Probability Generating Functions For Discrete Aggregates

Why Compose Generating Functions?

An aggregate loss combines two sources of randomness: the number of claims and the amount of each claim. When individual claim amounts lie on a lattice, a probability generating function (PGF) provides a compact way to represent that aggregate distribution. Its key operation is composition: substitute the severity PGF into the frequency PGF. The coefficients of the resulting function give the probabilities of the possible aggregate amounts.

This representation is useful for organizing calculations, especially when different claim counts can lead to the same total. A coefficient must include every such possibility. The method below makes those contributions visible without requiring a separate case for every combination of claim amounts.

Definition: For a nonnegative integer-valued random variable $K$, its probability generating function is $G_K(z)=E[z^K]$. If $P(K=k)=p_k$, then $G_K(z)=\sum_{k=0}^{\infty}p_kz^k$. The coefficient of $z^k$ is $P(K=k)$. In an insurance model, $N$ denotes the claim count, $X_i$ the amount of claim $i$, and $S=\sum_{i=1}^{N}X_i$ the aggregate loss.

The Compound Aggregate PGF

Suppose claim amounts are measured in a common lattice unit, so each $X_i$ is a nonnegative integer number of units. Let $G_N(s)=E[s^N]$ be the frequency PGF, and let $H_X(z)=E[z^{X_i}]$ be the severity PGF. Assume that, given the count, claim amounts are independent and identically distributed, and that the claim count is independent of the claim amounts. These assumptions are what allow one severity PGF to represent every claim.

Theorem: Under the assumptions above, the aggregate loss PGF is $$ G_S(z)=G_N\bigl(H_X(z)\bigr). $$ The coefficient of $z^k$ in $G_S(z)$ is $P(S=k)$, where $k$ is measured in lattice units.

The composition follows by conditioning on $N$. If there are $n$ claims, the aggregate is the sum of $n$ independent claim amounts, whose PGF is $H_X(z)^n$. Averaging that conditional PGF over the possible claim counts gives

G_S(z)=E\!\left[E[z^S\mid N]\right] =E[H_X(z)^N] =\sum_{n=0}^{\infty}P(N=n)H_X(z)^n =G_N\bigl(H_X(z)\bigr).

For $N=0$, the sum of claim amounts is zero, and the conditional PGF is $H_X(z)^0=1$. Thus the formula includes the no-claim outcome automatically. The formula does not require the severity to be strictly positive: if a claim can have amount zero, its probability appears as a constant term in $H_X(z)$. It does require integer-valued lattice units for coefficient extraction in this form. If amounts are multiples of a fixed unit, first express them as integer multiples of that unit.

Write the frequency PGF.
Express $G_N(s)$ using the specified distribution of the claim count.

Write the severity PGF.
Use one term $P(X_i=j)z^j$ for each possible claim amount $j$ in lattice units.

Compose.
Substitute $H_X(z)$ for the frequency argument $s$, and simplify as far as needed.

Extract and check.
Read aggregate probabilities from coefficients. When checking a finite list, remember that omitted higher powers may still have positive probability.

Worked Examples

Worked Example: Finding a Coefficient From Multiple Claim Counts

A hypothetical portfolio has claim count probabilities $P(N=n)=0.5(0.5)^n$ for $n=0,1,2,\ldots$. Each claim independently has severity one or two lattice units, each with probability $0.5$. Find the aggregate PGF and calculate $P(S=0)$, $P(S=1)$, $P(S=2)$, and $P(S=3)$.

Method. Construct the frequency and severity PGFs, compose them, and extract coefficients. This method avoids counting only one claim count when multiple counts can produce the requested total.

The frequency PGF is a geometric series:

G_N(s)=\sum_{n=0}^{\infty}0.5(0.5)^ns^n =\frac{0.5}{1-0.5s}, \qquad |s|<2.

The severity PGF is $H_X(z)=0.5z+0.5z^2$. Therefore the aggregate PGF is

G_S(z)=\frac{0.5}{1-0.5(0.5z+0.5z^2)} =\frac{0.5}{1-0.25z-0.25z^2}.

To obtain the requested coefficients, expand as a geometric series:

G_S(z)=0.5\sum_{n=0}^{\infty}\left[0.25(z+z^2)\right]^n.

For $z^0$, only $n=0$ contributes, so $P(S=0)=0.5$. For $z^1$, only $n=1$ contributes, with coefficient $0.5(0.25)=0.125$. For $z^2$, the $n=1$ term contributes $0.5(0.25)$, and the $n=2$ term contributes $0.5(0.25)^2$, since the coefficient of $z^2$ in $(z+z^2)^2$ is 1. Thus

$$ P(S=2)=0.5(0.25)+0.5(0.25)^2=0.125+0.03125=0.15625. $$

For $z^3$, the possible powers are $n=2$ and $n=3$: $n$ claims have total severity at least $n$ and at most $2n$, so both counts can produce three units. The coefficient of $z^3$ in $(z+z^2)^2$ is 2; the coefficient in $(z+z^2)^3$ is 1. Therefore

P(S=3)=0.5\left[2(0.25)^2+(0.25)^3\right] =0.5(0.125+0.015625) =0.0703125.

The direct probability check gives the same result: two claims contribute $P(N=2)P(X_1+X_2=3)=0.125(0.5)=0.0625$, while three one-unit claims contribute $P(N=3)(0.5)^3=0.0625(0.125)=0.0078125$. Their sum is $0.0703125$. The requested probabilities sum to $0.8515625$, which is less than one because aggregates above three units remain possible. Common mistake avoided: retaining the two-claim contribution but omitting the valid three-claim outcome.

Worked Example: A Binomial Count and Two Severity Sizes

A hypothetical coverage has $N\sim\operatorname{Binomial}(2,0.5)$. Each claim amount is either one or three thousand dollars, with probabilities $0.75$ and $0.25$, respectively. Find the aggregate PGF and the probabilities of aggregate losses of zero, one, two, three, four, and six thousand dollars.

Method. Measure amounts in thousands of dollars. The frequency PGF is $G_N(s)=(0.5+0.5s)^2$, and the severity PGF is $H_X(z)=0.75z+0.25z^3$. Composition gives

$$ G_S(z)=(0.5+0.5H_X(z))^2 =0.25+0.5H_X(z)+0.25H_X(z)^2. $$

The squared severity PGF is

$$ H_X(z)^2=(0.75z+0.25z^3)^2 =0.5625z^2+0.375z^4+0.0625z^6. $$

Substituting this expression gives

$$ G_S(z)=0.25+0.375z+0.125z^3+0.140625z^2+0.09375z^4+0.015625z^6. $$

The coefficient of $z^0$ is $0.25$; of $z^1$, $0.375$; of $z^2$, $0.140625$; of $z^3$, $0.125$; of $z^4$, $0.09375$; and of $z^6$, $0.015625$. There is no $z^5$ term, so a five-thousand-dollar aggregate is impossible. The coefficient check is $0.25+0.375+0.140625+0.125+0.09375+0.015625=1$. Thus all aggregate probabilities are accounted for. Common mistake avoided: treating the two-claim outcomes as only the equal-size pairs; the cross-pair contributes to four thousand dollars.

Worked Example: A Finite Frequency Distribution on a Dollar Lattice

For a hypothetical policy group, the claim count has probabilities $P(N=0)=0.2$, $P(N=1)=0.5$, and $P(N=2)=0.3$. Each claim independently pays either \$250 with probability $0.6$ or \$500 with probability $0.4$. Find the aggregate PGF, $P(S=\$500)$, and $P(S=\$750)$.

Method. Express the aggregate in units of \$250. Then the frequency PGF is $G_N(s)=0.2+0.5s+0.3s^2$, and the severity PGF is $H_X(z)=0.6z+0.4z^2$. The aggregate PGF is

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

Squaring the severity PGF gives $H_X(z)^2=0.36z^2+0.48z^3+0.16z^4$. Hence

$$ G_S(z)=0.2+0.3z+(0.2+0.108)z^2+0.144z^3+0.048z^4 =0.2+0.3z+0.308z^2+0.144z^3+0.048z^4. $$

A \$500 aggregate is two lattice units, so $P(S=\$500)=0.308$. It can arise from one \$500 claim, contributing $0.5(0.4)=0.2$, or two \$250 claims, contributing $0.3(0.6)^2=0.108$. A \$750 aggregate is three units, and can arise only from two claims of different sizes, so $P(S=\$750)=0.3[2(0.6)(0.4)]=0.144$. The full coefficient check is $0.2+0.3+0.308+0.144+0.048=1$. Common mistake avoided: assigning the \$500 aggregate only to the one-claim case and missing the two-claim combination.

Coefficient Checks and Exam Traps

A PGF is more than a compact notation: it is a bookkeeping device. The frequency PGF assigns a weight to each possible count, while the severity PGF raised to that count distributes the probability across all totals possible with that many claims. If the same total can be produced by several claim counts, its coefficient receives contributions from each one. The first worked example illustrates why checking only a single power in the expansion can produce an incomplete answer.

A useful overall check is $G_S(1)=1$, provided the frequency and severity distributions are proper probability distributions. Indeed, $H_X(1)=1$, so $G_S(1)=G_N(1)=1$. For a polynomial aggregate PGF, this means its coefficients sum to one. For an infinite series, it confirms the total probability, but does not by itself verify a particular coefficient. Check individual coefficients by identifying every claim count that can produce the requested aggregate and accounting for all relevant terms.

Do not confuse the two PGFs. The frequency PGF takes the frequency argument $s$; the severity PGF is substituted for that argument.
Keep the lattice unit explicit. If $z$ counts \$250 units, then $z^3$ represents \$750, not \$3.
Do not stop at the first valid claim count. A specified aggregate may be obtained from multiple counts or multiple severity combinations.
Check assumptions before composing. The composition uses independent, identically distributed claim amounts independent of the count. If those assumptions fail, this simple formula need not apply.

For a nonnegative integer-valued aggregate, the PGF can also encode moments when the relevant derivatives exist: $G_S'(1)=E[S]$ and $G_S''(1)=E[S(S-1)]$. These derivative identities follow by differentiating the probability series term by term. They are useful checks and summaries, but coefficient extraction remains the direct way to obtain a specific aggregate probability.

Key takeaway: For an independent random sum with iid lattice-valued severities, form the aggregate PGF by substituting the severity PGF into the frequency PGF. To find an aggregate probability, extract its coefficient only after considering every claim count that can produce that lattice total.

Check Your Understanding

For each problem, identify the frequency and severity PGFs, compose them, and use the relevant coefficient or coefficients.

A hypothetical contract has exactly two claims. Each claim is one unit with probability $0.4$ and two units with probability $0.6$. Write the aggregate PGF and find the probability of a three-unit aggregate. (Answer: $G_S(z)=0.16z^2+0.48z^3+0.36z^4$; $P(S=3)=0.48$.)
A claim count is zero with probability $0.7$ and one with probability $0.3$. A claim pays one unit with probability $0.8$ and four units with probability $0.2$. Find the aggregate PGF and the probability of a four-unit aggregate. (Answer: $G_S(z)=0.7+0.24z+0.06z^4$; $P(S=4)=0.0600$.)
A claim count has probabilities $0.2,0.5,0.3$ for zero, one, and two claims. Each claim is one unit with probability $0.6$ and two units with probability $0.4$. Find $P(S=2)$ and $P(S=3)$. (Answer: $P(S=2)=0.3080$; $P(S=3)=0.1440$.)
A hypothetical frequency model has $P(N=n)=0.75(0.25)^n$ for $n=0,1,2,\ldots$. Every claim is exactly two lattice units. Find the aggregate PGF and the probability of a six-unit aggregate. (Answer: $G_S(z)=\frac{0.75}{1-0.25z^2}$; $P(S=6)=0.01171875$.)
A claim count has PGF $G_N(s)=0.2+0.5s+0.3s^2$. Each claim independently has severity one or two units, each with probability $0.5$. Find $P(S=2)$ and $P(S=3)$, and explain why the three-unit coefficient has a two-claim contribution. (Answer: $P(S=2)=0.3250$; $P(S=3)=0.1500$, from two claims with different severities.)

What You'll Learn