The Zero-Loss Atom In Compound Models

Why a Compound Aggregate Can Be Zero

A claim count of zero always produces a zero aggregate. But a positive claim count does not necessarily produce a positive aggregate: if a severity distribution assigns positive probability to a zero payment, every reported claim in a period could have zero severity. The probability of zero aggregate loss must therefore account for both the claim count and the possibility of zero-valued claims.

The distinction matters when interpreting a fitted aggregate model or checking a calculation. The probability of no claims is determined by the frequency model alone. The probability of zero aggregate loss also depends on the severity distribution’s probability mass at zero.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be independent, identically distributed nonnegative claim severities, independent of $N$. Write $q_0=P(X_1=0)$, and define the compound aggregate by $S_N=\sum_{j=1}^{N}X_j$, with the empty sum $S_0=0$. The probability generating function (PGF) of $N$ is $G_N(s)=E[s^N]$. Then $P(S_N=0)=G_N(q_0)$, whereas $P(N=0)=G_N(0)$.

The assumptions are important. Severities are nonnegative, so positive severities cannot cancel one another to produce a zero sum; and the severities are independent, so conditional on $N=n$, the probability that all $n$ severities are zero is $q_0^n$. The claim count and severities are also independent, as in the standard compound model. If losses could be negative, or if severities were dependent, this result would not generally apply.

Deriving the Zero-Aggregate Formula

Condition first on the claim count. If $N=n$, the aggregate is zero exactly when every one of the $n$ severities is zero. For $n\geq 1$, independence gives probability $q_0^n$. When $n=0$, the aggregate is zero by the empty-sum convention, so its conditional probability is 1. This is represented by $q_0^0=1$ in the PGF sum.

\begin{aligned} P(S_N=0) &=\sum_{n=0}^{\infty}P(S_N=0\mid N=n)P(N=n)\\ &=\sum_{n=0}^{\infty}q_0^nP(N=n)\\ &=G_N(q_0). \end{aligned}

This derivation also shows why $P(S_N=0)$ is at least $P(N=0)$: the event of no claims is one way, but not the only way, to obtain a zero aggregate. The two probabilities are equal when $q_0=0$, meaning that each claim has strictly positive severity with probability 1.

Special case: strictly positive severity: If $P(X_1=0)=0$, then $P(S_N=0)=G_N(0)=P(N=0)$. If the severity distribution has a zero atom, use $G_N(q_0)$; do not substitute zero into the count PGF unless $q_0=0$.

Using Common Claim-Count Models

The calculation requires the frequency PGF, evaluated at the severity zero probability. The following standard PGFs are enough for many direct applications. For the negative binomial model shown, $N$ counts failures before the $r$th success, with success probability $p$.

Claim-count model	PGF $G_N(s)$	Zero aggregate probability
Poisson, mean $\lambda$	$\exp(\lambda(s-1))$	$\exp(\lambda(q_0-1))$
Binomial, $m$ trials and probability $\theta$	$(1-\theta+\theta s)^m$	$(1-\theta+\theta q_0)^m$
Negative binomial, failures before $r$ successes, probability $p$	$\left(\frac{p}{1-(1-p)s}\right)^r$	$\left(\frac{p}{1-(1-p)q_0}\right)^r$

For example, a Poisson frequency with mean $\lambda$ gives $P(S_N=0)=\exp[-\lambda(1-q_0)]$. When $q_0=0$, this becomes $\exp(-\lambda)$, the probability of no Poisson claims. When $q_0>0$, the zero aggregate probability is larger because some periods with claims can still have zero total severity.

Worked Example: Poisson Frequency With a Zero Severity Atom

A hypothetical policy portfolio has a Poisson claim count with mean $\lambda=2$ claims per period. Each claim has severity $X=0$ with probability $0.25$; conditional on $X>0$, severity is exponential with mean $\$800$. Find (a) the probability of zero aggregate loss and (b) the probability of no claims. Amounts are nominal; interest and mortality are not modeled.

Method. Evaluate the Poisson count PGF at $q_0=0.25$ for the aggregate probability. Evaluate it at zero for the no-claim probability.

(a) Zero aggregate. Here $G_N(s)=\exp(2(s-1))$. Therefore,

P(S_N=0)=G_N(0.25)=\exp(2(0.25-1)) =\exp(-1.5)=0.223130\ldots

Rounded to four decimal places, $P(S_N=0)=0.2231$.

(b) No claims.

P(N=0)=G_N(0)=\exp(2(0-1)) =\exp(-2)=0.135335\ldots

Rounded to four decimal places, $P(N=0)=0.1353$. The difference, $0.223130\ldots-0.135335\ldots=0.087795\ldots$, is the probability of having at least one claim but still having zero aggregate loss. The result is checked by noting that the aggregate-zero probability exceeds the no-claim probability, as it must when $q_0>0$. The common mistake avoided is using $\exp(-2)$ for both probabilities.

Worked Example: Binomial Frequency and Fixed Positive Payments

A hypothetical device warranty covers four devices. Each device independently generates a claim with probability $\theta=0.30$, so $N\sim\operatorname{Binomial}(4,0.30)$. For each claim, the payment is $\$500$ with probability $0.80$ and $\$0$ with probability $0.20$. Find the probability of zero aggregate payment and the probability of no claims. Payments are nominal; interest and mortality are not modeled.

Method. The severity zero probability is $q_0=0.20$, and the count PGF is $G_N(s)=(1-\theta+\theta s)^4$.

Zero aggregate. Substitute $s=q_0$, not $s=0$:

\begin{aligned} P(S_N=0) &=G_N(0.20)\\ &=(1-0.30+0.30(0.20))^4\\ &=(0.70+0.06)^4\\ &=0.76^4=0.33362176. \end{aligned}

Thus $P(S_N=0)=0.3336$, rounded to four decimal places. This includes both no claims and claims whose payments are all zero.

No claims. Set $s=0$:

$$ P(N=0)=G_N(0)=(1-0.30)^4=0.70^4=0.2401. $$

The difference is $0.33362176-0.2401=0.09352176$, or $0.0935$ to four decimal places. Since the aggregate-zero probability is greater than the no-claim probability, the result has the expected direction. The common mistake avoided is treating each covered device’s probability of a zero payment as if it were its probability of no claim.

Worked Example: Negative Binomial Parameterization and the Zero Atom

In a hypothetical portfolio, $N$ is the number of failures before $r=2$ successes, with success probability $p=0.50$. Each claim has severity $\$1{,}200$ with probability $0.80$ and zero with probability $0.20$. Find the probability of zero aggregate loss, the probability of no claims, and the probability of a positive claim count with zero aggregate loss. Amounts are nominal; interest and mortality are not modeled.

Method. For this stated negative binomial parameterization, $G_N(s)=\left(\frac{p}{1-(1-p)s}\right)^r$. Use the same PGF at $q_0=0.20$ and at zero.

Zero aggregate.

\begin{aligned} P(S_N=0) &=G_N(0.20)\\ &=\left(\frac{0.50}{1-(1-0.50)(0.20)}\right)^2\\ &=\left(\frac{0.50}{1-0.10}\right)^2\\ &=\left(\frac{0.50}{0.90}\right)^2\\ &=\left(\frac{5}{9}\right)^2 =\frac{25}{81}=0.308641975\ldots \end{aligned}

Therefore $P(S_N=0)=0.3086$, rounded to four decimal places.

No claims.

P(N=0)=G_N(0) =\left(\frac{0.50}{1-(1-0.50)(0)}\right)^2 =(0.50)^2=0.2500.

The probability of at least one claim but zero aggregate is $0.308641975\ldots-0.2500=0.058641975\ldots$, or $0.0586$ to four decimal places. Adding this to the no-claim probability recovers the zero-aggregate probability: $0.2500+0.058641975\ldots=0.308641975\ldots$. The common mistake avoided is using the wrong negative binomial parameterization or mis-evaluating the denominator: $1-(1-0.50)(0.20)=0.90$, giving $0.3086$, not $0.3543$.

Common Checks and Exam Traps

A useful way to organize a calculation is to identify the two distinct inputs before substituting anything: the claim-count PGF and the severity zero probability. Then evaluate the same PGF twice if the question asks for both zero aggregate and no claims.

Identify the severity atom.
Set $q_0=P(X=0)$. Do not confuse it with the probability that a claim occurs.

Write the frequency PGF.
Check the count model and its parameterization before inserting values.

Evaluate at the correct argument.
Use $G_N(q_0)$ for zero aggregate and $G_N(0)$ for no claims.

Check the result.
For $0\leq q_0\leq1$, $G_N(q_0)\geq G_N(0)$, because each term $P(N=n)q_0^n$ is nonnegative and the $n=0$ term is unchanged.

The most frequent conceptual error is setting the aggregate-zero probability equal to the no-claim probability even when $q_0>0$. The algebraic version of the same error is evaluating $G_N(0)$ instead of $G_N(q_0)$. Conversely, when every claim has strictly positive severity, using a positive $q_0$ would overstate the probability of a zero aggregate. The limiting checks are simple: if $q_0=0$, the two probabilities coincide; if $q_0=1$, every claim has zero severity and $P(S_N=0)=G_N(1)=1$.

Key takeaway: Under the standard compound-model assumptions, zero aggregate loss occurs when every severity in the realized claim count is zero. Therefore $P(S_N=0)=G_N(P(X=0))$, while the probability of no claims remains $G_N(0)$. This distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical model, calculate the requested probabilities. Give probabilities to four decimal places.

A Poisson claim count has mean $0.8$, and severity is zero with probability $0.10$. Find the probability of zero aggregate and the probability of no claims. (Answer: $P(S_N=0)=0.4868$, $P(N=0)=0.4493$)
A count is binomial with $m=5$ and $\theta=0.40$. Severity is zero with probability $0.25$. Find the zero-aggregate probability and the no-claim probability. (Answer: $P(S_N=0)=0.1681$, $P(N=0)=0.0778$)
A Poisson count has mean $3$, and severity is zero with probability $0.40$. Find the probability of a positive claim count but zero aggregate loss. (Answer: $0.1155$)
$N$ is the number of failures before $r=3$ successes, with success probability $p=0.60$. Severity is zero with probability $0.10$. Find the zero-aggregate probability and the probability of a positive claim count but zero aggregate. (Answer: $P(S_N=0)=0.2441$, $P(N>0,S_N=0)=0.0281$)
A hypothetical frequency distribution has $P(N=0)=0.10$, $P(N=1)=0.30$, and $P(N=2)=0.60$. Each severity is independently zero with probability $0.20$. Find the zero-aggregate probability and the probability of a positive claim count but zero aggregate. (Answer: $P(S_N=0)=0.1840$, $P(N>0,S_N=0)=0.0840$)

Claim-count model	PGF \(G_N(s)\)	Zero aggregate probability
Poisson, mean \(\lambda\)	\(\exp(\lambda(s-1))\)	\(\exp(\lambda(q_0-1))\)
Binomial, \(m\) trials and probability \(\theta\)	\((1-\theta+\theta s)^m\)	\((1-\theta+\theta q_0)^m\)
Negative binomial, failures before \(r\) successes, probability \(p\)	\(\left(\frac{p}{1-(1-p)s}\right)^r\)	\(\left(\frac{p}{1-(1-p)q_0}\right)^r\)

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