poisson process help applying equations. 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.
 A: 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:
$$P \left[T_1 < s\mid N_t=1\right]=\frac{\lambda s e^{-\lambda s} \;e^{-\lambda (t-s)}}{\lambda te^{-\lambda t}} =\frac s t$$
A: 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?
