= Solution
After the third arrival, the future interarrival times are <independent> of the first three interarrival times and again generate a rate-$\lambda$ <Poisson process>. Conditional on $T_3=t<2$, exactly three arrivals have occurred by time $t$, and the number of further arrivals before two is Poisson with mean $\lambda(2-t)$. Consequently, in the regular conditional-distribution sense at $t=1$,
$$
\boxed{N_2\mid(T_3=1)\ \sim\ 3+\operatorname{Poisson}(\lambda).}
$$
Explicitly its conditional <mass> is zero for $k<3$ and is
$$
\mathbb P(N_2=k\mid T_3=1)=e^{-\lambda}\frac{\lambda^{k-3}}{(k-3)!},\qquad k\geq3.
$$
This conditions on the third arrival time itself, rather than merely on the event that the third arrival has happened by time one.
Back to article page
