Frequency Moments From Probability Generating Functions

Why Use a Probability Generating Function?

A claim-frequency model describes the random number of claims \(N\) in a specified period. Once a model has been selected, the mean and variance summarize its central tendency and spread. A probability generating function (PGF) provides a systematic way to obtain these quantities, as well as factorial moments, directly from the probabilities of the count.

The method is especially useful because the same derivative rules apply to binomial, Poisson, and negative binomial counts. It also helps prevent a frequent calculation error: the second derivative at 1 is not the variance. It gives the second factorial moment, which must be converted to a variance using the mean.

The ordinary second moment and the variance can be recovered from these derivatives. Since \(N^2=N(N-1)+N\), the second raw moment is \(E[N^2]=G_N''(1)+G_N'(1)\). Therefore,

These formulas require the stated moments to exist. The binomial, Poisson, and negative binomial models considered here have finite moments of every order for their usual parameter values.

Why the Derivative Rules Work

Write \(p_k=P(N=k)\). The PGF is the power series \(G_N(z)=\sum_{k=0}^{\infty}p_k z^k\). Differentiating term by term where the series and derivatives converge gives

At \(z=1\), each factor \(z^{k-1}\) or \(z^{k-2}\) equals 1. Thus the first derivative sums \(kP(N=k)\), which is \(E[N]\), and the second sums \(k(k-1)P(N=k)\), which is \(E[N(N-1)]\). The variance formula then follows from \(N^2=N(N-1)+N\) and \(\operatorname{Var}(N)=E[N^2]-(E[N])^2\). This is why the second derivative alone cannot be reported as either the variance or the ordinary second moment.

Binomial Claim Counts

Suppose \(N\) is the number of claims among \(n\) independent exposures, each of which produces one claim with probability \(p\). Then \(N\sim\operatorname{Bin}(n,p)\), and its PGF is

Differentiating and evaluating at 1 gives

Consequently, \(E[N]=np\), \(E[N(N-1)]=n(n-1)p^2\), and \(\operatorname{Var}(N)=np(1-p)\). The second factorial moment counts ordered pairs of distinct claim-producing exposures in expectation; this explains why its factor is \(n(n-1)\), not \(n\).

Poisson Claim Counts

For a Poisson count with parameter \(\lambda>0\), the probability of \(k\) claims is \(e^{-\lambda}\lambda^k/k!\), for \(k=0,1,2,\ldots\). Its PGF follows by summing the probability series:

Differentiating gives \(G_N'(z)=\lambda e^{\lambda(z-1)}\) and \(G_N''(z)=\lambda^2e^{\lambda(z-1)}\). Thus \(E[N]=\lambda\), \(E[N(N-1)]=\lambda^2\), and

The equality of the Poisson mean and variance is a feature of this model, not a general relationship between a count's mean and variance.

Negative Binomial Claim Counts

Use the following parameterization throughout this section. Let \(N\) count failures before the \(r\)th success in independent Bernoulli trials, with success probability \(p\) and failure probability \(q=1-p\). For a positive integer \(r\), the PGF is

This defines a count of failures, not a count of trials including the final success. Parameterizations of the negative binomial differ, so identifying what \(N\) counts is essential before using a formula.

Differentiating the PGF yields

At \(z=1\), \(1-q=p\), so

Using the derivative-to-variance identity gives \(\operatorname{Var}(N)=rq/p^2\). The second raw moment is \(r(r+1)q^2/p^2+rq/p\). These results also apply for the standard extension to positive real \(r\), when that negative binomial distribution is used.

Moment Relationships and an Exam Trap

For any count with a finite second moment, the conversions to remember are

The PGF makes the first two factorial moments convenient to obtain, while many questions ask for variance or the ordinary second moment. Keep the requested quantity visible at each step and convert only after identifying the derivatives.

Common mistake. In the binomial example with \(n=10\) and \(p=0.30\), an incorrect calculation might report \(E[N(N-1)]=np(1-p)=2.1\). That expression is the variance, not the second factorial moment. The correction is

A check using the variance identity confirms the correction: \(8.1+3-3^2=2.1\). Substituting \(2.1\) for the factorial moment instead would not reproduce the variance.