Worked Chain: Binomial Frequency With Discrete Severity

From Ground-Up Losses to an Exact Aggregate Distribution

A compound loss calculation links two sources of randomness: how many claims occur and how much is paid for each claim. When the number of claims has a binomial distribution and ground-up losses take only finitely many values, the aggregate distribution can be calculated exactly. The main work is to apply the contract terms to each possible loss, then combine the resulting payments with the frequency model without mixing up conditional and unconditional probabilities.

The setting here is a fixed number of independent exposure opportunities. Each opportunity produces a claim with the same probability, and claim severities are independent and identically distributed, independently of the claim count. This discrete calculation is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $N\sim\operatorname{Bin}(n,q)$ be the number of claims from $n$ independent exposure opportunities, each with claim probability $q$. Let $X_j$ be the ground-up severity of claim $j$, with the $X_j$ independent and identically distributed and independent of $N$. For a per-claim deductible $d$ and a maximum insurer payment $u$ per claim, define the covered payment $Y_j=\min\{(X_j-d)_+,u\}$, where $(z)_+=\max(z,0)$. The aggregate covered loss is $S=\sum_{j=1}^{N}Y_j$, with $S=0$ when $N=0$.

In this definition, $u$ is the cap on the insurer’s payment after applying the deductible. Thus a ground-up loss greater than $d+u$ still produces a payment of $u$, not a payment of $u-d$. If the contract instead states a cap on ground-up loss, the payment formula must reflect that different wording.

Conditioning on the Claim Count

For each possible value $k$ of $N$, calculate the distribution of the sum of $k$ covered claim payments. Then weight that conditional distribution by the probability that the claim count is $k$. For a binomial frequency model, $P(N=k)=\binom{n}{k}q^k(1-q)^{n-k}$, for $k=0,\ldots,n$.

P(S=s)=\sum_{k=0}^{n}P(N=k)\,P\left(\sum_{j=1}^{k}Y_j=s\right).

The term for $k=0$ is a point mass at zero: its conditional aggregate payment is zero with probability one. For $k\geq 1$, the conditional distribution is the $k$-fold convolution of the covered payment distribution. The severity probabilities in that convolution are conditional on a claim occurring; the probability $q$ belongs in the binomial frequency probabilities, not in each conditional severity probability.

There is also an equivalent exposure-by-exposure calculation. One exposure contributes no payment either because no claim occurs or because its claim has zero covered payment. If $G_Y(z)=E[z^Y]$ is the probability generating function of a covered claim payment, the aggregate probability generating function is $[1-q+qG_Y(z)]^n$. This follows because an exposure has generating function $1-q+qG_Y(z)$, and the $n$ exposure contributions are independent. Coefficients of the expanded polynomial give the exact aggregate probabilities when payments lie on an integer-valued grid.

Why the conditioning formula holds: The events $N=0,\ldots,N=n$ are mutually exclusive and exhaustive. By the law of total probability, $P(S=s)=\sum_{k=0}^{n}P(S=s\mid N=k)P(N=k)$. Given $N=k$, the aggregate is the sum of $k$ independent covered severities, so its distribution is their $k$-fold convolution. For $k=0$, the sum is defined to be zero. These statements give the conditioning formula above.

Worked Calculations

Worked Example: Two Binomial Exposure Opportunities

A hypothetical policy portfolio has $n=2$ independent exposure opportunities, each with claim probability $q=0.5$. Conditional on a claim, the ground-up loss $X$ is $\$70$ or $\$130$, each with probability $0.5$. The policy has a per-claim deductible of $\$30$ and a maximum insurer payment of $\$100$ per claim. Find the exact distribution of aggregate covered loss $S$.

Method: transform the severity, then condition on $N$. For $X=\$70$, the payment is $\min(70-30,100)=\$40$. For $X=\$130$, it is $\min(130-30,100)=\$100$. Thus a covered claim payment is $\$40$ or $\$100$, each with probability $0.5$.

The binomial probabilities are $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$. Given $N=0$, $S=0$. Given $N=1$, $S$ is $\$40$ or $\$100$, each with conditional probability $0.5$. Given $N=2$, the two independent payments produce:

$\$80$ with probability $0.25$, $\$140$ with probability $0.50$, and $\$200$ with probability $0.25$. Weighting by the frequency probabilities gives $P(S=\$80)=0.25(0.25)=0.0625$, $P(S=\$140)=0.25(0.50)=0.1250$, and $P(S=\$200)=0.25(0.25)=0.0625$.

For $N=1$, the unconditional contributions are $P(S=\$40)=0.50(0.50)=0.2500$ and $P(S=\$100)=0.50(0.50)=0.2500$. Including the no-claim case, the aggregate distribution is:

Aggregate loss $S$	Probability
$\$0$	0.2500
$\$40$	0.2500
$\$80$	0.0625
$\$100$	0.2500
$\$140$	0.1250
$\$200$	0.0625

Check. The probabilities sum to $0.2500+0.2500+0.0625+0.2500+0.1250+0.0625=1.0000$. Also $E[N]=2(0.5)=1$, and the mean covered payment per claim is $(40+100)/2=\$70$, so $E[S]=\$70$. The distribution gives $40(0.25)+80(0.0625)+100(0.25)+140(0.125)+200(0.0625)=\$70$. The check confirms the result. The method avoids applying the claim probability a second time to the conditional severity probabilities.

Worked Example: A Deductible Can Produce Zero Payment

A hypothetical insurer observes $n=2$ independent exposure opportunities with $q=0.5$. Conditional on a claim, $X$ is $\$40,\$90,$ or $\$160$, with probabilities $0.25,0.50,$ and $0.25$, respectively. The per-claim deductible is $\$60$, and the maximum insurer payment is $\$80$. Find the exact aggregate distribution.

Method: retain zero-payment claims in the severity distribution. Applying $Y=\min\{(X-60)_+,80\}$ gives payment $\$0$ for $X=\$40$, $\$30$ for $X=\$90$, and $\$80$ for $X=\$160$. Conditional on a claim, these payments have probabilities $0.25,0.50,$ and $0.25$.

For one exposure, the unconditional payment is zero with probability $0.5+0.5(0.25)=0.625$: either no claim occurs, or a claim falls below the deductible. The exposure produces $\$30$ with probability $0.5(0.50)=0.25$, and $\$80$ with probability $0.5(0.25)=0.125$.

The two exposure contributions are independent. Convolving their distributions gives $P(S=0)=0.625^2=0.390625$; $P(S=\$30)=2(0.625)(0.25)=0.312500$; and $P(S=\$60)=0.25^2=0.062500$. It also gives $P(S=\$80)=2(0.625)(0.125)=0.156250$, $P(S=\$110)=2(0.25)(0.125)=0.062500$, and $P(S=\$160)=0.125^2=0.015625$.

Check. The six probabilities sum to $1.000000$. The mean payment per exposure is $30(0.25)+80(0.125)=\$17.50$, so $E[S]=2(\$17.50)=\$35$. The aggregate distribution gives $30(0.3125)+60(0.0625)+80(0.15625)+110(0.0625)+160(0.015625)=\$35$. The result is consistent.

Worked Example: An Exact Aggregate Tail Probability

A hypothetical block has $n=3$ independent exposure opportunities, with claim probability $q=0.5$ for each. Conditional on a claim, the ground-up loss is $\$80$ or $\$130$, each with probability $0.5$. A deductible of $\$50$ applies, followed by a maximum insurer payment of $\$100$ per claim. Calculate $P(S>\$100)$.

Method: convolve the unconditional contribution from each exposure. The covered claim payments are $\$30$ and $\$80$, each with conditional probability $0.5$. For one exposure, the unconditional contribution is zero with probability $0.5$, $\$30$ with probability $0.25$, and $\$80$ with probability $0.25$.

Classify the outcomes for three exposures by the counts $a$, $b$, and $c$ of $\$30$, $\$80$, and zero contributions, where $a+b+c=3$. Each particular ordered combination with those counts has probability $0.25^{a+b}0.5^c$, and there are $3!/(a!b!c!)$ such combinations. The aggregate amount is $30a+80b$.

The outcomes exceeding $\$100$ are: one $\$30$ and one $\$80$, with the third contribution zero; two $\$80$ contributions and one zero; and the all-nonzero three-exposure outcomes with at least two $\$80$ contributions. Their probabilities are, respectively, $6(0.25)(0.25)(0.5)=0.1875$, $3(0.25)^2(0.5)=0.09375$, $3(0.25)^3=0.046875$, and $0.25^3=0.015625$. The third term is the two-$\$30$, one-$\$80$ outcome, and the last is three $\$80$ payments.

Therefore $P(S>\$100)=0.1875+0.09375+0.046875+0.015625=0.34375$. The combination of one $\$30$ and one $\$80$ contributes because its total is $\$110$, not $\$100$. Check: the boundary outcome of two $\$30$ payments and one zero is exactly $\$60$, while one $\$30$ and one $\$80$ is $\$110$; the strict threshold is applied to the aggregate, not to individual claims.

Checks and Common Errors

Two checks are especially useful. First, probabilities across all possible aggregate amounts must sum to one. Second, the aggregate mean must agree with the frequency-severity calculation. Under the independence assumptions here, if $Y$ denotes the payment conditional on a claim, then $E[S]=E[N]E[Y]=nqE[Y]$. A zero covered payment is allowed in $Y$; it contributes zero to the mean but does not alter the stated claim frequency.

Common trap: Do not set the probability of zero aggregate payment equal to $P(N=0)$ when a claim can fall below the deductible. In the second example, that incorrect calculation gives $P(S=0)=P(N=0)=0.25$. In fact, each exposure has probability $0.625$ of producing zero payment, so $P(S=0)=0.625^2=0.390625$. The error is treating claims with no covered payment as impossible rather than as zero-valued payments.

A second frequent error is to apply the claim probability $q$ twice. When conditioning on $N=k$, use the severity probabilities conditional on a claim; the binomial probability already accounts for whether claims occur. Alternatively, work exposure by exposure with unconditional contributions, in which case the probability $q$ is included in those contribution probabilities. Mixing these two approaches generally produces an incorrect distribution.

Check Your Understanding

For each problem, transform the ground-up losses before combining them with the binomial frequency. Treat the listed claim severities as conditional on a claim.

A hypothetical portfolio has $n=2$, $q=0.3$, and covered claim payments of $\$40$ and $\$100$, each with probability $0.5$. Find $P(S=\$140)$. (Answer: 0.0450)
For $n=2$ and $q=0.4$, ground-up severity is $\$50$ or $\$100$, equally likely. A deductible of $\$50$ and a maximum payment of $\$100$ apply. Find $P(S=\$50)$. (Answer: 0.3200)
There are $n=3$ exposures with $q=1/3$. Conditional on a claim, the covered payment is $\$20$ or $\$60$, each with probability $0.5$. Find $P(S=\$80)$. (Answer: 0.1111)
For $n=2$ and $q=0.5$, ground-up severity is $\$40,\$80,$ or $\$120$, with probabilities $0.25,0.50,$ and $0.25$. Apply a deductible of $\$50$ and a maximum payment of $\$70$. Find $P(S=\$100)$. (Answer: 0.0625)
There are $n=3$ exposures with $q=0.2$. Conditional on a claim, a hypothetical coverage produces a payment of $\$25$ or $\$75$, equally likely. Find $P(S>\$100)$. (Answer: 0.0310)

Aggregate loss \(S\)	Probability
\(\$0\)	0.2500
\(\$40\)	0.2500
\(\$80\)	0.0625
\(\$100\)	0.2500
\(\$140\)	0.1250
\(\$200\)	0.0625

What You'll Learn