Use Probability Generating Functions for Covered Aggregate Loss

Why Use a Probability Generating Function?

When a claim-count model is combined with discrete claim payments, the aggregate covered loss can have many possible values. A probability generating function (PGF) provides a compact way to describe that distribution. It is especially useful when a contract changes the severity distribution by creating a zero payment, capping a payment, or mapping ground-up losses onto a payment lattice. The PGF can give individual aggregate probabilities and moments without listing every possible combination of claims.

The construction depends on the model: the number of claims must be independent of the claim payments, and the payment amounts must be identically distributed and independent of one another. Coverage is applied to each claim before the aggregate is formed. These assumptions are stated explicitly in the results and examples below.

Definition: For a nonnegative integer-valued random variable $W$, its probability generating function is $G_W(z)=E[z^W]$, for values of $z$ where the expectation exists. If $P(W=j)=p_j$, then $G_W(z)=\sum_{j=0}^{\infty}p_jz^j$. The coefficient of $z^j$ is $P(W=j)$, and the probabilities represented by the coefficients sum to 1.

For monetary losses on a lattice, express amounts in a chosen unit, such as one dollar or one thousand dollars. The PGF variable then counts those units. For example, if the unit is $500, a covered payment of $1,000 is represented by the exponent 2. A payment of zero contributes $z^0=1$; it must not be omitted merely because the insurer pays nothing on that claim.

From a Covered Claim to the Aggregate PGF

Let $N$ be the number of claims, with PGF $G_N(z)$. Let $Y_1,Y_2,\ldots$ be the covered payments per claim, expressed as nonnegative integer numbers of the chosen monetary unit. Assume that, conditional on the claim count, the payments are iid with PGF $G_Y(z)$, and that $N$ is independent of the payments. Define the aggregate covered loss by $S=\sum_{j=1}^{N}Y_j$, with $S=0$ when $N=0$.

Theorem: Under the independence and identical-distribution assumptions just stated, the aggregate PGF is $$ G_S(z)=G_N(G_Y(z)). $$ The coefficient of $z^s$ in this function is $P(S=s)$, where $s$ is measured in the selected monetary unit.

To see why the formula holds, condition on $N=n$. Independence makes the PGF of the sum of $n$ payments the product of their PGFs, $G_Y(z)^n$. Averaging this conditional PGF over the possible values of $N$ gives

G_S(z)=\sum_{n=0}^{\infty}P(N=n)G_Y(z)^n =G_N(G_Y(z)).

The result is a composition: substitute the payment PGF for the argument of the frequency PGF. A standard example is Poisson frequency, $N\sim\operatorname{Poisson}(\lambda)$, for which $G_N(z)=\exp(\lambda(z-1))$. The compound aggregate PGF is therefore $\exp(\lambda(G_Y(z)-1))$.

Moments From the PGF

For an integer-valued $W$, differentiation of its PGF gives $G_W'(1)=E[W]$ and $G_W''(1)=E[W(W-1)]$, when the corresponding moments are finite. Since $W^2=W(W-1)+W$, it follows that

E[W]=G_W'(1),\qquad \operatorname{Var}(W)=G_W''(1)+G_W'(1)-\bigl(G_W'(1)\bigr)^2.

Applying the chain rule to $G_S(z)=G_N(G_Y(z))$ gives $G_S'(1)=G_N'(1)G_Y'(1)$. A second differentiation gives $G_S''(1)=G_N''(1)(G_Y'(1))^2+G_N'(1)G_Y''(1)$. Equivalently, in terms of the claim count and the payment amount,

E[S]=E[N]E[Y],\qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2.

These formulas require the stated independence and iid payment assumptions and finite second moments. The PGF composition also requires a discrete payment distribution on the selected lattice. For dependent claim payments or severity distributions that are not represented on a lattice, this particular coefficient-based approach does not directly apply.

Worked Example: A Binomial Count With Deductible-Limit Payments

A hypothetical portfolio has two independent exposure units. Each produces a claim with probability 0.5, so $N\sim\operatorname{Binomial}(2,0.5)$. Conditional on a claim, ground-up loss $X$ is $1,000, $2,000, or $4,000 with probabilities 0.2, 0.5, and 0.3, respectively. The policy has per-claim deductible $d=\$1{,}000$ and maximum payment $u=\$2{,}000$, so the payment is $Y=\min((X-d)_+,u)$. Find the aggregate PGF, $P(S\le \$1{,}000)$, and the mean and variance of $S$.

Method and covered severity. Measure payments in units of $1,000. The three ground-up losses produce payments of $0, $1,000, and $2,000. Thus the payment-unit random variable has probabilities $P(Y=0)=0.2$, $P(Y=1)=0.5$, and $P(Y=2)=0.3$, and

$$ G_Y(z)=0.2+0.5z+0.3z^2. $$

The zero-payment outcome remains in this PGF because $N$ counts claims, not positive covered payments. The count PGF is $G_N(z)=(0.5+0.5z)^2$. Therefore

G_S(z)=\left(0.5+0.5(0.2+0.5z+0.3z^2)\right)^2 =0.25(1.2+0.5z+0.3z^2)^2.

Probability. Expanding the square, the coefficients of $z^0$ and $z^1$ are

P(S=0)=0.25(1.2)^2=0.36,\qquad P(S=1)=0.25\bigl(2(1.2)(0.5)\bigr)=0.30.

Consequently, $P(S\le \$1{,}000)=0.36+0.30=0.66$. The probability is expressed for aggregate payments of zero or one unit of $1,000.

Moments. The payment mean is $E[Y]=0(0.2)+1(0.5)+2(0.3)=1.1$ units. Also $E[Y^2]=0+1(0.5)+4(0.3)=1.7$, so $\operatorname{Var}(Y)=1.7-(1.1)^2=0.49$. For the binomial count, $E[N]=2(0.5)=1$ and $\operatorname{Var}(N)=2(0.5)(0.5)=0.5$. Hence

E[S]=1(1.1)=1.1\text{ thousand dollars}=\$1{,}100, \qquad \operatorname{Var}(S)=1(0.49)+0.5(1.1)^2=1.095 \text{ (thousand dollars)}^2.

Check. Expanding the full PGF gives coefficients $0.36, 0.30, 0.2425, 0.075,$ and $0.0225$, which sum to 1. The coefficient probabilities for zero and one unit agree with the direct calculation. The common mistake avoided here is applying the deductible to the aggregate rather than separately to each claim.

Worked Example: A Compound Poisson Probability Through Two Units

A hypothetical claim process has $N\sim\operatorname{Poisson}(0.75)$. Each claim produces a covered payment of one unit with probability $2/3$, or two units with probability $1/3$, independently of the count and other payments. Find $P(S\le 2)$ and the first two moments of the aggregate, where one unit is $1,000.

Method. The payment PGF is $G_Y(z)=\frac23z+\frac13z^2$. Thus the compound Poisson PGF is

G_S(z)=\exp\left(0.75\left(\frac23z+\frac13z^2-1\right)\right) =e^{-0.75}\exp(0.5z+0.25z^2).

To obtain the coefficients through $z^2$, use $\exp(0.5z)=1+0.5z+0.125z^2+\cdots$ and $\exp(0.25z^2)=1+0.25z^2+\cdots$. Multiplication gives coefficients $1$, $0.5$, and $0.125+0.25=0.375$ for powers zero, one, and two, respectively. Therefore

P(S\le 2)=e^{-0.75}(1+0.5+0.375) =1.875e^{-0.75}\approx 0.8857.

The approximation is rounded to four decimal places. In particular, $1.875e^{-0.75}=0.8856872828\ldots$, which rounds to $0.8857$.

For the payment, $E[Y]=\frac23(1)+\frac13(2)=\frac43$ units, and $E[Y^2]=\frac23(1)+\frac13(4)=2$ units squared. For a Poisson count with mean $\lambda$, the compound aggregate has mean $\lambda E[Y]$ and variance $\lambda E[Y^2]$. Here these are $0.75(4/3)=1$ unit and $0.75(2)=1.5$ units squared: $E[S]=\$1{,}000$ and $\operatorname{Var}(S)=1.5$ million dollars squared.

Check and trap. The probability $P(S=2)$ includes either one two-unit payment or two one-unit payments; the coefficient $0.375$ includes both contributions. Treating every claim as a one-unit payment would omit the two-unit possibility and give the wrong aggregate distribution. The probabilities through two units sum to $0.8857$, leaving a nonnegative probability of approximately $0.1143$ above two units.

Worked Example: Positive Payments After a Deductible

A hypothetical insurer covers three independent exposure units. Each has claim probability 0.4. Conditional on a claim, the ground-up loss is $400 or $900 with equal probability. The policy has deductible $d=\$400$ and per-claim limit $u=\$500$. Find the probability the aggregate covered payment is at least $1,000, its mean, and its variance.

Method and PGF. The contract pays $Y=\min((X-400)_+,500)$, so the payment is zero for a $400 loss and $500 for a $900 loss. The severity PGF in $500 units is $G_Y(z)=0.5+0.5z$. The claim count has PGF $G_N(z)=(0.6+0.4z)^3$; hence

G_S(z)=G_N(G_Y(z)) =\left(0.6+0.4(0.5+0.5z)\right)^3 =(0.8+0.2z)^3.

This expansion is the PGF of a binomial count with three trials and probability 0.2 of a $500 payment per exposure unit. The aggregate is at least $1,000 exactly when there are two or three such payments:

P(S\ge \$1{,}000) =\binom32(0.2)^2(0.8)+(0.2)^3 =3(0.04)(0.8)+0.008 =0.1040.

For the moment calculations, $E[Y]=0.5$ units of $500 and $\operatorname{Var}(Y)=0.25$ units squared. Also $E[N]=3(0.4)=1.2$ and $\operatorname{Var}(N)=3(0.4)(0.6)=0.72$. Thus

E[S]=1.2(0.5)=0.6\text{ units}=\$300, $$ $$ \operatorname{Var}(S)=1.2(0.25)+0.72(0.5)^2 =0.30+0.18=0.48\text{ units}^2 =0.48(500)^2=\$120{,}000.

Check and trap. The expanded PGF is $0.512+0.384z+0.096z^2+0.008z^3$; its coefficients sum to 1, and the last two sum to $0.104$. A common error is to use 0.4 as the probability of a positive payment. Only half of the claims exceed the deductible, so the probability of a payment per exposure unit is $0.4(0.5)=0.2$.

Practical Checks and Limits

PGFs are useful when the requested probability is a coefficient or a sum of a few coefficients, and when derivatives provide a convenient moment check. They also give a clear way to preserve zero payments arising from deductibles. They do not remove the need to identify the correct frequency basis: if the count is the number of claims, retain zero-payment claims in $G_Y$; if the count is explicitly the number of positive covered payments, use the conditional distribution given a positive payment and the corresponding count model.

Before interpreting a coefficient, verify the monetary unit and the exponent. A coefficient of $z^3$ means three selected units of aggregate payment, not three claims. Also verify that the coefficients are nonnegative and sum to 1 when the PGF has been fully expanded. For a compound Poisson model, the mean and variance are particularly convenient checks: $\lambda E[Y]$ and $\lambda E[Y^2]$, respectively. The variance uses the payment second moment, not just the payment variance.

The composition method and coefficient extraction here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. They build on the earlier material on covered claim counts and policy-level coverage changes, while relying on the stated independence and lattice assumptions.

Check Your Understanding

For each problem, construct or interpret the relevant PGF before using the final answer as a check.

A hypothetical claim count is Poisson with mean 0.4. Each claim has covered payment $1,000 with probability 0.5 and zero otherwise. Find the probability of no aggregate payment. (Answer: $e^{-0.2}\approx 0.8187$.)
There are four independent exposure units, each with claim probability 0.25. Conditional on a claim, payment is $2,000 with probability 0.5 and zero otherwise. Find the mean aggregate payment. (Answer: $\$1{,}000$.)
A Poisson claim count has mean 0.6. Independent payments are $1,000 or $3,000 with equal probabilities. Find the probability that the aggregate payment is at most $1,000. (Answer: $1.3e^{-0.6}\approx 0.7135$.)
Two exposure units each have claim probability 0.3. Conditional on a claim, payment is $4,000 with probability 0.75 and zero otherwise. Find the probability the aggregate is exactly $4,000. (Answer: $2(0.225)(0.775)=0.34875$, or $0.3488$ to four decimals.)
A Poisson count has mean 0.5. Each claim independently pays $2,000 with probability 0.8 or $5,000 with probability 0.2. Find the mean and variance of aggregate payment. (Answer: Mean $\$1{,}300$; variance $4.1$ million dollars squared.)

What You'll Learn