Model claim frequencies using Poisson processes.

A health insurer logs claims as they arrive through the day. The arrival times look random, but the count over any hour is remarkably predictable. The Poisson process formalizes this and underpins almost every frequency model on MAS-I.

The three axioms. A counting process $\{N(t):t\ge 0\}$ is a homogeneous Poisson process with rate $\lambda >0$ if:

• $N(0)=0$.
• Independent increments: counts in disjoint intervals are independent.
• Stationary increments: the distribution of $N(t+s)-N(s)$ depends only on the length $t$, and equals $\text{Poisson}(\lambda t)$.

KEY: Independent plus stationary plus Poisson-distributed increments is the working definition. Some textbooks state two extra "small interval" conditions (one event with probability $\lambda h+o(h)$, two or more with $o(h)$); they are equivalent.

The probability of exactly $k$ claims in an interval of length $t$ is given below.

$$P(N(t)=k)=\frac{e^{-\lambda t}(\lambda t{)}^k}{k!},k=0,1,2,\dots$$

Mean and variance both equal $\lambda t$. That equality is the first thing diagnostic plots check.

Inter-arrival and waiting times. Let $T_i$ be the gap between the $(i-1)$-th and $i$-th claim. The $T_i$ are iid $\text{Exponential}(\lambda )$, each with mean $1/\lambda$ and variance $1/{\lambda }^2$.

Common mistakes

• Using $\text{Var}(X)$ instead of $E[X^2]$ in the compound Poisson variance. The correct formula is $\text{Var}(S)=\lambda tE[X^2]$. Plugging in variance alone understates the answer.
• Forgetting that thinned streams are independent. Candidates often assume large-claim and small-claim counts must be negatively correlated because they share a parent process. Under independent classification, they are exactly independent Poisson processes.
• Treating an NHPP rate as if stationary. The count over $[a,b]$ depends on ${\int }_a^b\lambda (s)ds$, not on $\lambda$ evaluated at the midpoint or endpoint. Always integrate.

Bottom line

• Homogeneous Poisson process with rate $\lambda$: $N(t)\sim \text{Poisson}(\lambda t)$ with independent stationary increments, $N(0)=0$, and mean and variance both equal to $\lambda t$.
• Inter-arrival times are iid $\text{Exponential}(\lambda )$ with mean $1/\lambda$; the memoryless property means past waiting carries no information about future waiting.
• Waiting time to the $n$-th claim is $\text{Gamma}(n,\lambda )$ with mean $n/\lambda$.
• Superposition: the sum of independent Poisson processes with rates ${\lambda }_1,{\lambda }_2$ is Poisson with rate ${\lambda }_1+{\lambda }_2$; independence is required.

Exam shortcut

If a problem gives exponential gaps, immediately convert to a Poisson count over the relevant interval; the two views unlock different formulas for the same probability. When asked about a sub-population (large losses, bodily-injury, female drivers), apply thinning first to get a clean Poisson sub-process, then answer inside that sub-process. For aggregate-loss questions, memorize $E[S]=\lambda tE[X]$ and $\text{Var}(S)=\lambda tE[X^2]$; reaching for the variance of $X$ is the single most common arithmetic trap.

The full lesson (about 1,888 words, 13 min read) adds 2 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• A1

Browse all free MAS-I lessons or jump into free MAS-I practice questions.