poisson process help applying equations.

Question

Let $\{N (t), t ≥ 0\}$ be a Poisson process with rate $λ.$ Let $T_1 ={}$time of the first event, $T_n ={}$elapsed time between the $(n − 1)$-th and the $n$-th event, $S_n ={}$the time of the $n$-th event, $n = 1,2,\ldots$ Find $P[T_1 <s\mid N(t)=1],$ for $s≤t.$ Really stuck here and don't know where to begin.

Answer 1

We have $0<s<t.$ \begin{align} & \Pr(T_1<s\mid N(t)=1) \\[8pt] = {} & \frac{\Pr(T_1<s\ \&\ N(t)=1)}{\Pr(N(t)=1)} \\[8pt] = {} & \frac{\Pr(N(s)=1\ \&\ N(t)-N(s)=0)}{\Pr(N(t)=1)} \\[8pt] = {} & \frac{\Pr(N(s)=1) \cdot \Pr(N(t)-N(s)=0)}{\Pr(N(t)=1)} \end{align} Can you do the rest?

Answer 2

It is like calculating the conditional distribution of the first arrival time assuming an event has happened. Apply definition of conditional expectation:

$$P \left[T_1 < s\mid N_t=1\right]=\frac{P \left[N_s=1,\,N_{t-s}=0\right]}{P \left[N_t=1\right]}$$

The increments are independent, so the right hand side simplifies:

$$P \left[T_1 < s\mid N_t=1\right]=\frac{P \left[N_s=1\right]\, P \left[N_{t-s}=0\right]}{P \left[N_t=1\right]}$$

Now plug in the Poisson density (pmf), $p_n\left(t\right)=\frac{\left(\lambda t\right)^n}{n!}e^{-\lambda t}$, and simplify to see that the the conditional arrival time is uniform distributed:

$$P \left[T_1 < s\mid N_t=1\right]=\frac{\lambda s e^{-\lambda s} \;e^{-\lambda (t-s)}}{\lambda te^{-\lambda t}} =\frac s t$$