Zero Payments, Positive-Payment Counts, and Severity

Why Count Positive Payments Separately?

An insurer may receive many claims but make payments on only some of them. A deductible can eliminate small claims, an exclusion can produce no payment, and a coverage rule can otherwise map a claim to zero. In a frequency-severity model, it is therefore important to distinguish the original claim count from the positive-payment count. The latter is the count of claims that actually contribute a positive amount to covered aggregate loss.

This distinction does not change the aggregate payment: adding zero-valued payments does not change a sum. It does change the frequency model and the severity distribution used to describe that sum. The calculation requires a consistent claim-level model, in which claims have independent, identically distributed severities and each claim's payment is determined independently of other claims.

Definition: Let $N$ be the number of original claims. For claim $i$, let $X_i$ be its loss and let $Y_i=g(X_i)$ be the insurer's payment under a specified per-claim coverage rule. Define $K$ as the number of positive payments: $$ K=\sum_{i=1}^{N} I(Y_i>0). $$ Write $p_0=P(Y=0)$ and $q=P(Y>0)=1-p_0$. When $q>0$, the positive-payment severity $Y^+$ has the conditional distribution of $Y$ given $Y>0$.

Here $N$ counts claims before zero payments are removed. In contrast, $K$ counts only claims with $Y_i>0$. If the model has $N=0$, then $K=0$; if $N>0$, it is still possible that $K=0$ because every claim produces a zero payment. When $q=0$, the positive-payment severity is undefined, because no payment can be positive.

Conditional Severity and the Positive-Payment Count

The zero-payment probability is obtained from the payment rule, not automatically from the original loss distribution. For example, if $X$ is nonnegative and a per-claim deductible $d$ is applied, the payment is zero whenever $X\le d$. If there is also a payment limit $u$, a common rule is $Y=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. Under this rule, the positive-payment probability is $q=P(X>d)$, provided a loss above the deductible produces a positive payment.

For any nonnegative payment $Y$ with distribution function $F_Y$, the conditional distribution of a positive payment is $$ F_{Y^+}(y)=P(Y\le y\mid Y>0)=\frac{F_Y(y)-p_0}{q},\qquad y\ge 0,\quad q>0. $$ The subtraction removes the probability mass at zero; division by $q$ renormalizes the remaining probability to one. This formula also handles positive payment atoms, including an atom at a policy limit.

The positive-payment count depends on the original frequency model. Given $N=n$, each of the $n$ claims independently produces a positive payment with probability $q$. Thus $K\mid N=n$ has a binomial distribution with parameters $n$ and $q$. For an arbitrary original count distribution with probability generating function $G_N(s)=E[s^N]$, this gives $$ G_K(z)=G_N(1-q+qz). $$ This identity is useful because the positive-payment count need not have the same distributional form as the original count in every frequency model.

Why Poisson Frequency Stays Poisson

Suppose $N\sim\operatorname{Poisson}(\lambda)$, and each claim independently produces a positive payment with the same probability $q$. The generating-function identity gives

G_K(z)=G_N(1-q+qz) =\exp\{\lambda[(1-q+qz)-1]\} =\exp\{\lambda q(z-1)\}.

The final expression is the probability generating function of a Poisson random variable with parameter $\lambda q$. Therefore $K\sim\operatorname{Poisson}(\lambda q)$. This is a thinning result: each original claim is independently retained in the positive-payment count with probability $q$.

The severity marks also have to be updated. Conditional on being counted in $K$, a payment has distribution $Y^+$, not the unconditional distribution of $Y$, which includes zero. Under the independent, identically distributed claim model, the positive payments are independent with this conditional distribution, and their count is independent of their positive-payment severities. Consequently, the covered aggregate payment can be represented either as a sum over all original claims or as a sum over positive payments:

S=\sum_{i=1}^{N}Y_i \ \overset{d}{=}\ \sum_{j=1}^{K}Y_j^+, \qquad K\sim\operatorname{Poisson}(\lambda q).

The sum on the right is zero when $K=0$. This is an equivalent representation of the same covered aggregate loss, not an assertion that the original number of claims was $K$.

Map a claim to its payment.
Apply the contract rule to the original loss to obtain $Y$, including its possible value of zero.

Find the positive-payment probability.
Calculate $p_0=P(Y=0)$ and $q=1-p_0$.

Update the frequency and severity.
Thin the original count using $q$, and condition the payment distribution on $Y>0$.

Check the model assumptions.
Use this independent thinning representation only when claim payment outcomes are independent across claims under the stated model.

Worked Applications

Worked Example: A Poisson Count With a Deductible and Limit

A hypothetical insurer receives $N\sim\operatorname{Poisson}(12)$ claims. Claim severity $X$, in dollars, is independent across claims and has the distribution $P(X=80)=0.20$, $P(X=250)=0.50$, and $P(X=700)=0.30$. The per-claim payment is $Y=\min((X-200)_+,400)$. Find the distribution of the positive-payment count and the conditional positive-payment severity. Also find the probability of no positive payments. Round the final probability to four decimal places.

Method and payment map. First apply the $200 deductible and then the $400 payment limit. For the three possible losses:

If $X=80$, then $Y=0$, since $80\le 200$. If $X=250$, then $Y=\min(250-200,400)=50$. If $X=700$, then $Y=\min(700-200,400)=400$. Thus $p_0=0.20$ and $q=1-0.20=0.80$.

Positive-payment count. The original count is Poisson with parameter 12, and each claim independently produces a positive payment with probability 0.80. Therefore

K\sim\operatorname{Poisson}(12(0.80)) =\operatorname{Poisson}(9.6).

Conditional severity. The unconditional probabilities of payments $50 and $400 are 0.50 and 0.30. Conditional on a positive payment, divide each by $q=0.80$:

P(Y^+=50)=\frac{0.50}{0.80}=0.625,\qquad P(Y^+=400)=\frac{0.30}{0.80}=0.375.

These probabilities sum to $0.625+0.375=1$. Finally, $P(K=0)=e^{-9.6}$, which rounds to 0.0001 to four decimal places. The method avoids treating the 12 original claims per expected count as though all 12 generate payments; the positive-payment count has Poisson parameter 9.6.

Worked Example: Binomial Exposure Count With Claim Payments

A hypothetical portfolio has six independent exposure units. Each unit generates one claim with probability 0.40, so the original count is $N\sim\operatorname{Binomial}(6,0.40)$. Conditional on a claim, its severity is $100, $300, or $700 with probabilities 0.25, 0.45, and 0.30, respectively. The per-claim payment is $Y=\min((X-200)_+,400)$. Find the distribution of $K$, the conditional positive-payment severity, and $P(K=2)$.

Method and payment map. A loss of $100 produces zero payment. A loss of $300 produces a $100 payment, and a loss of $700 produces a $400 payment after applying the deductible and limit. Given that a claim occurs, the probability of a positive payment is

$$ q=0.45+0.30=0.75. $$

For each exposure unit, the probability of a positive payment is the probability of a claim times the probability that the claim produces a positive payment:

$$ 0.40(0.75)=0.30. $$

The six exposure units act independently, so $K\sim\operatorname{Binomial}(6,0.30)$. Conditional on a positive payment, the severities are

P(Y^+=100)=\frac{0.45}{0.75}=0.60,\qquad P(Y^+=400)=\frac{0.30}{0.75}=0.40.

The requested count probability is

P(K=2)=\binom{6}{2}(0.30)^2(0.70)^4 =15(0.09)(0.2401)=0.324135.

Thus $P(K=2)=\mathbf{0.3241}$ to four decimal places. The conditional severity probabilities sum to $0.60+0.40=1$, and the probability $0.30$ used in the count model correctly includes both claim occurrence and positive payment given a claim. Multiplying the original claim count by 0.75 alone would omit the 0.40 claim-occurrence probability for each exposure.

Worked Example: Exponential Loss With a Deductible and a Payment Cap

A hypothetical insurer's original claim count is $N\sim\operatorname{Poisson}(10)$. Independent claim severities $X$ follow an exponential distribution with mean $1,000, so $P(X>x)=e^{-x/1000}$ for $x\ge0$. The payment is $Y=\min((X-500)_+,1500)$. Find the positive-payment count distribution, the conditional probability that a positive payment reaches the cap, and the conditional probability that a positive payment is less than $1,000. Give probabilities to four decimal places and the Poisson parameter to four significant figures.

Method and positive-payment probability. A payment is positive exactly when $X>500$. Hence

q=P(X>500)=e^{-500/1000}=e^{-0.5}=0.6065307\ldots.

Poisson thinning gives $K\sim\operatorname{Poisson}(10e^{-0.5})$, whose parameter is 6.065 to four significant figures.

Conditional positive-payment severity. Given $X>500$, the excess $Z=X-500$ has an exponential distribution with mean $1,000. Indeed, for $z\ge0$,

P(Z>z\mid X>500) =\frac{P(X>500+z)}{P(X>500)} =\frac{e^{-(500+z)/1000}}{e^{-500/1000}} =e^{-z/1000}.

The positive payment is $Y^+=\min(Z,1500)$. It reaches the $1,500 cap when $Z\ge1500$, so the conditional probability of a payment at the cap is

P(Y^+=1500)=P(Z\ge1500)=e^{-1500/1000}=e^{-1.5}=0.2231.

A positive payment is less than $1,000 exactly when $Z<1000$; the cap does not affect that event. Therefore

P(Y^+<1000)=1-e^{-1000/1000}=1-e^{-1}=0.6321.

The cap probability is an atom in the conditional payment distribution; it is not a continuous payment value to be spread over an interval. The calculation conditions on a positive payment first, which is why these probabilities use the excess distribution given $X>500$.

Common Exam Trap: Mixing the Count and Severity Adjustments

The frequency and severity adjustments are related but distinct. The positive-payment probability changes how many claims are counted; conditioning on positive payment changes the distribution of each counted payment. Applying only one of these adjustments does not give the correct compound representation.

For the first worked example, an incorrect count calculation would use $K\sim\operatorname{Poisson}(12)$, as if every original claim produced a positive payment. The correction is $K\sim\operatorname{Poisson}(12(0.80))=\operatorname{Poisson}(9.6)$. An incorrect severity calculation would retain the unconditional probability $P(Y=50)=0.50$. Among positive payments, the correct probability is $0.50/0.80=0.625$, because the zero-payment probability must be removed and the remaining probabilities renormalized.

Do not automatically multiply probabilities from separate coverage steps. If a claim must first be retained and then must generate a positive payment, the overall probability is $P(\text{retained and positive})$. It equals a product only when the relevant events have the required independence, or when the second probability is explicitly conditional on retention. Similarly, if payment outcomes are dependent across claims, or if an aggregate-level deductible determines payments using the total loss, independent claim-by-claim thinning generally does not apply.

Key takeaway: For independent claim-level payments, calculate $q=P(Y>0)$, thin the original frequency using $q$, and use the payment distribution conditional on $Y>0$ as the severity for counted payments. For Poisson original frequency, the positive-payment count is Poisson with parameter $\lambda q$.

Check Your Understanding

For each problem, distinguish the original count from the positive-payment count and condition the severity where requested.

A hypothetical insurer has $N\sim\operatorname{Poisson}(5)$ claims. Claim losses are $100, $400, and $900 with probabilities 0.30, 0.40, and 0.30. A $300 per-claim deductible applies with no limit. Find the distribution of $K$ and the conditional probabilities of the positive payment amounts. (Answer: $K\sim\operatorname{Poisson}(3.5)$; $Y^+=\$100$ with probability $4/7=0.5714$, and $Y^+=\$600$ with probability $3/7=0.4286$.)
Eight independent exposure units each generate a claim with probability 0.25. Given a claim, the payment is positive with probability 0.60, independently across units. Find the distribution of $K$ and $P(K=1)$. (Answer: $K\sim\operatorname{Binomial}(8,0.15)$; $P(K=1)=8(0.15)(0.85)^7=0.3847$.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(7)$. The payment per claim is zero with probability 0.20, $200 with probability 0.30, and $500 with probability 0.50. Find $K$, $P(K=0)$, and the conditional probabilities of the positive payment amounts. (Answer: $K\sim\operatorname{Poisson}(5.6)$; $P(K=0)=e^{-5.6}=0.0037$; $P(Y^+=\$200)=0.375$ and $P(Y^+=\$500)=0.625$.)
Claim frequency is Poisson with parameter 4, and severity is exponential with mean $200. A $100 deductible and a $300 payment cap apply, so $Y=\min((X-100)_+,300)$. Find the positive-payment count parameter and the conditional probability of payment at the cap. (Answer: $K\sim\operatorname{Poisson}(4e^{-0.5})$, with parameter 2.426; $P(Y^+=\$300)=e^{-1.5}=0.2231$.)
Original claims follow a Poisson distribution with parameter $\lambda$. Each claim is independently retained with probability 0.80, and, conditional on retention, produces a positive payment with probability 0.75. Find the positive-payment count distribution. State what additional severity distribution is needed to complete a positive-payment compound representation. (Answer: $K\sim\operatorname{Poisson}(0.60\lambda)$; the required severity is the payment distribution conditional on both retention and a positive payment.)

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