Nonhomogeneous Poisson Process

Study guide: Exam MAS-I

A Nonhomogeneous Poisson Process (NHPP) is a Poisson Process whose rate changes with time: events arrive at intensity $\lambda(t)$ rather than at a constant $\lambda$ . Counts over an interval are still Poisson, but the mean is the integrated rate over that interval.

$m(t) = \int_0^{t} \lambda(u)\,du$

$N(b) - N(a) \sim \text{Poi}\!\left(\int_a^b \lambda(u)\,du\right)$

Independent increments still hold, but stationary increments do not: two intervals of the same length can have different expected counts (their counts share a distribution only when the integrated rates over them are equal)
$m(t)$ is called the mean value function; $E[N(t)] = \text{Var}(N(t)) = m(t)$
When $\lambda(t) > 0$ , so that $m$ is strictly increasing, an NHPP becomes a rate- $1$ homogeneous process under the time change $s = m(t)$ — this is what makes seasonal or trending claim arrivals tractable
Thinning a homogeneous process with a time-varying probability $p(t)$ produces an NHPP with rate $\lambda p(t)$ ; conversely, splitting an NHPP by event type gives independent NHPPs
Actuarial uses: hurricane arrivals concentrated in one season, claim reporting patterns that decay after the accident date, and exposure growth over a policy year
The waiting time to the first event has survival function $P(T_1 > t) = e^{-m(t)}$ , so the Hazard Rate of $T_1$ is exactly $\lambda(t)$

Expected Claims Under a Seasonal Rate Claims arrive as an NHPP with $\lambda(t) = 3 + 2t$ per month, where $t$ is measured in months from the start of the year. Find the expected number of claims in the first four months and the probability of no claims in month 1.

Answer $m(4) = \int_0^4 (3 + 2u)\,du = \left[3u + u^2\right]_0^4 = 12 + 16 = 28$ So $N(4) \sim \text{Poi}(28)$ and $E[N(4)] = 28$ . For month 1, the integrated rate is $\int_0^1 (3+2u)\,du = 3 + 1 = 4$ , so $P(N(1) = 0) = e^{-4} \approx 0.0183$

Counts on a Later Interval With the same rate $\lambda(t) = 3 + 2t$ , find the distribution of the number of claims between $t = 4$ and $t = 6$ .

Answer $\int_4^6 (3 + 2u)\,du = \left[3u + u^2\right]_4^6 = (18 + 36) - (12 + 16) = 26$ The count is $\text{Poi}(26)$ : $26/2 = 13$ expected claims per month on $(4, 6]$ , against $28/4 = 7$ per month on $(0, 4]$ , because the rate is rising. Note this is a two-month window carrying nearly the same expected count as the first four months combined.