Properties of sum of Poisson processes Let $N_1$ and $N_2$ be a independent Poisson processes with intensities $\lambda_1=1$ and $\lambda_2=4$. Let $N=N_1+N_2$ and $S_n$ be a moment of $n$ event. I need to calculate the following:

*

*$P(N_1(3)-N_1(1)=2)$

*$P(N_2(5)=3|N(5)=5)$

*$E(S_3)$

*$E(S_3|N_2(2)=1)$

*$E(N(5)-N(3)|N_2(2)=3)$
These are my attempts to solve it:
$$(1)\ P(N_1(3)-N_1(1)=2)=e^{-1(3-1)}\cdot\frac{(3-1)^2}{2!}=2e^{-2}$$
$$(2)\ P(N_2(5)=3|N(5)=5)=P(N_1(5)=2\ and\ N_2(5)=3)=P(N_1(5)=2)P(N_2(5)=3)$$
$$(3)\ E(S_3)=E(\sum_{i=1}^4T_i)=4E(T_1)=\frac{4}{4}=1$$
$$(5)\ E(N(5)-N(3)|N_2(2)=3)=E(N(5)-N(3))=e^{-5(5-3)}\frac{(5-3)^5}{5!}=e^{-10}\frac{32}{120}$$
Is that correct? Could you give me any hints for fourth point?
 A: For 1. we have
$$
\mathbb P(N_1(3)-N_1(1)=2) = \mathbb P(N_1(2)=2) = e^{-2\lambda_1}\frac{(2\lambda_1)^2}{2!} = e^{-2}\frac 42 =2e^{-2}.
$$
For 2. we have
\begin{align}
\mathbb P(N_2(5)=3\mid N(5)=5) &= \frac{\mathbb P(N_2(5)=3,N(5)=5)}{\mathbb P(N(5)=5)}\\
&= \frac{\mathbb P(N_2(5)=3,N_1(5)=2)}{\mathbb P(N(5)=5)}\\
&= \frac{e^{-5\lambda_2}\frac{(5\lambda_2)^3}{3!}e^{-5\lambda_1}\frac{(5\lambda_1)^2}{2!} }{e^{-5(\lambda_1+\lambda_2)}\frac{(5(\lambda_1+\lambda_2))^5}{5!}}
\end{align}\
&= \frac{128}{625}.
For 3. we know that $S_3$ has Erlang distribution with parameters $k=3$ and $\lambda = \lambda_1+\lambda_2=5$, and hence we compute
$$
\mathbb E[S_3] = \int_0^\infty \frac{(5x)^3 e^{-5 x}}{2!}\ \mathsf dx = \frac 35.
$$
For 4. we use the independence of the processes $N_1$ and $N_2$ along with the memoryless process to deduce that
\begin{align}
\mathbb E[S_3\mid N_2(2)=1] &= 2 + \mathbb E[S_2] = 2 + \int_0^\infty (5x)^2 e^{-5 x}\ \mathsf dx = 2+\frac25 = \frac{12}5.
\end{align}
For 5. your observation that $N(5)-N(3)$ is independent of $N_2(2)$ is correct, but the computation is rather
$$
\mathbb E[N(5)-N(3)] = (1+4)(5-3) = 10.
$$
