Let $N(t), t\geq 0$ be a Poisson process with intensity $\lambda = 2$. Let $S_{n}$ denote the time of the $n^{th}$ event. Find $E(S_{6}|N(2)=4)$.
I have been told that $E(S_{6}|N(2)=4) = 2 + E(S_{6}-S_{4}) = 2+ E(S_{6}) -E(S_{4})$.
How is it that $E(S_{6}|N(2)=4) = 2 + E(S_{6}-S_{4})$?
How do we know that $2 + E(S_{6}-S_{4}) = 2+ E(S_{6}) -E(S_{4})$, given that we have not been told if the $S_{n}$ are independent?.
For your second question, $E[X-Y] = E[X] - E[Y]$ always holds, even if $X$ and $Y$ are dependent.
I think the first step is not correct (although it leads you to the right answer).
Here is how I would think of it. Conditioned on $N(2) = 4$, you can think of the Poisson process starting anew at time $2$ (independent increments property). So the sixth event in the full process is actually the second event after time $2$, and thus $E[S_6 \mid N(2) = 4] = 2 + E[S_6 - 2 \mid N(2) = 4] = 2 + E[\tilde{S}_2] = 2 + 2 / \lambda$, where I am using $\tilde{S}_2$ to denote the time of the second event in a separate Poisson process with intensity $\lambda$.