Poisson Factorial Moments

Introduction

A claim count is often summarized by its mean and variance, but some questions concern combinations of claims: how many distinct pairs of claims are expected, or what is the probability that at least one pair occurs? For a Poisson count, factorial moments provide a direct route to these quantities. They also make it straightforward to recover ordinary moments such as the mean and variance.

The key distinction is between the number of claims and the number of ways to select claims from the count. If \(N\) claims occur, there are \(N(N-1)\) ordered pairs of distinct claims, but only \(N(N-1)/2\) unordered pairs. Keeping that distinction explicit prevents a common factor-of-two error.

The calculations here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The focus is on deriving and using moments under the Poisson model, not on selecting a frequency model for a particular portfolio.

Poisson Counts and Factorial Moments

Let \(N\) be the number of claims in a specified, fixed exposure period. Assume \(N\) has a Poisson distribution with parameter \(\lambda\), where \(\lambda\geq 0\) is the expected number of claims during that period. Its probability mass function is

The parameter \(\lambda\) is tied to the stated exposure period. If a question gives an annual claim count, for example, the resulting moments and probabilities refer to a year unless the model is explicitly adjusted to another period.

For example, \((N)_1=N\), \((N)_2=N(N-1)\), and \((N)_3=N(N-1)(N-2)\). These expressions count ordered selections of distinct claims: if \(N=4\), then \((N)_2=4\cdot3=12\) ordered pairs. The falling factorial is zero when \(N<r\), because there are not enough claims to select \(r\) distinct claims.

Deriving the Poisson Factorial-Moment Formula

For a Poisson count, the factorial moments have a particularly simple form. The derivation starts from the definition of expectation and the Poisson probability mass function. For integer \(r\geq1\), terms with \(n<r\) contribute zero, so

Set \(j=n-r\). As \(n\) runs from \(r\) to infinity, \(j\) runs from 0 to infinity. The remaining sum is the power series for \(e^\lambda\), which gives

The cancellation in the derivation explains why falling factorials are useful: the \(n!\) in the Poisson probability mass function cancels the \(n!\) in \((n)_r\), leaving a shifted exponential series. The result applies to a Poisson count with the stated parameter; it should not be transferred to a different count model merely because that model has the same mean.

From Factorial Moments to Ordinary Moments

An ordinary moment uses powers such as \(N^2\) or \(N^3\), rather than falling factorials. Algebraic identities convert between the two forms. For the second moment,

Taking expectations and applying the Poisson factorial-moment formula gives the second raw moment. Subtracting the square of the mean then gives the variance.

For a higher-order example, expand \(N(N-1)(N-2)=N^3-3N^2+2N\), or rearrange this identity to obtain \(N^3=(N)_3+3(N)_2+(N)_1\). Therefore,

These identities are useful when a question asks for a raw moment directly, or when a moment expression must be simplified before a numerical value is calculated. Always identify whether the requested quantity is factorial, raw, or central; these are different summaries.

Claim Pairs: Counts and Probabilities

For a realized count \(N=n\), the number of distinct unordered pairs of claims is \(\binom{n}{2}\). Define \(K=\binom{N}{2}\) as the random number of such pairs. Since \(\binom{N}{2}=N(N-1)/2=(N)_2/2\), its expectation is

The factor \(1/2\) converts ordered pairs into unordered pairs. In contrast, \(E[N(N-1)]=\lambda^2\) is the expected number of ordered pairs of distinct claims.

An expected pair count is not itself a probability. For example, \(K\) can take values 0, 1, 3, 6, and so on; it is not restricted to 0 or 1. The event that at least one pair exists is exactly the event \(N\geq2\). Its probability must be calculated from the count probabilities:

The probability of exactly one unordered pair is different: this occurs when \(N=2\), because two claims make one pair, whereas three claims make three pairs. Thus \(P(K=1)=P(N=2)=e^{-\lambda}\lambda^2/2\).

Common Exam Traps and a Reliable Workflow

Pair-related questions can use similar-looking expressions while asking for different quantities. The following checks help keep the calculations aligned with the wording.

• Confirm the moment type. \(E[N(N-1)]\) is the second factorial moment, while \(E[N^2]\) is the second raw moment. For Poisson \(N\), the former is \(\lambda^2\), but the latter is \(\lambda^2+\lambda\).
• Check whether pairs are ordered. \(N(N-1)\) counts ordered pairs; \(\binom{N}{2}\) counts unordered pairs. Divide by 2 for the unordered count.
• Separate expectation from probability. \(E[\binom{N}{2}]=\lambda^2/2\) is an average number of pairs, not the probability that any pair occurs.
• Translate the event into count values. At least one pair means \(N\geq2\). Exactly one pair means \(N=2\), not \(N\leq2\) or \(N\geq2\).
• Use the parameter for the stated period. A monthly parameter cannot be used for a quarterly event without first expressing the model for that quarter.

A practical sequence is to identify the random count and its Poisson parameter, translate the requested quantity into a formula in \(N\), and then decide whether that formula calls for a factorial moment or for individual probabilities. After calculating, verify that the units and interpretation match the question.