The problem of the distribution function of arrival epochs $S_i$ and time $t$ difference is in the Poisson process.

Question

Let $N(t)$ be a Poisson process, and the rate of arrival is 1. We know that at time $t=1$, $N(t)=3$.
(a) Find the distribution function of $S_4-1$.
(b) Find the $E[(S_4-1)|N(1)=3]$
(c) Find the $E[(S_4-S_3)|N(1)=3]$
I know that the distribution function of interval time $X_i$ can be expressed as:
$Pr\{X>x\}=F_X^{c}(x)=exp(-\lambda x)$.
So the answer to question A is this? $Pr\{S_4>X>1\}$
This is a hard problem for me and I hope to get your help. I was also confused about (b) and (c) and didn't know how to do it. It looks like in (c), the difference between $S_4$ and $S_3$ is $X_4$.

Answer 1

The important concept to grasp is that if you know $N(1)=3$, then you can treat the process from time $t=1$ onward as a brand new Poisson process with the same rate. (That is, at time $t=1$, just start a brand new Poisson process, and this will have the same distribution as the original process conditioned on $N(1)=3$.) This is a consequence of the independent increments property of Poisson processes.

Consequently, for (a) note that $S_4-1$ is the time between $t=1$ and the first arrival after $t=1$, so in the "new" Poisson process, this is simply "the time of the first arrival," whose distribution you know to be exponential with rate $\lambda=1$. This should allow you to immediately solve (b) as well.

For (c), it suffices to compute $E[S_3 \mid N(1)=3]$. Show that $$P(S_3 < t \mid N(1) = 3) = \frac{P(S_3 < t, N(1)=3)}{P(N(1)=3)} = \begin{cases} \frac{(e^{-\lambda t}(\lambda t)^3/3!)(e^{-\lambda(1-t)})}{e^{-\lambda} \lambda^3/ 3!} = t^3& 0 \le t \le 1 \\ 0 & \text{otherwise}\end{cases}$$ and compute $E[S_3 \mid N(1)=3] = \int_0^1 P(S_3 \ge t \mid N(1)=3) \, dt$.