Probability Generating Functions For Aggregate Loss

Why Generating Functions Help With Aggregate Loss

For an aggregate loss, the claim count is random and each claim can have a random severity. Directly listing every possible claim count and every combination of severities can become cumbersome, even when each severity has only a few possible values. A probability generating function packages the probabilities of a nonnegative integer-valued variable into a power series. For aggregate loss, composing the frequency and severity generating functions provides a systematic way to obtain that series.

This tutorial considers a random claim count $N$, independent and identically distributed nonnegative integer claim severities $X_1,X_2,\ldots$, and aggregate loss $S=\sum_{j=1}^{N}X_j$. The empty sum is defined to be zero when $N=0$. The focus is on deriving and using the generating-function identity, not on approximating the aggregate distribution.

Definition: If $W$ is a random variable taking values in $\{0,1,2,\ldots\}$, its probability generating function is $$ G_W(z)=E[z^W]=\sum_{k=0}^{\infty}P(W=k)z^k. $$ The coefficient of $z^k$ is $P(W=k)$. In particular, $G_W(0)=P(W=0)$, and $G_W(1)=1$. The series is well-defined for $|z|\leq 1$.

Write $G_N$ for the generating function of the claim count and $G_X$ for that of one claim severity. The severities are assumed to have the same distribution and to be mutually independent, and $N$ is assumed independent of the severity sequence. These conditions are what allow the generating functions to combine in the simple form below.

The Compound Aggregate Generating Function

Theorem: Let $N$ be a nonnegative integer-valued random variable. Let $X_1,X_2,\ldots$ be independent, identically distributed nonnegative integer-valued severities, independent of $N$, and define $S=\sum_{j=1}^{N}X_j$, with $S=0$ if $N=0$. Then $$ G_S(z)=G_N\bigl(G_X(z)\bigr),\qquad |z|\leq 1. $$

The expression is a composition: substitute the entire severity generating function into the argument of the frequency generating function. This differs from multiplying $G_N(z)$ and $G_X(z)$. To see why composition is correct, first fix a claim count $N=n$. Conditional on that count, the aggregate is the sum of $n$ independent severities, so its generating function is $G_X(z)^n$. Averaging over the possible counts gives the frequency-weighted sum of these powers.

Condition on the claim count.
For a fixed count $n$, the aggregate is $X_1+\cdots+X_n$. Independence gives $E[z^{X_1+\cdots+X_n}]=G_X(z)^n$. When $n=0$, the empty sum is zero and its generating function is $1=G_X(z)^0$.

Average over the count distribution.
Each possible $n$ contributes its conditional generating function weighted by $P(N=n)$.

Recognize the frequency generating function.
The weighted sum is precisely $G_N$ evaluated at $G_X(z)$.

\begin{aligned} G_S(z) &=E[z^S]\\ &=\sum_{n=0}^{\infty}P(N=n)E[z^{X_1+\cdots+X_n}\mid N=n]\\ &=\sum_{n=0}^{\infty}P(N=n)\bigl(G_X(z)\bigr)^n\\ &=G_N\bigl(G_X(z)\bigr). \end{aligned}

For $|z|\leq 1$, the severity generating function has magnitude at most one, so the frequency series can be evaluated at $G_X(z)$. If zero severity is possible, a fixed aggregate value can receive contributions from arbitrarily large claim counts, because some claims may contribute zero. The coefficient sum still converges: its terms are nonnegative probability contributions whose total cannot exceed one.

Extracting Aggregate Probabilities

Once $G_S(z)$ is written as a power series, the probability of an aggregate amount $s$ is its coefficient of $z^s$, commonly written $[z^s]G_S(z)$. One useful way to organize the extraction is to expand the frequency generating function as $G_N(w)=\sum_{n\geq0}P(N=n)w^n$, then substitute $G_X(z)$:

P(S=s)=[z^s]G_S(z) =\sum_{n=0}^{\infty}P(N=n)[z^s]\bigl(G_X(z)\bigr)^n.

The factor $[z^s](G_X(z))^n$ is the probability that $n$ independent severities sum to $s$. This coefficient form is especially useful when the count distribution has only a few possible values, or when the generating function simplifies to a rational expression with an easy coefficient recurrence.

Worked Example: A Bernoulli Count With Zero-Severity Possibility

A hypothetical coverage has $N\sim\operatorname{Binomial}(1,0.4)$: there is at most one reported claim, with claim probability $0.4$. Given a claim, its nonnegative integer severity has probabilities $P(X=0)=0.2$, $P(X=1)=0.5$, and $P(X=2)=0.3$. The count and severity are independent. Find the aggregate distribution. Use generating functions and check that the probabilities sum to one.

Method. Form each generating function, compose them, and read the coefficients. This method avoids treating a zero-valued claim as though no claim had occurred; those are distinct events in the underlying model, even though both produce zero aggregate loss.

The claim-count generating function is $G_N(z)=0.6+0.4z$. The severity generating function is $G_X(z)=0.2+0.5z+0.3z^2$. Therefore,

\begin{aligned} G_S(z) &=G_N(G_X(z))\\ &=0.6+0.4(0.2+0.5z+0.3z^2)\\ &=0.6+0.08+0.20z+0.12z^2\\ &=0.68+0.20z+0.12z^2. \end{aligned}

Thus $P(S=0)=0.68$, $P(S=1)=0.20$, and $P(S=2)=0.12$. The zero-aggregate probability includes both no claim, with probability $0.6$, and a claim of zero severity, with probability $0.4(0.2)=0.08$. The check is $0.68+0.20+0.12=1.00$, as required.

Worked Example: Binomial Frequency and Two Positive Severities

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$, so there are at most two claims and each claim occurs with probability $0.5$. Independent of the count, claim severities are independent and satisfy $P(X=1)=0.25$ and $P(X=2)=0.75$. Find $P(S=2)$, $P(S=3)$, and $P(S=4)$, and verify the full aggregate distribution.

Method. Substitute the severity generating function into the binomial count generating function, then multiply out to the needed powers. This avoids confusing the number of claims with the dollar amount of aggregate loss.

Here $G_N(z)=(0.5+0.5z)^2$ and $G_X(z)=0.25z+0.75z^2$. Composition gives

\begin{aligned} G_S(z) &=(0.5+0.5G_X(z))^2\\ &=(0.5+0.125z+0.375z^2)^2\\ &=0.25+0.125z+0.390625z^2+0.09375z^3+0.140625z^4. \end{aligned}

For example, the coefficient of $z^2$ is $0.390625$: it combines the possibility of one claim of severity 2 and the possibility of two claims, each of severity 1. The requested probabilities are $P(S=2)=0.390625$, $P(S=3)=0.09375$, and $P(S=4)=0.140625$. The remaining coefficients are $P(S=0)=0.25$ and $P(S=1)=0.125$. Their sum is $0.25+0.125+0.390625+0.09375+0.140625=1$.

Using a Rational Generating Function

A count distribution with infinitely many possible values can still yield a convenient aggregate generating function. For example, if $N$ has a geometric distribution on $\{0,1,2,\ldots\}$, with $P(N=n)=(1-r)r^n$ for $0\leq r<1$, then

G_N(z)=\frac{1-r}{1-rz}.

After composition, multiplying by the denominator often gives a recurrence for the aggregate probabilities. This approach extracts coefficients without having to enumerate every possible claim count separately.

Worked Example: Coefficients Under Geometric Frequency

A hypothetical book of small equipment policies has a claim count with $P(N=n)=0.6(0.4)^n$ for $n=0,1,2,\ldots$. Claims have independent severities $P(X=1)=0.5$ and $P(X=2)=0.5$, independent of the count. Find $P(S=0),\ldots,P(S=4)$. Give probabilities to five decimal places.

Method. Compose the geometric frequency generating function with the severity generating function, then use the denominator to obtain a coefficient recurrence. This avoids truncating the possible claim counts, which are unbounded.

The severity generating function is $G_X(z)=0.5z+0.5z^2$. Since $r=0.4$, the aggregate generating function is

G_S(z)=\frac{0.6}{1-0.4(0.5z+0.5z^2)} =\frac{0.6}{1-0.2z-0.2z^2}.

Write $G_S(z)=\sum_{k=0}^{\infty}p_kz^k$, where $p_k=P(S=k)$. Multiplying by the denominator gives $(1-0.2z-0.2z^2)\sum_{k=0}^{\infty}p_kz^k=0.6$. Equating coefficients yields $p_0=0.6$, $p_1=0.2p_0$, and $p_k=0.2p_{k-1}+0.2p_{k-2}$ for $k\geq2$. Thus,

\begin{aligned} p_0&=0.60000,\\ p_1&=0.2(0.60000)=0.12000,\\ p_2&=0.2(0.12000)+0.2(0.60000)=0.14400,\\ p_3&=0.2(0.14400)+0.2(0.12000)=0.05280,\\ p_4&=0.2(0.05280)+0.2(0.14400)=0.03936. \end{aligned}

The requested probabilities are therefore $0.60000, 0.12000, 0.14400, 0.05280,$ and $0.03936$, respectively. Their sum is $0.95616$, not one, because positive probability remains at aggregate amounts greater than 4. The first five coefficients are probabilities, not a complete distribution.

Common Mistakes and Model Checks

A particularly important trap is multiplying the two generating functions instead of composing them. In the Bernoulli example, that incorrect operation would give $G_N(z)G_X(z)$, whose constant coefficient is $0.6(0.2)=0.12$. The correct zero-aggregate probability is $0.68$, since the aggregate is zero whenever there is no claim or the single claim has zero severity. Multiplication does not represent a random sum with a random number of terms.

Other useful checks are:

Check the support. If each positive claim is at least 1 and the count is at most $m$, aggregate values above $m$ cannot occur.
Check the value at one. A valid frequency and severity generating function each equals 1 at $z=1$, so $G_S(1)=G_N(G_X(1))=G_N(1)=1$.
Keep zero severity distinct from zero claim count. They both contribute to $S=0$, but they arise from different model events.
Do not mistake a partial coefficient list for the whole distribution. With unbounded frequency or severity, unlisted aggregate values may carry positive probability.
Check assumptions before composing. The formula in this tutorial uses independent, identically distributed severities that are independent of $N$. If severities differ by claim position or depend on the count, the simple composition need not apply as written.

Key takeaway: For a random sum of independent, identically distributed nonnegative integer severities, the aggregate generating function is the frequency generating function evaluated at the severity generating function. Aggregate probabilities are then obtained by extracting coefficients of the resulting series.

Check Your Understanding

For each problem, state or use the generating function and identify the requested coefficient or probability.

A hypothetical contract has $N\sim\operatorname{Binomial}(1,0.3)$. Given a claim, severity has $P(X=1)=0.6$ and $P(X=2)=0.4$, independently of $N$. Find the probabilities of aggregate losses 0, 1, and 2. (Answer: $P(S=0)=0.7000,\ P(S=1)=0.1800,\ P(S=2)=0.1200$.)
Suppose $N\sim\operatorname{Binomial}(2,0.4)$, while each independent severity is 1 or 2 with probabilities 0.5 and 0.5. Find the full aggregate distribution. (Answer: $P(S=0)=0.3600,\ P(S=1)=0.2400,\ P(S=2)=0.2800,\ P(S=3)=0.0800,\ P(S=4)=0.0400$.)
A count has $P(N=n)=0.75(0.25)^n$ for $n=0,1,2,\ldots$, and every claim has deterministic severity 1. Find $P(S\leq2)$. (Answer: $0.9844$, to four decimal places.)
A claim count has probabilities $P(N=0)=0.2,\ P(N=1)=0.5,\ P(N=2)=0.3$. Independent severities have $P(X=1)=0.25$ and $P(X=2)=0.75$. Find $P(S=3)$. (Answer: $0.1125$.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.8)$. Independent severities satisfy $P(X=0)=0.25$ and $P(X=1)=0.75$. Find the zero-aggregate probability, to four decimal places. (Answer: $P(S=0)=e^{-0.6}=0.5488$.)

What You'll Learn