Poisson Distribution: Derivation, Properties, and Insurance Applications

Definition and Derivation

The Poisson distribution models the count of events occurring in a fixed interval when events happen independently at a constant average rate. If X ~ Poisson(lambda), the PMF is P(X = k) = e^(-lambda) * lambda^k / k! for k = 0, 1, 2, ... Both the mean and variance equal lambda.

The Poisson arises as a limit of the Binomial(n, p) when n is large, p is small, and lambda = np is moderate. This approximation is useful when computing exact binomial probabilities is impractical. For Exam P, if n >= 50 and p <= 0.05, the Poisson approximation is generally appropriate.

The conditions for a Poisson process are: (1) events occur one at a time, (2) the occurrence of an event in one interval is independent of occurrences in any non-overlapping interval, and (3) the expected number of events in an interval is proportional to the length of the interval. When these conditions hold, the count of events in a fixed interval follows a Poisson distribution.

Key Properties

The MGF of X ~ Poisson(lambda) is M(t) = exp(lambda(e^t - 1)). A critical property: the sum of independent Poisson random variables is Poisson. If X ~ Poisson(lambda1) and Y ~ Poisson(lambda2) independently, then X + Y ~ Poisson(lambda1 + lambda2). This makes the Poisson distribution particularly convenient for combining claim counts from independent lines of business.

The Poisson distribution is right-skewed, with skewness = 1/sqrt(lambda). As lambda increases, the distribution becomes approximately normal.

A useful recursive formula for computing Poisson probabilities is P(X = k+1) = (lambda / (k+1)) * P(X = k), starting from P(X = 0) = e^(-lambda). This is efficient for building a probability table without computing factorials. On Exam P, this recursion can save time when you need several consecutive probabilities.

The factorial moments of the Poisson have an elegant form: E[X(X-1)(X-2)...(X-k+1)] = lambda^k. In particular, E[X(X-1)] = lambda^2, which gives E[X^2] = lambda^2 + lambda. This confirms Var(X) = lambda^2 + lambda - lambda^2 = lambda.

Poisson Thinning and Splitting

If X ~ Poisson(lambda) and each event is independently classified as type A with probability p or type B with probability 1-p, then the number of type A events follows Poisson(lambda * p) and the number of type B events follows Poisson(lambda * (1-p)). Moreover, these two counts are independent. This is called the thinning or splitting property.

Example: An insurance company receives claims at a Poisson rate of 10 per week. Each claim is independently classified as "large" (probability 0.2) or "small" (probability 0.8). The number of large claims per week follows Poisson(2), and the number of small claims follows Poisson(8), independently of each other.

This property extends to more than two categories. If events are classified into k types with probabilities p1, p2, ..., pk (summing to 1), the counts follow independent Poisson distributions with parameters lambda * p1, lambda * p2, ..., lambda * pk. This is extremely useful on Exam P for problems involving compound classification.

Conditional Distributions

If X ~ Poisson(lambda1) and Y ~ Poisson(lambda2) are independent, and S = X + Y ~ Poisson(lambda1 + lambda2), then the conditional distribution of X given S = n is Binomial(n, lambda1/(lambda1 + lambda2)). This result connects the Poisson and binomial distributions and occasionally appears on Exam P.

The intuition is that given the total count, each event is independently assigned to X or Y with probabilities proportional to their rates. This is analogous to the thinning property viewed from the opposite direction.

Insurance Applications

In insurance, the Poisson distribution is the standard model for claim frequency. If an insurer expects 3 claims per month on average, the probability of 0 claims is e^(-3) = 0.0498, and the probability of more than 5 claims is 1 - P(X <= 5). The Poisson assumption underlies many actuarial models including aggregate loss distributions and credibility theory. Exam P problems often combine Poisson frequency with a severity distribution to compute expected total losses.

A common Exam P problem type: if claims arrive at a Poisson rate and each claim amount is independent with some distribution, find E[total losses] and Var[total losses]. If N ~ Poisson(lambda) is the claim count and X1, X2, ... are iid claim amounts, then total losses S = X1 + X2 + ... + X_N. By the Wald equation, E[S] = E[N] * E[X] = lambda * E[X]. For variance, Var[S] = lambda * E[X^2]. This is a special case of the compound Poisson model.

Common Pitfalls

A frequent error is using the Poisson distribution when the variance does not approximately equal the mean. If observed data show variance much larger than the mean (overdispersion), the Poisson model is inappropriate. In such cases, a negative binomial or mixed Poisson model might be better.

Another mistake is applying the Poisson distribution to counts that have an upper bound. The Poisson has infinite support (X can take any non-negative integer value), so it is suitable for modeling events with no theoretical maximum. If the count is bounded (for example, the number of defective items in a fixed batch), the binomial distribution is more appropriate.

Students also sometimes confuse the rate parameter with the probability. If lambda = 3, this does not mean P(X = 1) = 3. Lambda is the expected number of events, and P(X = 1) = 3 * e^(-3) = 0.1494.

Study Tips

For Exam P, memorize P(X = 0) = e^(-lambda) and use the recursive formula to compute subsequent probabilities. Know the splitting property and the connection to the binomial via conditioning. Practice problems where Poisson counts are combined or where you must determine the distribution of a sum of Poisson variables. Keep in mind that the Poisson is the "default" claim frequency model in actuarial science, so it appears in many problem contexts. On exam day, if you see a problem involving event counts in a fixed interval with independent occurrences, think Poisson first.