Exact Aggregate Probabilities by Discrete Convolution

From Covered Payments to an Aggregate Distribution

When individual covered payments take only a few possible amounts, an aggregate probability can often be calculated exactly by listing the ways those payments can add to the target total. The calculation has two parts: first find the distribution of total payment given the number of claims; then average those conditional probabilities over the frequency distribution. Keeping these parts separate helps prevent a frequent error: using the probability of a claim count as though it were already the probability of an aggregate amount.

The payment distribution used here is the distribution after any applicable contract terms have been applied. Thus, a deductible, limit, or coinsurance feature is already reflected in each possible covered payment. The tutorial concerns the exact aggregation of that discrete payment distribution; it does not derive the contract transformation itself.

Definition: Let $N$ be the number of claims, and let $Y_1,Y_2,\ldots$ be independent, identically distributed covered payments, independent of $N$. The aggregate covered loss is $S=\sum_{i=1}^{N}Y_i$, with $S=0$ when $N=0$. Suppose each payment is a nonnegative integer multiple of a unit $c$, and define $h_j=P(Y_i=jc)$ for $j=0,1,2,\ldots$. The $n$-fold convolution $h_j^{*n}$ is the probability that $n$ independent payments sum to $jc$. By convention, $h_0^{*0}=1$ and $h_j^{*0}=0$ for $j>0$.

The unit $c$ is a convenient scale, not a claim that every possible payment must be a whole dollar. For example, payments in multiples of $50 can be represented with $c=\$50$. In the examples below, the monetary unit is stated explicitly.

Conditional Convolution and Frequency Weighting

For a fixed number of claims, enumerate the combinations of payments that produce the target total. For two payments, for instance, the probability of total $jc$ is the sum of $h_kh_{j-k}$ over all possible $k$. Repeating this operation gives the conditional aggregate distribution for any fixed claim count.

h_j^{*n}=\sum_{k=0}^{j}h_kh_{j-k}^{*(n-1)} \quad (n\geq 1), \qquad P(S=jc)=\sum_{n=0}^{\infty}P(N=n)h_j^{*n}.

The first equality builds the distribution for $n$ payments from the distribution for $n-1$ payments and one new payment. The second equality follows by partitioning the event $S=jc$ according to the possible values of $N$. These events are mutually exclusive, so their probabilities add. Given $N=n$, independence and the specified payment distribution make the conditional probability of the target total exactly $h_j^{*n}$.

The frequency probabilities in the sum must match the stated model. For a binomial count $N\sim\operatorname{Binomial}(m,q)$, the support is $0,\ldots,m$ and $P(N=n)=\binom{m}{n}q^n(1-q)^{m-n}$. For a Poisson count with parameter $\lambda$, $P(N=n)=e^{-\lambda}\lambda^n/n!$. For a negative binomial count parameterized by a positive integer $r$ and success probability $p$, let $N$ count failures before the $r$th success; then $P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n$. Other negative binomial parameterizations exist, so the definition in the problem must be checked before applying this formula.

Key result: Under the assumptions of independent claim count and independent, identically distributed discrete payments, the aggregate probability at $jc$ is a frequency-weighted sum of the conditional $n$-payment probabilities. If every payment is at least $c>0$, then $h_j^{*n}=0$ for $n>j$; only finitely many claim counts can contribute to $P(S=jc)$. If a zero payment is possible, arbitrarily many claims may still produce the same total, so this finite bound does not hold for an unbounded frequency model.

For binomial frequency, there are only finitely many count values in any case. For Poisson or negative binomial frequency with a zero-payment atom, the sum over claim counts may be infinite even at a fixed aggregate amount. It is not exact to stop at an arbitrary claim count and ignore the remaining terms. A calculation must either include the remaining contribution or use a valid transformation of the model, such as correctly accounting for the count of positive payments. This issue is distinct from rounding intermediate probabilities.

Worked Calculations

Worked Example: Poisson Frequency and a Target Total

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ claims. Each claim independently produces a covered payment of $1 or $2, each with probability $0.5$. Find the probability that aggregate covered loss is $3. The stated payments are the final covered amounts; there are no further payment modifications.

Method. Use units of $c=\$1$, so $h_1=h_2=0.5$. Since every payment is at least $1, at most three claims can produce a total of $3. One claim cannot produce $3. For two claims, the possible ordered pairs are $(1,2)$ and $(2,1)$, so $h_3^{*2}=2(0.5)(0.5)=0.5$. For three claims, all three payments must be $1, so $h_3^{*3}=(0.5)^3=0.125$.

The relevant Poisson probabilities are $P(N=2)=e^{-1}/2!=0.1839397$ and $P(N=3)=e^{-1}/3!=0.0613132$. Thus,

P(S=\$3) =P(N=2)h_3^{*2}+P(N=3)h_3^{*3} =(0.1839397)(0.5)+(0.0613132)(0.125) =0.0996341.

The exact expression is $\frac{13}{48}e^{-1}$, which is approximately 0.0996 to four decimal places. The method avoids the mistake of considering only the two-claim combinations: three $1 payments also produce $3, and their probability must be included. As a check, both contributing terms are nonnegative, and the result is less than $P(N\geq 2)$, as required for this event.

Worked Example: Binomial Frequency and the Full Aggregate Distribution

A hypothetical benefit program has $N\sim\operatorname{Binomial}(2,0.5)$ claims. Each claim independently results in a covered payment of $1 or $2, each with probability $0.5$. Find the probability mass function of aggregate covered loss.

Method. The claim-count probabilities are $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$. Conditional on one claim, the aggregate is $1 or $2 with probability $0.5$ each. Conditional on two claims, the sums $2, $3, and $4 have probabilities $0.25, 0.50,$ and $0.25$, respectively: the middle sum arises from the two ordered pairs $(1,2)$ and $(2,1)$.

Weight each conditional distribution by its claim-count probability. For aggregate $1, only $N=1$ contributes. For $2, both $N=1$ and $N=2$ contribute. The complete calculation is:

\begin{aligned} P(S=\$0)&=0.25,\\ P(S=\$1)&=(0.50)(0.50)=0.25,\\ P(S=\$2)&=(0.50)(0.50)+(0.25)(0.25)=0.3125,\\ P(S=\$3)&=(0.25)(0.50)=0.125,\\ P(S=\$4)&=(0.25)(0.25)=0.0625. \end{aligned}

The probabilities sum to $0.25+0.25+0.3125+0.125+0.0625=1.0000$, so the distribution is complete. In particular, $P(S=\$2)=0.3125$. This method avoids assigning the entire probability $P(N=1)$ to an aggregate of $2: when $N=1$, only half the payments are $2, and two claims can also sum to $2.

Worked Example: Negative Binomial Frequency With a Target Total

A hypothetical portfolio has a negative binomial claim count $N$, defined as the number of failures before the second success, with $r=2$ and $p=0.5$. Each claim independently produces a covered payment of $1 or $2, each with probability $0.5$. Find $P(S=\$3)$.

Method. Under the stated parameterization, $P(N=n)=\binom{n+1}{n}(0.5)^2(0.5)^n$. The needed count probabilities are $P(N=2)=3(0.5)^4=0.1875$ and $P(N=3)=4(0.5)^5=0.125$. One claim cannot produce $3, and four or more claims cannot produce $3 because each payment is at least $1.

For two claims, the ordered pairs $(1,2)$ and $(2,1)$ give $h_3^{*2}=2(0.5)(0.5)=0.5$. For three claims, only three $1 payments give a total of $3, so $h_3^{*3}=(0.5)^3=0.125$. Therefore,

\begin{aligned} P(S=\$3) &=(0.1875)(0.5)+(0.125)(0.125)\\ &=0.09375+0.015625\\ &=0.109375=\frac{7}{64}. \end{aligned}

The answer is 0.1094 to four decimal places. This calculation uses the negative binomial probabilities for the specified failure-count convention; substituting probabilities from a different negative binomial parameterization would give incorrect weights. The two terms account for all possible claim counts and payment combinations, so no omitted count remains.

Checks and Common Traps

Before accepting an aggregate probability calculation, check the payment scale, the frequency parameterization, and the set of claim counts that can reach the target. If the payment distribution has a positive minimum, the target total gives a useful finite upper bound on the count. If zero payments are possible, that shortcut must not be used for an unbounded frequency model.

Forgetting frequency weights. A conditional probability $h_j^{*n}$ is not an aggregate probability by itself; it must be multiplied by $P(N=n)$.
Counting unordered pairs only once. When claim payments are independent and claims are distinguishable, $(1,2)$ and $(2,1)$ are separate outcomes. Their probabilities both contribute.
Omitting a feasible count. A target total of $3 can arise from two payments of $1 and $2, and also from three payments of $1. Check all feasible counts.
Truncating when zero payments are possible. A large claim count can still yield a small aggregate if enough claims pay zero. A finite count bound based only on the target total is then invalid.
Changing the frequency convention silently. For a negative binomial model, state what $N$ counts and how its parameters define $P(N=n)$ before weighting the convolutions.

Key takeaway: First calculate the payment-sum probabilities conditional on each possible claim count; then weight them by that count’s probability. A complete exact calculation includes every feasible contribution, respects the stated frequency parameterization, and verifies that any full probability mass function sums to one.

Check Your Understanding

For each problem, use the stated covered-payment distribution and frequency model. Amounts are in dollars unless noted.

A hypothetical account has $N\sim\operatorname{Binomial}(2,0.3)$, and each claim pays exactly $2. Find $P(S=\$4)$. (Answer: 0.0900)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$. Each payment is $1 or $2 with probability $0.5$ each. Find $P(S=\$2)$. (Answer: $1.5e^{-2}$, approximately 0.2030)
A hypothetical count is negative binomial, counting failures before the first success, with success probability $p=0.5$. Each claim pays $1 or $2 with probability $0.5$ each. Find $P(S=\$1)$. (Answer: 0.1250)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(1.5)$. Each payment is $1 with probability $0.4$ and $2 with probability $0.6$. Find $P(S=\$2)$, including every feasible claim count. (Answer: $1.08e^{-1.5}$, approximately 0.2410)
A hypothetical contract has $N\sim\operatorname{Binomial}(3,0.25)$. Independently for each claim, the covered payment is $0 with probability $0.2$ and $2 with probability $0.8$. Find $P(S=0)$. (Answer: 0.5120)

What You'll Learn