The Zero-Claim Probability Mass

Why Aggregate Loss Can Be Exactly Zero

An aggregate loss model combines a random number of claims with the amounts paid on those claims. The value zero is special: if every claim payment is strictly positive, the aggregate can be zero only when there are no claims. That makes the probability at zero particularly direct to calculate, even when the positive claim amounts have a complicated distribution.

The key condition is not that claim amounts have a particular average or distribution. It is that every claim amount is positive with probability one. If a claim can instead produce a zero payment, one or more claims may occur while the aggregate remains zero; the calculation must then include those possibilities.

Definition: Let $N$ be a finite, nonnegative integer-valued claim count, and let $Z_1,Z_2,\ldots$ be the claim severities or payments. Define aggregate loss by $S=\sum_{j=1}^{N}Z_j$, with the empty sum defined as zero when $N=0$. If $P(Z_j>0)=1$ for every claim that can occur, then $S=0$ if and only if $N=0$. Consequently, $P(S=0)=P(N=0)$.

The event equivalence is exact. When $N=0$, the sum has no terms and equals zero. When $N\geq 1$, it contains at least one strictly positive term, so the sum is strictly positive. No assumption that the severities are identically distributed or independent is needed for this event argument; the count must be finite and each included payment must be strictly positive.

Calculating the Zero-Claim Probability

Once the event equivalence applies, the severity model drops out of the calculation. Find the probability that the count is zero under the specified frequency distribution. The model’s parameterization matters, especially for the negative binomial distribution.

Claim count model	Parameters and convention	Zero-claim probability
Poisson	$N\sim\operatorname{Poisson}(\lambda)$, where $\lambda$ is the mean count	$P(N=0)=e^{-\lambda}$
Binomial	$N\sim\operatorname{Binomial}(m,q)$, with $m$ opportunities and claim probability $q$ per opportunity	$P(N=0)=(1-q)^m$
Negative binomial	$N$ is the number of failures before $r$ successes, with success probability $p$ on each trial	$P(N=0)=p^r$

These formulas are the probabilities of zero counts, not general formulas for an aggregate distribution. They give $P(S=0)$ only under the strictly positive payment condition. The negative binomial convention in the table is explicit: zero failures means the first $r$ trials are all successes, with probability $p^r$. Other parameterizations may describe a different count, so identify the convention before substituting values.

Worked Example: Zero Aggregate Loss with a Poisson Count

A hypothetical repair portfolio has a Poisson number of claims per year with mean $\lambda=3$. Each reported claim produces a strictly positive payment; for simplicity, suppose each payment is exactly $400. Find the probability of zero aggregate payment and the probability of at least one payment.

Method. Since every claim payment is positive, zero aggregate payment occurs exactly when $N=0$. The fixed payment amount does not affect this event.

For a Poisson count, $P(N=0)=e^{-\lambda}$. Substituting $\lambda=3$ gives $P(S=0)=e^{-3}=0.0497871\ldots$, or 0.0498 to four decimal places.

The event of at least one payment is the complement: $P(S>0)=1-P(S=0)=1-0.0497871\ldots=0.9502129\ldots$, or 0.9502 to four decimal places.

Check. The probabilities of zero payment and positive payment sum to $0.0498+0.9502=1.0000$ after rounding. A common mistake would be to use the $400 payment as though its amount determined the probability of zero; the event depends on whether the count is zero, not on the size of a positive payment.

Worked Example: Zero Aggregate Loss with a Binomial Count

A hypothetical equipment warranty covers six independent units for a year. Each unit generates a claim with probability $q=0.25$, independently of the other units. Any claim has a strictly positive payment. Find the probability that no payment is made, then find the probability that payments are made on exactly one or more units.

Method. The number of claims has a binomial distribution with $m=6$ opportunities and claim probability $q=0.25$. A zero aggregate payment means all six units avoid a claim.

The probability of no claim on one unit is $1-q=0.75$. By independence, the probability of no claims on all six units is $P(S=0)=P(N=0)=(1-q)^m=(0.75)^6$.

Step by step, $0.75^2=0.5625$, $0.75^3=0.421875$, $0.75^4=0.31640625$, $0.75^5=0.2373046875$, and $0.75^6=0.177978515625$. Thus the probability of zero payment is 0.1780 to four decimal places.

The probability of at least one payment is $1-0.177978515625=0.822021484375$, or 0.8220 to four decimal places. The two events are complements, so their probabilities sum to 1 before rounding.

Check. There are six opportunities, not an unlimited number of claims. Using a Poisson zero-count formula here would replace the stated binomial model with a different model and would not give the exact answer.

Worked Example: Zero Aggregate Loss with a Negative Binomial Count

A hypothetical portfolio has a negative binomial claim count $N$, defined as the number of failures before $r=4$ successes, where each trial has success probability $p=0.8$. Each claim produces a strictly positive payment. Find the probability of zero aggregate payment and the probability of at least one claim.

Method. Under this convention, $N=0$ means that the first four trials are all successes. The probability of that event is $p^r$.

Substitute the parameters: $P(S=0)=P(N=0)=0.8^4$. Since $0.8^2=0.64$, $0.8^3=0.512$, and $0.8^4=0.4096$, the probability of zero aggregate payment is 0.4096.

The probability of at least one claim is the complement, $P(N\geq1)=1-0.4096=0.5904$. Because all claim payments are positive, this is also $P(S>0)$.

Check. The two probabilities add to $0.4096+0.5904=1$. In this parameterization, $p$ is the success probability, so the probability of zero failures is $p^r$, not $(1-p)^r$.

Why the Positive-Severity Condition Matters

If a claim can have a zero payment, the equivalence $S=0$ if and only if $N=0$ no longer holds. For example, a claim may be reported but produce no payment under a hypothetical contract. Then the aggregate is zero both when no claims occur and when claims occur but every one has zero severity.

Suppose the severities are independent and identically distributed, independent of $N$, and each has probability $c=P(Z_j=0)$ of being zero. Conditional on $N=n$, the aggregate is zero exactly when all $n$ severities are zero, which has probability $c^n$. This includes $n=0$, for which $c^0=1$ and the empty sum is zero. Averaging over the claim count gives

P(S=0)=\sum_{n=0}^{\infty}P(N=n)c^n=E[c^N].

When $c=0$, only the $n=0$ term contributes, and the expression reduces to $P(N=0)$. When $c>0$, positive claim counts can also contribute. This distinction is essential: the probability of no claims and the probability of no aggregate payment are not automatically the same.

Derivation for the negative binomial case: If $N$ counts failures before $r$ successes with success probability $p$, then $P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n$, for $n=0,1,\ldots$. Therefore, when the severity-zero probability is $c$, $$ P(S=0)=\sum_{n=0}^{\infty}\binom{n+r-1}{n}p^r\big((1-p)c\big)^n =p^r\big(1-(1-p)c\big)^{-r} =\left(\frac{p}{1-(1-p)c}\right)^r. $$ The middle equality follows from the binomial-series identity $\sum_{n=0}^{\infty}\binom{n+r-1}{n}x^n=(1-x)^{-r}$, valid for $|x|<1$. Here $0\leq(1-p)c<1$, so the series converges.

Worked Example: Negative Binomial Count with Zero-Payment Claims

A hypothetical contract portfolio has $N$ distributed as the number of failures before $r=4$ successes, with $p=0.8$. Each claim independently produces a zero payment with probability $0.1$, or a $500 payment with probability $0.9$. Severities are independent of the count. Find the probability of zero aggregate payment.

Method. Given $N=n$, zero aggregate payment requires all $n$ claims to have zero payment, so $P(S=0\mid N=n)=0.1^n$. Averaging over $N$ gives $P(S=0)=E[0.1^N]$.

For this negative binomial convention, substitute $c=0.1$, $p=0.8$, and $r=4$ into the derived expression: $P(S=0)=\left(0.8/[1-(1-0.8)(0.1)]\right)^4$.

The denominator is $1-(0.2)(0.1)=1-0.02=0.98$. Thus the ratio is $0.8/0.98=80/98=40/49$, and $P(S=0)=(40/49)^4=2{,}560{,}000/5{,}764{,}801=0.4440743\ldots$. The probability of zero aggregate payment is 0.4441 to four decimal places.

Check. The result exceeds $P(N=0)=0.8^4=0.4096$, as it must: zero aggregate payment includes the zero-claim event plus additional cases in which one or more claims occur but all have zero payment.

Common Exam Trap

A frequent error is to use $P(N=0)$ as the probability of zero aggregate loss without checking whether a claim payment can be zero. In the preceding negative binomial example, the incorrect calculation would be $P(S=0)\overset{\text{wrong}}{=}P(N=0)=0.8^4=0.4096$. That omits all cases with $N\geq1$ in which every claim has zero payment.

The correction is to condition on the claim count: $P(S=0)=E[0.1^N]=(40/49)^4=0.4441$. Conversely, if the problem explicitly guarantees strictly positive severity, adding zero-payment claim cases would be a mistake: then $S=0$ occurs only when $N=0$. State the severity condition before choosing the calculation.

Key takeaway: For a finite claim count and strictly positive claim payments, $P(S=0)=P(N=0)$. For independent severities with zero-payment probability $c$, use $P(S=0)=E[c^N]$; the count distribution and its parameterization then determine the calculation.

Check Your Understanding

For each problem, first decide whether strictly positive severities make zero aggregate loss equivalent to zero claims. Answers are rounded to four decimal places unless given exactly.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2.4)$, and every claim payment is strictly positive. Find the probability of zero aggregate payment. (Answer: $e^{-2.4}=0.0907$)
Seven independent policies each have a claim probability of $0.2$, and all claim payments are positive. Find the probability that the aggregate payment is zero. (Answer: $0.8^7=0.2097$)
Let $N$ be the number of failures before $r=2$ successes, with success probability $p=0.6$. If every claim has a positive payment, find $P(S=0)$. (Answer: $0.6^2=0.3600$)
Let $N\sim\operatorname{Binomial}(3,0.5)$. Each claim independently has zero payment with probability $0.2$, and otherwise has a positive payment. Find the probability of zero aggregate payment. (Answer: $(0.5+0.5(0.2))^3=0.6^3=0.2160$)
Let $N$ count failures before $r=3$ successes with $p=0.7$, and let each independent severity be zero with probability $0.25$. Find $P(S=0)$. (Answer: $\left(\frac{0.7}{1-0.3(0.25)}\right)^3=\left(\frac{28}{37}\right)^3\approx0.4334$)

Claim count model	Parameters and convention	Zero-claim probability
Poisson	\(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda\) is the mean count	\(P(N=0)=e^{-\lambda}\)
Binomial	\(N\sim\operatorname{Binomial}(m,q)\), with \(m\) opportunities and claim probability \(q\) per opportunity	\(P(N=0)=(1-q)^m\)
Negative binomial	\(N\) is the number of failures before \(r\) successes, with success probability \(p\) on each trial	\(P(N=0)=p^r\)

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