Why Convolution Is Useful
When the number of claims is fixed, an aggregate loss is a sum of individual claim amounts. If claim severities take only a few possible values, we can calculate the distribution of that sum directly. The key operation is convolution: for each possible total, add the probabilities of all combinations of component losses that produce it.
Repeating that operation one claim at a time is recursive convolution. The method is transparent and works for discrete distributions even when the possible amounts have gaps. It also provides a useful check on calculations from other aggregate-loss methods. The examples use hypothetical insurance portfolios; no interest or mortality assumptions are needed because the calculations concern claim amounts only.
How the Recursion Works
To calculate the distribution for \(n\) claims, take each possible amount \(jh\) for one claim and add it to every total \(rh\) from the distribution for \(n-1\) claims. The resulting total is \((j+r)h\), and the probability contributed by that pair is \(p_jp^{(n-1)}_r\). If multiple pairs produce the same total, add their contributions.
The support must be managed as carefully as the probabilities. If the one-claim support is \(J\), then the possible totals for \(n\) claims belong to the \(n\)-fold sumset \(J+\cdots+J\). They need not fill every value between the smallest and largest totals. An absent amount has probability zero; keeping it in a table can be useful, but it must not receive probability just because it lies between two attainable amounts.
This identity follows by partitioning the event \(S_n=kh\) according to the value of \(X_n\). For each possible \(jh\), the remaining sum \(S_{n-1}\) must equal \((k-j)h\). Independence gives \(P(X_n=jh,\ S_{n-1}=(k-j)h)=p_jp^{(n-1)}_{k-j}\). The events for different values of \(j\) are disjoint, so summing their probabilities gives the recurrence.
List every attainable severity and its probability. Check that the probabilities sum to 1.
Place probability 1 at total loss 0. Convolve this table with the severity table once for each claim.
For each new claim amount, shift the existing totals and multiply their probabilities. Add contributions that land at the same total.
Confirm that probabilities are nonnegative, sum to 1, and occur only at attainable totals.
Worked Tables for Fixed Claim Counts
Worked Example: Three Claims With Two Possible Severities
A hypothetical small equipment policy has exactly three claims. Each claim is independently \$1 with probability 0.6 or \$3 with probability 0.4. Find the aggregate-loss distribution.
Method. Use recursive convolution, starting with one claim and convolving by the one-claim severity distribution twice more. The severity probabilities sum to \(0.6+0.4=1\).
For one claim, the totals are 1 and 3, with probabilities 0.6 and 0.4. For two claims, the possible totals and contributions are:
- Total 2: \(1+1\), with probability \(0.6(0.6)=0.36\).
- Total 4: \(1+3\) or \(3+1\), with probability \(0.6(0.4)+0.4(0.6)=0.48\).
- Total 6: \(3+3\), with probability \(0.4(0.4)=0.16\).
Now convolve the two-claim table with the one-claim table. For example, total 5 can arise from a previous total of 2 and a new loss of 3, or a previous total of 4 and a new loss of 1. Its probability is \(0.36(0.4)+0.48(0.6)=0.432\). The complete calculation is:
| Total loss | Contributions | Probability |
|---|---|---|
| 3 | \(0.36(0.6)\) | 0.216 |
| 5 | \(0.36(0.4)+0.48(0.6)\) | 0.432 |
| 7 | \(0.48(0.4)+0.16(0.6)\) | 0.288 |
| 9 | \(0.16(0.4)\) | 0.064 |
The probabilities sum to \(0.216+0.432+0.288+0.064=1.000\). The support also checks: three claims, each between \$1 and \$3, produce totals from \$3 to \$9, and only the odd totals listed are attainable.
Common mistake avoided: For total 5, counting only one ordering gives \(0.6^2(0.4)=0.144\), which omits the other ordering. The correct probability is \(3(0.6^2)(0.4)=0.432\), as the convolution table shows.
Worked Example: Zero Severities and Gaps in the Support
A hypothetical coverage has two independent claims. Each claim payment is \$0 with probability 0.2, \$2 with probability 0.5, or \$5 with probability 0.3. Find the distribution of the total payment.
Method. Convolve the one-claim table with itself. The probabilities sum to \(0.2+0.5+0.3=1\). The possible sums come from the pairs \(0+0\), \(0+2\), and so on. Since the severities are 0, 2, and 5, the attainable totals are 0, 2, 4, 5, 7, and 10.
| Total payment | Contributions | Probability |
|---|---|---|
| 0 | \(0.2(0.2)\) | 0.04 |
| 2 | \(0.2(0.5)+0.5(0.2)\) | 0.20 |
| 4 | \(0.5(0.5)\) | 0.25 |
| 5 | \(0.2(0.3)+0.3(0.2)\) | 0.12 |
| 7 | \(0.5(0.3)+0.3(0.5)\) | 0.30 |
| 10 | \(0.3(0.3)\) | 0.09 |
The check is \(0.04+0.20+0.25+0.12+0.30+0.09=1.00\). In particular, total 5 has two ordered ways to occur, but total 6 has none: no pair selected from \(\{0,2,5\}\) sums to 6.
Common mistake avoided: Treating every integer from 0 through 10 as an attainable total would incorrectly assign probability to 1, 3, 6, 8, and 9. These values are between the minimum and maximum, but they are not in the two-claim sumset.
When the Claim Count Is Random
If the claim count \(N\) is random, first construct the fixed-count distributions and then weight each one by the probability of its count. Write \(p_n=P(N=n)\). Assuming the severities are independent of \(N\) and have the same distribution for each claim, total probability gives
The \(n=0\) term matters: with no claims, the aggregate is zero. If \(N\) has finite support, the displayed sum has only finitely many nonzero terms. If its support is infinite, do not silently discard terms in an exact calculation; any truncation needs an explicit approximation and an assessment of the omitted probability.
Worked Example: Mixing Fixed-Count Tables
A hypothetical service plan has \(P(N=0)=0.25\), \(P(N=1)=0.50\), and \(P(N=2)=0.25\) claims in a period. Claim amounts are independent of the count and are \$1 with probability 0.75 or \$4 with probability 0.25. Find the aggregate-loss distribution.
Method. Construct the distributions for zero, one, and two claims, then multiply each probability by the corresponding probability of \(N\). The count probabilities sum to \(0.25+0.50+0.25=1\), and the severity probabilities sum to \(0.75+0.25=1\).
For \(N=0\), the total is 0 with probability 1. For \(N=1\), the total is 1 with probability 0.75 and 4 with probability 0.25. For \(N=2\), convolution gives total 2 with probability \(0.75^2=0.5625\), total 5 with probability \(2(0.75)(0.25)=0.375\), and total 8 with probability \(0.25^2=0.0625\).
Weight each row by its count probability. Thus, for example, \(P(S=1)=0.50(0.75)=0.375\), and \(P(S=5)=0.25(0.375)=0.09375\).
| Total loss | Weighted contributions | Probability |
|---|---|---|
| 0 | \(0.25(1)\) | 0.25 |
| 1 | \(0.50(0.75)\) | 0.375 |
| 2 | \(0.25(0.5625)\) | 0.140625 |
| 4 | \(0.50(0.25)\) | 0.125 |
| 5 | \(0.25(0.375)\) | 0.09375 |
| 8 | \(0.25(0.0625)\) | 0.015625 |
The probabilities sum to \(0.25+0.375+0.140625+0.125+0.09375+0.015625=1.000000\). Total 4 comes only from one claim of \$4; total 5 and total 8 can occur only with two claims. The support therefore agrees with the count-specific tables.
Common mistake avoided: The probability of total 0 is not just the probability of no claims. A claim can also have zero severity in some models, so both sources of zero aggregate loss would need to be included. In this example, severities are strictly positive, so \(P(S=0)=P(N=0)=0.25\).
Checks and Common Exam Traps
Direct convolution is straightforward, but a small support or bookkeeping error can change several probabilities. Use these checks as part of the calculation, not just as a final formality:
- Check each severity table first. The one-claim probabilities must sum to 1. A missing probability will propagate through every recursive step.
- Count all ordered combinations. The combinations \(a+b\) and \(b+a\) are distinct outcomes when claims are labeled, even though they produce the same total. Their probabilities both contribute.
- Track attainable totals, not just a range. Gaps in the support remain gaps unless some combination actually fills them.
- Include the zero-claim atom. When \(N=0\), the aggregate is zero with probability 1 conditional on that count.
- Check the final sum. A probability total different from 1 signals an omitted, duplicated, or incorrectly weighted contribution.
Recursive convolution is most useful when the severity support and claim count are small enough for an explicit table. It makes the origins of each aggregate probability visible, including ties where several combinations produce the same total. The same support discipline is important in larger computations, even when calculations are automated.
Check Your Understanding
For each problem, construct the relevant convolution table and check that the probabilities sum to 1.
- Two independent claims each equal \$1 with probability 0.8 or \$4 with probability 0.2. What is the distribution of their total? (Answer: \(P(S=2)=0.64,\ P(S=5)=0.32,\ P(S=8)=0.04\).)
- Two independent payments are \$0, \$3, or \$6 with probabilities 0.2, 0.5, and 0.3, respectively. Give the attainable total payments and their probabilities. (Answer: \(P(S=0)=0.04,\ P(S=3)=0.20,\ P(S=6)=0.37,\ P(S=9)=0.30,\ P(S=12)=0.09\).)
- Exactly three independent claims each equal \$2 with probability 0.6 or \$7 with probability 0.4. Find all possible totals and their probabilities. (Answer: \(P(S=6)=0.216,\ P(S=11)=0.432,\ P(S=16)=0.288,\ P(S=21)=0.064\).)
- A claim count has probabilities \(P(N=0)=0.4,\ P(N=1)=0.4,\ P(N=2)=0.2\). Each claim is \$2 with probability 0.75 or \$5 with probability 0.25, independently of the count. Find the aggregate distribution. (Answer: \(P(S=0)=0.4,\ P(S=2)=0.3,\ P(S=4)=0.1125,\ P(S=5)=0.1,\ P(S=7)=0.075,\ P(S=10)=0.0125\).)
- Two independent claims each equal \$0, \$2, or \$5 with probabilities 0.1, 0.6, and 0.3. What is the probability that their total is at least \$7? (Answer: \(P(S\geq 7)=0.45\), from totals 7 and 10.)