Exact Convolution for Discrete Aggregate Losses

Why Exact Convolution Matters

When claim payments take values on a finite set, an aggregate-loss distribution can be calculated directly rather than approximated. The calculation has two parts: first find the distribution of total payment for each possible claim count; then weight each of those distributions by the probability of that claim count. The key operation is discrete convolution, which combines the possible payment amounts and their probabilities.

The zero-loss outcome deserves particular attention. The aggregate loss is zero if there are no claims, but it can also be zero when claims occur and their modeled payments are zero. A complete calculation preserves every route to a zero total and checks that the resulting probabilities sum to one. These techniques are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $N$ be the number of claims, with frequency probabilities $p_n=P(N=n)$. Suppose the claim payments $Y_1,Y_2,\ldots$ are independent and identically distributed, are independent of $N$, and have a discrete probability mass function $f(y)=P(Y_i=y)$. The aggregate loss is $S=\sum_{i=1}^{N}Y_i$, with $S=0$ when $N=0$. The distribution of $S$ is obtained by convolving the payment distribution for each fixed claim count and mixing over $N$.

For a fixed count $n$, write $f^{*n}(s)=P(Y_1+\cdots+Y_n=s)$. By convention, $f^{*0}(0)=1$ and $f^{*0}(s)=0$ for $s\ne 0$: with no claims, the aggregate payment is exactly zero. The recursion for the fixed-count distribution is

f^{*n}(s)=\sum_y f^{*(n-1)}(s-y)f(y),\qquad n\ge 1.

The sum runs over the possible payment values $y$. A term contributes only when $s-y$ is a possible total for the first $n-1$ payments. Once the fixed-count distributions are calculated, conditioning on $N$ gives

P(S=s)=\sum_{n=0}^{\infty}p_n f^{*n}(s).

If the frequency distribution has a finite maximum count, this mixture is a finite sum. If it has unbounded support, the expression is an infinite sum; calculating only a finite number of terms is then a numerical truncation, not an exact finite mixture.

Why the Convolution Formula Holds

For $n$ payments, let the total after $n-1$ payments be $T_{n-1}=Y_1+\cdots+Y_{n-1}$. To have $T_{n-1}+Y_n=s$, the last payment must equal some possible value $y$, while the earlier payments must total $s-y$. Independence gives

P(T_{n-1}+Y_n=s\mid Y_n=y) =P(T_{n-1}=s-y) =f^{*(n-1)}(s-y).

The mutually exclusive possible values of $Y_n$ partition the event that the total is $s$. Multiplying by $P(Y_n=y)=f(y)$ and summing over those values proves the convolution formula. Similarly, conditioning on the mutually exclusive values of $N$ proves the aggregate mixture formula.

Convolution also provides a useful probability check. Summing the convolution over every possible total $s$ gives

\sum_s f^{*n}(s) =\sum_y f(y)\sum_s f^{*(n-1)}(s-y) =1\cdot 1=1.

Thus each fixed-count distribution should sum to one, and a mixture weighted by frequency probabilities that also sum to one should sum to one. A discrepancy usually signals an omitted total, an incorrect combination of terms, or a missing frequency probability.

A Reliable Calculation Sequence

For severity values that are integer multiples of a common unit $h$, it is often convenient to work with indices rather than dollar amounts. For example, if payments are in multiples of $500, a payment of $1,500 has index 3. Convolution then adds indices; convert back to dollars when presenting the aggregate distribution. The following sequence keeps the frequency and severity calculations separate.

State the assumptions and supports.
List the possible claim counts, their probabilities, and the possible payment amounts with their probabilities. Confirm that payments are independent and identically distributed and independent of the claim count.

Start at zero claims.
Set the fixed-count distribution for $n=0$ to probability one at aggregate loss zero. Do not discard this case.

Convolve to the largest needed count.
For each new payment, combine every existing total with every possible payment amount, multiplying the two probabilities and adding products that lead to the same total.

Mix over claim counts.
At each aggregate total, multiply each fixed-count probability by the probability of that count and add the contributions.

Audit the result.
Check that the support contains all attainable totals, that zero loss includes every possible source, and that all aggregate probabilities sum to one.

Worked Convolution Examples

Worked Example: A Binomial Count and Two Payment Amounts

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$ claims. Each claim payment is independently $1 or $2, each with probability 0.5, independently of $N$. Find the exact aggregate-loss distribution.

Method. First calculate the fixed-count distributions, then weight them by the binomial probabilities. The count probabilities are $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$.

For zero claims, the total is $0 with probability 1. For one claim, the total is $1 with probability 0.5 and $2 with probability 0.5. For two claims, enumerate the ordered pairs of payments. Each pair has probability $0.5(0.5)=0.25$:

\begin{aligned} P(Y_1+Y_2=2)&=0.25,\\ P(Y_1+Y_2=3)&=0.25+0.25=0.50,\\ P(Y_1+Y_2=4)&=0.25. \end{aligned}

Now mix the fixed-count probabilities. At $0, only $N=0$ contributes, so $P(S=0)=0.25$. At $1, only $N=1$ contributes, so $P(S=1)=0.50(0.50)=0.25$. At $2, the one-claim and two-claim cases both contribute:

$$ P(S=2)=0.50(0.50)+0.25(0.25)=0.25+0.0625=0.3125. $$

The remaining totals are $P(S=3)=0.25(0.50)=0.125$ and $P(S=4)=0.25(0.25)=0.0625$. Therefore the aggregate distribution is:

Aggregate loss $s$	0	1	2	3	4
$P(S=s)$	0.2500	0.2500	0.3125	0.1250	0.0625

Check and trap avoided. The probabilities sum to $0.2500+0.2500+0.3125+0.1250+0.0625=1.0000$. A common error is to use the two-claim distribution as the aggregate distribution and omit the $N=0$ and $N=1$ cases. Mixing must use all possible claim counts.

Worked Example: Unequal Payment Probabilities

A hypothetical warranty portfolio has $N\sim\operatorname{Binomial}(2,\frac13)$ claims. Conditional on a claim, the payment is $1 with probability $\frac14$ and $2 with probability $\frac34$. Assume payments are independent of one another and of $N$. Find the aggregate distribution.

The count probabilities are $P(N=0)=\frac49$, $P(N=1)=\frac49$, and $P(N=2)=\frac19$. The one-claim payment distribution is already specified. For two claims, the possible totals and probabilities are

\begin{aligned} P(Y_1+Y_2=2)&=\frac14\cdot\frac14=\frac1{16},\\ P(Y_1+Y_2=3)&=2\left(\frac14\cdot\frac34\right)=\frac38,\\ P(Y_1+Y_2=4)&=\frac34\cdot\frac34=\frac9{16}. \end{aligned}

The factor 2 for a total of $3 accounts for the two distinct ordered outcomes: ($1,$2) and ($2,$1). Weighting by the frequency probabilities gives the aggregate masses. At $0, $P(S=0)=\frac49$. At $1, $P(S=1)=\frac49\cdot\frac14=\frac19$. At $2, there are two contributions:

P(S=2)=\frac49\cdot\frac34+\frac19\cdot\frac1{16} =\frac13+\frac1{144} =\frac{49}{144}.

At $3, $P(S=3)=\frac19\cdot\frac38=\frac1{24}$. At $4, $P(S=4)=\frac19\cdot\frac9{16}=\frac1{16}$. There are no other attainable totals. In decimals, to four places, the distribution is:

Aggregate loss $s$	0	1	2	3	4
$P(S=s)$	0.4444	0.1111	0.3403	0.0417	0.0625

Check and trap avoided. The exact probabilities sum to $\frac{64+16+49+6+9}{144}=1$. Rounding the displayed decimals gives 1.0000. Do not forget to combine both sources of a $2 total: one claim paying $2 and two claims each paying $1.

Worked Example: Zero Payment Does Not Mean Zero Claims

A hypothetical covered-loss model has $N\sim\operatorname{Binomial}(2,0.5)$. Each claim independently produces a payment of $0 with probability 0.20, $2 with probability 0.50, or $4 with probability 0.30. A zero payment is possible under the modeled coverage. Find the aggregate distribution, including the probability of zero aggregate payment.

The frequency probabilities are 0.25, 0.50, and 0.25 for $N=0,1,2$, respectively. For $N=2$, convolve the payment distribution:

\begin{aligned} P(S=0\mid N=2)&=0.20^2=0.04,\\ P(S=2\mid N=2)&=2(0.20)(0.50)=0.20,\\ P(S=4\mid N=2)&=0.50^2+2(0.20)(0.30)=0.25+0.12=0.37,\\ P(S=6\mid N=2)&=2(0.50)(0.30)=0.30,\\ P(S=8\mid N=2)&=0.30^2=0.09. \end{aligned}

For the aggregate distribution, the probability at zero has three contributions: no claims, one claim with zero payment, and two claims both with zero payment. Thus

$$ P(S=0)=0.25+0.50(0.20)+0.25(0.04)=0.25+0.10+0.01=0.36. $$

For the positive totals, $P(S=2)=0.50(0.50)+0.25(0.20)=0.30$; $P(S=4)=0.50(0.30)+0.25(0.37)=0.2425$; $P(S=6)=0.25(0.30)=0.0750$; and $P(S=8)=0.25(0.09)=0.0225$. The complete distribution is:

Aggregate loss $s$	0	2	4	6	8
$P(S=s)$	0.3600	0.3000	0.2425	0.0750	0.0225

Check and trap avoided. The probabilities sum to $0.3600+0.3000+0.2425+0.0750+0.0225=1.0000$. The incorrect calculation $P(S=0)=P(N=0)=0.25$ ignores zero payments when claims occur; the correct mass at zero is 0.36.

Common Errors and Final Checks

Exact convolution is manageable when the supports are listed systematically. Before accepting a result, check the modeling assumptions as well as the arithmetic. If payments are dependent, have different distributions by claim number, or depend on the frequency count, the independent, identically distributed setup used here does not apply as written. Likewise, a finite table of fixed-count totals is exact only when all relevant claim counts and payment possibilities have been included.

Omitting the zero-claim case. The $n=0$ distribution puts probability one at aggregate loss zero, and its contribution is weighted by $P(N=0)$.
Omitting zero-valued payments. If the payment model allows a zero payment, positive claim counts can also contribute to aggregate loss zero.
Combining probabilities incorrectly. When different payment combinations give the same total, add their probabilities. For two independent payments, the outcomes in opposite orders are distinct combinations.
Mixing before calculating fixed-count totals. Keep the conditional total distributions distinct until each has been convolved; then apply the frequency weights.
Checking only rounded probabilities. Use exact fractions or unrounded decimals for the sum-to-one audit, then round displayed probabilities only at the final step.

Key takeaway: For each possible claim count, convolve the payment distribution to obtain the total-payment distribution. Then mix those distributions using the frequency probabilities, retaining every contribution to zero and verifying that the aggregate probabilities sum to one.

Check Your Understanding

For each problem, calculate the aggregate distribution and check that its probabilities sum to one.

A hypothetical contract has $N\sim\operatorname{Binomial}(1,0.4)$. A claim payment is $2 with probability 0.25 and $6 with probability 0.75. Give the aggregate probabilities at each attainable total. (Answer: $P(S=0)=0.6000,\ P(S=2)=0.1000,\ P(S=6)=0.3000$.)
Let $N\sim\operatorname{Binomial}(2,0.5)$, and let independent claim payments be $1 or $3, each with probability 0.5. Find all aggregate probabilities. (Answer: $P(S=0)=0.2500,\ P(S=1)=0.2500,\ P(S=2)=0.0625,\ P(S=3)=0.2500,\ P(S=4)=0.1250,\ P(S=6)=0.0625$.)
A portfolio has $P(N=0)=0.20,\ P(N=1)=0.50,\ P(N=2)=0.30$. Each claim payment is $2 or $4, each with probability 0.5. Find the aggregate distribution. (Answer: $P(S=0)=0.2000,\ P(S=2)=0.2500,\ P(S=4)=0.3250,\ P(S=6)=0.1500,\ P(S=8)=0.0750$.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$. Each payment is $500 with probability 0.80 and $1,000 with probability 0.20. Find the distribution in dollars, including the zero-loss probability. (Answer: $P(S=0)=0.2500,\ P(S=500)=0.4000,\ P(S=1{,}000)=0.2600,\ P(S=1{,}500)=0.0800,\ P(S=2{,}000)=0.0100$.)

What You'll Learn