Suppose customer arrivals at a post office are modeled by a Poisson process N with intensity λ > 0. Let T1 be the time of the first arrival. Let t > 0. Suppose we learn that by time t there has been precisely one arrival, in other words, Nt = 1. What is the distribution of T1 under this new information? In other words, find the conditional probability P (T1 ≤ s|Nt = 1) for all s ≥ 0.

Step-by-step explanation:

We need to find the conditional probability P (T1 < s|N (t) = 1) for all s ≥ 0

P (time of the first person's arrival < s till time t exactly 1 person has arrived)

= P (time of the first person's arrival < s, till time t exactly 1 person has arrived) / P (exactly 1 person has arrived till time t)

{ As till time t, we know that exactly 1 person has arrived, thus relevant values of s : 0 < s < t }

P (time of the first person arrival < s, till time t exactly 1 person has arrived) / P (exactly 1 person has arrived till time t)

= P (exactly 1 person has arrived till time s) / P (exactly 1 person has arrived till time t)

P (exactly x person has arrived till time t) ~ Poisson (kt) where k = lambda

Therefore,

P (exactly 1 person has arrived till time s) / P (exactly 1 person has arrived till time t)

= [ kse-ks/1! ] / [ kte-kt/1! ]

= (s/t) e-k (s-t)