Intensity Function

Study guide: Exam MAS-I

The Intensity Function $\lambda(t)$ is the instantaneous event rate of a Nonhomogeneous Poisson Process: at time $t$ the probability of one event in a short interval of length $h$ is approximately $\lambda(t)\,h$ . Its integral, the mean value function $m(t)$ , is the expected number of events by time $t$ .

$m(t) = \int_0^t \lambda(s)\, ds$

$N(t+s) - N(s) \sim \text{Poi}\big(m(t+s) - m(s)\big)$

Formally (Ross): $P\big(N(t+h) - N(t) = 1\big) = \lambda(t)h + o(h)$ and $P\big(N(t+h) - N(t) \geq 2\big) = o(h)$ . $\lambda(t)$ is a rate, in events per unit time, not a probability — it can exceed $1$ .
A constant intensity $\lambda(t) = \lambda$ gives $m(t) = \lambda t$ and the ordinary Poisson Process. Expected counts are areas under the intensity curve, so a question giving $\lambda(t)$ piecewise is an integration exercise.
The first event time has survival function $e^{-m(t)}$ , so $\lambda(t)$ is exactly the Hazard Rate of the first arrival — the same mathematics as a Survival Model with force of mortality $\mu(t)$ .
Given $N(t) = n$ , the $n$ event times are independent with density $\lambda(s)/m(t)$ on $(0, t)$ . The number falling in $(a, b]$ is therefore binomial with $p = \big(m(b) - m(a)\big)/m(t)$ .
If $\lambda(t) \leq \lambda$ for all $t$ , the process can be built by thinning a rate- $\lambda$ process, keeping an event at time $t$ with probability $\lambda(t)/\lambda$ .
Uses: seasonal catastrophe arrivals, exposure growing through a policy year, claim reporting that tails off after the accident date.

A Triangular Storm Season During a six-month season, storms arrive with intensity $\lambda(t) = t$ for $0 \leq t \leq 3$ and $\lambda(t) = 6 - t$ for $3 \leq t \leq 6$ ( $t$ in months, storms per month).

(a) Find the expected number of storms in the season. (b) Find the probability of no storm in month 1. (c) Find the probability of exactly two storms in months 3 and 4 ( $2 < t \leq 4$ ).
Answer (a) $m(6)$ is the area of a triangle with base $6$ and height $3$ : $\tfrac{1}{2}(6)(3) = 9$ storms.

(b) $m(1) = \int_0^1 s\,ds = 0.5$ , so $P(N(1) = 0) = e^{-0.5} = 0.6065$ .

(c) Integrate over the window, splitting at the peak:
 $\begin{align*} m(4) - m(2) &= \int_2^3 s\,ds + \int_3^4 (6 - s)\,ds \\ &= 2.5 + 2.5 \\ &= 5 \end{align*}$ 
 $\begin{align*} P\big(N(4) - N(2) = 2\big) &= \frac{e^{-5}\,5^2}{2!} \\ &= 0.0842 \end{align*}$ 
The two peak months carry $5$ of the season's $9$ expected storms.

Where the Season's Storms Fell Using the same intensity, the season produced exactly $3$ storms. Find the expected number that arrived in the first two months, and the probability that none did.
Answer Given $N(6) = 3$ , each storm independently falls in $(0, 2]$ with probability
 $\begin{align*} p &= \frac{m(2)}{m(6)} \\ &= \frac{2}{9} \end{align*}$ 
The expected number is $3 \times 2/9 = 0.667$ , and $P(\text{none}) = (7/9)^3 = 0.4705$ . Note that the first two months are a third of the season but carry only $2/9$ of its intensity.