Consider a Poisson process $\{X_t\}_{t\ge 0}$ with parameter $\lambda$, and $W_5$, the random variable of the 5th event. Compute $E[W_5|X_t=3]$.
$\bf Idea$
I think that it can be computed by the sustitution of $W_5$, by the inter arrival times in each state ($S_i$) and the fact that ($X_t=3$ if and only if $S_3\le t$):
$E[W_5|X_t=3]=E[S_1+S_2+S_3+S_4+S_5|S_3\le t]$
And by independency of the inter arrival times:
$E[W_5|X_t=3]=E[S_1]+E[S_2]+E[S_3|S_3\le t]+E[S_4]+E[S_5]=\frac{4}{\lambda}+E[S_3|S_3\le t]$
I wondering if it's correct and how can I compute the final answer?
$X_t=3$ is not equivalent to $S_3 \le t$. First, you probably meant $S_1+S_2+S_3 \le t$, but even then that is not enough; you also need $S_1+\cdots+S_4> t$.
The easier approach is to use the independent increments property of the Poisson process to note that for any $s>t$, the following are independent: $X_s-X_t$ (the number of arrivals by time $s$, after $t$) and $X_t$. In particular, this implies $X_s-X_t$ can be interpreted as another Poisson process with the same rate.
Then $E[W_5 \mid X_t=3]$ is simply $t$ plus the waiting time of the second arrival after time $t$ (which is $2/\lambda$).