Discrete Severity Convolution

Why Discrete Severity Convolution Matters

An aggregate loss is the sum of claim amounts. Even when each claim amount has a simple discrete distribution, the possible totals can be more numerous than the possible amounts for one claim. Convolution provides a systematic way to find the probability of each total: list the ways the claims can produce it, multiply the probabilities for each independent combination, and add those products.

This tutorial considers a fixed number of claims with discrete severities. That setting isolates the severity calculation: there is no uncertainty about how many claims occur. The central issue is how individual severity distributions combine into a distribution for their sum.

Definition: Let $X_1$ and $X_2$ be independent discrete claim severities with probability mass functions $p_1(k)=P(X_1=k)$ and $p_2(k)=P(X_2=k)$. The sum $S_2=X_1+X_2$ has probability mass function $$ P(S_2=s)=\sum_k p_1(k)p_2(s-k). $$ A term is zero if either severity value is outside its support. The formula applies when the severities are independent; without independence, their joint probabilities are needed instead.

Two Severities: Add the Ways to Reach Each Total

For a given total $s$, each possible value $k$ of the first severity requires the second severity to equal $s-k$. Independence makes the probability of that particular pair equal to $p_1(k)p_2(s-k)$. Different pairs that produce the same total are mutually exclusive, so their probabilities are added.

The sum in the formula is taken over the support of $X_1$, or equivalently over all possible integer values if probabilities outside each support are defined to be zero. This convention avoids accidentally omitting a combination at the edge of a support or adding a nonexistent combination.

Why the convolution formula holds: The event $X_1+X_2=s$ can be partitioned into disjoint events $X_1=k,\ X_2=s-k$, one for each possible $k$. Therefore, $$ \begin{aligned} P(X_1+X_2=s) &=\sum_k P(X_1=k,\ X_2=s-k)\\ &=\sum_k P(X_1=k)P(X_2=s-k), \end{aligned} $$ where the second equality follows from independence. The sum is a probability because its terms correspond to disjoint ways of obtaining exactly $s$.

Worked Example: Convolving Two Different Severity Distributions

A hypothetical portfolio has exactly two claims. The first severity $X_1$, in thousands of dollars, is 1 with probability 0.50 and 3 with probability 0.50. Independently, the second severity $X_2$ is 0 with probability 0.25 and 2 with probability 0.75. Find the distribution of $S_2=X_1+X_2$ and $P(S_2\leq 3)$.

Method. Apply the two-distribution convolution, retaining each pair of severity values before combining pairs that give the same total. The common mistake this avoids is treating the two ways to obtain a total of 3 as if they were one outcome.

The four independent pairs and their probabilities are:

$X_1=1,\ X_2=0$: total 1, probability $0.50(0.25)=0.125$.
$X_1=1,\ X_2=2$: total 3, probability $0.50(0.75)=0.375$.
$X_1=3,\ X_2=0$: total 3, probability $0.50(0.25)=0.125$.
$X_1=3,\ X_2=2$: total 5, probability $0.50(0.75)=0.375$.

The two pairs producing 3 are distinct, so their probabilities add:

P(S_2=1)=0.125,\qquad P(S_2=3)=0.375+0.125=0.500,\qquad P(S_2=5)=0.375.

Thus $P(S_2\leq 3)=0.125+0.500=0.625$. Check: the masses sum to $0.125+0.500+0.375=1.000$, as required.

Extending Convolution to a Fixed Number of Claims

Suppose $X_1,\ldots,X_n$ are independent discrete severities and $S_n=X_1+\cdots+X_n$. If all claims have the same severity distribution, the distribution of $S_n$ can be built one claim at a time. This repeated calculation is called iterated convolution. It avoids trying to list every possible $n$-tuple at once.

Recursive convolution: For independent, identically distributed severities with mass function $p(k)=P(X_j=k)$, define $f_n(s)=P(S_n=s)$. Set $f_0(0)=1$ and $f_0(s)=0$ for $s\ne 0$. Then, for $n\geq 1$, $$ f_n(s)=\sum_k f_{n-1}(s-k)p(k). $$ The starting value $f_0(0)=1$ represents a sum of zero claims equal to zero with certainty. Each subsequent step convolves the previous aggregate distribution with the severity distribution of one more independent claim.

The recurrence follows by the same disjoint-event argument as the two-claim formula: for the total to equal $s$, the first $n-1$ claims must total $s-k$ and the final claim must equal $k$. These events are disjoint over $k$; independence gives probability $f_{n-1}(s-k)p(k)$ for each one.

For nonidentical but independent severities, the same step-by-step method works with the mass function of the next claim at each step. The assumption of identical distributions is needed for using the same $p(k)$ in every step, not for convolution itself. If severities are dependent, multiplying the separate probabilities is not justified.

Start with the one-claim distribution.
Write down each possible severity and its probability. Check that the probabilities sum to 1.

Convolve to add one claim.
For each possible new total, add the products of probabilities for all pairs that produce that total.

Repeat for the required claim count.
Use the current aggregate distribution as one input and the next independent severity distribution as the other.

Check the result.
Confirm that all possible totals are represented and the final probabilities sum to 1, allowing only for stated rounding.

Worked Example: Three Identically Distributed Claims

A hypothetical contract has exactly three independent claims. Each claim severity, in thousands of dollars, is 1 with probability 0.60 and 2 with probability 0.40. Find the aggregate distribution and the probability that the total is at least 5.

Method. First convolve two severities, then convolve that result with the third severity. This recursive method avoids listing all eight three-claim combinations individually. The common mistake it avoids is stopping at the two-claim distribution when the question asks for three claims.

For the first two claims, total 2 occurs when both are 1, total 3 occurs in either order, and total 4 occurs when both are 2:

\begin{aligned} P(S_2=2)&=0.60(0.60)=0.36,\\ P(S_2=3)&=0.60(0.40)+0.40(0.60)=0.48,\\ P(S_2=4)&=0.40(0.40)=0.16. \end{aligned}

Now add the third claim. For a total of 3, the first two claims must total 2 and the third must be 1. For a total of 4, there are two contributing cases: $(S_2=2,X_3=2)$ and $(S_2=3,X_3=1)$. Continuing in the same way gives:

\begin{aligned} P(S_3=3)&=0.36(0.60)=0.216,\\ P(S_3=4)&=0.36(0.40)+0.48(0.60)=0.144+0.288=0.432,\\ P(S_3=5)&=0.48(0.40)+0.16(0.60)=0.192+0.096=0.288,\\ P(S_3=6)&=0.16(0.40)=0.064. \end{aligned}

Therefore $P(S_3\geq 5)=0.288+0.064=0.352$. Check: $0.216+0.432+0.288+0.064=1.000$, and the possible totals run from the minimum $1+1+1=3$ to the maximum $2+2+2=6$.

Overlapping Totals and Support Checks

Convolution is especially easy to misapply when different combinations overlap at the same total. A useful calculation table lists each pair, its total, and its product probability. Only after all pairs have been considered should equal totals be combined. The support of the sum consists of totals that can actually be formed from values in the input supports; not every integer between the minimum and maximum must necessarily have positive probability.

A second check is normalization: the probabilities in a complete aggregate distribution must sum to 1. A total below 1 can signal an omitted combination; a total above 1 can signal that a combination was counted more than once. This check does not prove that each individual mass is correct, so also check the pairings for totals that have multiple contributing combinations.

Worked Example: Unequal Supports With Shared Aggregate Totals

A hypothetical two-claim portfolio has independent severities, in thousands of dollars. $X_1$ is 0, 2, or 4 with probabilities 0.20, 0.50, and 0.30, respectively. $X_2$ is 1 with probability 0.70 or 3 with probability 0.30. Find the aggregate distribution and $P(S_2\geq 5)$.

Method. Enumerate the six independent pairs and combine equal totals. This explicitly handles overlapping totals, the common mistake being to assign separate final masses to the two different ways of obtaining 3.

If $X_1=0$, the totals are 1 with probability $0.20(0.70)=0.14$, and 3 with probability $0.20(0.30)=0.06$.
If $X_1=2$, the totals are 3 with probability $0.50(0.70)=0.35$, and 5 with probability $0.50(0.30)=0.15$.
If $X_1=4$, the totals are 5 with probability $0.30(0.70)=0.21$, and 7 with probability $0.30(0.30)=0.09$.

Combine the contributions to totals 3 and 5:

\begin{aligned} P(S_2=1)&=0.14,\\ P(S_2=3)&=0.06+0.35=0.41,\\ P(S_2=5)&=0.15+0.21=0.36,\\ P(S_2=7)&=0.09. \end{aligned}

The requested probability is $P(S_2\geq 5)=0.36+0.09=0.45$. Check: $0.14+0.41+0.36+0.09=1.00$. Totals 2, 4, and 6 have zero probability because no available pair produces them.

Common Exam Traps

Multiplying without checking independence. The product $P(X_1=k)P(X_2=s-k)$ is valid only when the severities are independent. Otherwise use the joint probability for that pair.
Forgetting to add overlapping combinations. If multiple pairs produce the same aggregate amount, their probabilities all contribute to that mass.
Using the wrong number of claims. Repeated convolution is needed once for each claim. A distribution for two claims is not the distribution for three.
Including impossible totals. A total between the minimum and maximum may still be unattainable. Assign positive probability only when at least one combination produces it.
Reading a tail event as an exact event. $P(S_n\geq c)$ requires adding the masses at every supported total at least $c$, not just the mass at $c$.

Convolution turns a list of individual claim probabilities into an aggregate distribution, with no need for a special formula for each support. The key discipline is to identify every contributing combination, apply independence only where it is justified, and check the completed probability mass function.

Check Your Understanding

For each problem, treat the stated severities as independent. Amounts are in the units given in the question.

Two claims have severities $X_1=0$ with probability 0.40 and $X_1=2$ with probability 0.60; $X_2=1$ with probability 0.25 and $X_2=3$ with probability 0.75. Find the aggregate distribution. (Answer: $P(S_2=1)=0.10,\ P(S_2=3)=0.45,\ P(S_2=5)=0.45$.)
Two identically distributed claims each have severity 1 with probability 0.30 and 4 with probability 0.70. Find the aggregate distribution. (Answer: $P(S_2=2)=0.09,\ P(S_2=5)=0.42,\ P(S_2=8)=0.49$.)
Three claims each have severity 0 or 2, each with probability 0.50. Find the probability that the aggregate is at least 4. (Answer: $0.5000$.)
Two claims have independent severities $X_1=1$ with probability 0.20 or 2 with probability 0.80, and $X_2=0$ with probability 0.60 or 3 with probability 0.40. Find the probability that the aggregate is at least 4. (Answer: $0.4000$.)
Three claims each have severity 1 with probability 0.25 or 2 with probability 0.75. Find the aggregate distribution and the probability that the total is at least 5. (Answer: $P(S_3=3)=0.015625,\ P(S_3=4)=0.140625,\ P(S_3=5)=0.421875,\ P(S_3=6)=0.421875;\ P(S_3\geq5)=0.84375$.)

What You'll Learn