cov(N_t, N_{3t}|n_{5t})=Cov(N_t, N_{3t}-N_t|n_{5t})+Var(N_t|n_{5t})=\frac{2}{25}n_{5t}$$
Finally,
$$E[Cov(N_t, N_{3t}|N_{5t})]=\frac{2}{25} E[N_{5t}]=\frac{2 \lambda t}{5}$$
I am not a native English speaker and studying statistics by myself. So please feel free to correct my answer. Thank you. 
