Let $N(t)$ be a Poisson Process of rate $\lambda$, and $S_n$ the time of the $n$'th occurrence, find $E[S_4|N(2)=3]$ and $E[N(2)|S_4=2]$.
If I'm right, for $E[N(2)|S_4=2]$, it's very easily $=4$.
For $E[S_4|N(2)=3]$ = $\frac{1}{\lambda} + E[max\{U_1, U_2, U_3\}]$ where $U_1, U_2, U_3 \text{~} Unif[0,2]$
My calculate yields $E[max\{U_1, U_2, U_3\}]=\frac{3}{2}$. Therefore, $E[S_4|N(2)=3]$ = $\frac{1}{\lambda} + \frac{3}{2}$.
Can someone please verify if I am correct? Thank you in advance!
If $M(t)$ is defined by: $$M\left(t\right)=N\left(t+2\right)-N\left(2\right)$$ then also $M(t)$ is a Poisson process of rate $\lambda$.
If further we define $R_{n}$ as the time of the $n$-th occurrence then under condition $N\left(2\right)=3$: $$R_{n}=S_{n+3}-2$$ and consequently: $$S_{4}=2+R_{1}$$
So we conclude that: $$\mathbb{E}\left[S_{4}\mid N\left(2\right)=3\right]=\mathbb{E}\left[2+R_{1}\right]=2+\frac{1}{\lambda}$$
We could say that at point $t=2$ we have observed $3$ occurrences, and then - memoryless - the process just goes on.
The waiting time from $2$ to the next occurrence is $\frac1{\lambda}$ and does not depend on $\mathbb E[S_3\mid N(2)=3]$ (as you seem to think).