Interarrival Time

Study guide: Exam MAS-I

The Interarrival Time $T_n$ is the waiting time between the $(n-1)$ -th and $n$ -th event of a counting process. For a Poisson Process with rate $\lambda$ , the interarrival times are independent and identically distributed exponential variables with mean $1/\lambda$ — and that property is an equivalent definition of the Poisson process.

$T_n \stackrel{\text{iid}}{\sim} \text{Exp}(\lambda), \qquad E[T_n] = \frac{1}{\lambda}$

$S_n = T_1 + \cdots + T_n \sim \text{Gamma}(n, \lambda)$

The counting view and the waiting-time view are two descriptions of the same process: $N(t) \ge n \iff S_n \le t$
Memoryless: $P(T > s + t \mid T > s) = P(T > t)$ . Time already spent waiting tells you nothing about the wait remaining, so there is no "due for a claim" effect
The arrival time $S_n$ of the $n$ -th event is Gamma (Erlang) with shape $n$ and rate $\lambda$ , mean $n/\lambda$ and variance $n/\lambda^2$
Given $N(t) = n$ , the $n$ arrival times are distributed as the Order Statistics of $n$ independent $\text{Uniform}(0, t)$ draws — arrivals are "scattered at random" over the interval
For a Nonhomogeneous Poisson Process the interarrival times are neither identically distributed nor exponential; only after the time change $s = m(t)$ do they become $\text{Exp}(1)$
Competing risks: if two independent Poisson streams have rates $\lambda_1$ and $\lambda_2$ , the wait for the first event of either is $\text{Exp}(\lambda_1 + \lambda_2)$ , and it is of type 1 with probability $\lambda_1/(\lambda_1 + \lambda_2)$

Waiting for the Third Claim Claims arrive at $\lambda = 4$ per month. Find the expected time until the third claim and the probability that the next claim takes more than one month.

Answer $E[S_3] = \frac{3}{\lambda} = \frac{3}{4} = 0.75 \text{ months}$ $P(T > 1) = e^{-\lambda t} = e^{-4} \approx 0.0183$ Equivalently, $P(T > 1) = P(N(1) = 0)$ — no claims in the month is the same event as a wait longer than a month.

Memorylessness in Practice A claims department has waited three weeks with no large claim, in a process with $\lambda = 1$ large claim per month. What is the probability it waits at least another month?

Answer $P(T > 0.75 + 1 \mid T > 0.75) = P(T > 1) = e^{-1} \approx 0.368$ The three quiet weeks are irrelevant: the remaining wait is again exponential with mean one month. Any argument that a large claim is "overdue" contradicts the Poisson assumption.