I have the following problem:
Compute $E[W_6|X(4)=5]$, where X(t) is a Poisson Process and $W_n$ the time of the nth occurrence.
I know that $E[W_6|X(4)=5]=\int_4^\infty tf_{W_6|B}(t|B) dt$, where $B=[X(4)=5]$.
To solve the related problem Compute $E[W_2|X(4)=5]$ I use that for a Poisson Process $X(t)$ with $\lambda>0$, $f_{W_1,...,W_n}(t_1,...,t_n|A)=\frac{n!}{t^n}$ if $0<t_1<...<t_n<t$, so
$E[W_2|X(4)=5]=\frac{5!}{4^5}\int_0^4\int_0^{t_2}t_2 dt_1dt_2$.
But what about the case that happens beyond what we know so far? How do you compute $f_{W_6|B}(t|B)$ where $B=[X(4)=5]$.
One way to approach this: $E(W_6\mid X(4)=5)$ equals $4$ plus whatever time it takes after time $t=4$ (at which time we know we have exactly $5$ events) to get the next event (i.e. the sixth event).
Because of the memorylessness of the Poisson process, this expected extra time after $t=4$ is just the expected inter-arrival time of the process (the time between any two consecutive events). This inter-arrival time has an exponential distribution with parameter $\lambda$ and mean $1/\lambda$. Thus,
$$E(W_6\mid X(4)=5) = 4 + 1/\lambda.$$