Exact Convolution of Independent Discrete Losses

Why Exact Convolution Is Useful

When component losses are discrete and their distributions are known, an aggregate distribution can be calculated directly rather than approximated. This is useful when a risk measure depends on the full distribution, such as a selected loss quantile, and the set of possible component outcomes is small enough to enumerate. Exact calculation also makes the role of each combination of component losses visible.

The essential assumption is independence. If the losses are dependent, multiplying their marginal probabilities does not in general give the probability of a joint outcome. In that case, the joint distribution or an appropriate dependence model is needed. This tutorial concerns exact convolution of independent discrete losses.

Definition: A discrete loss $X$ has probability mass function $p_X(x)=P(X=x)$. For independent discrete losses $X$ and $Y$, their aggregate is $S=X+Y$. Its probability mass function is the convolution $$ p_S(s)=P(S=s)=\sum_x p_X(x)p_Y(s-x), $$ where the sum is over the possible values $x$ of $X$; a term is zero if $s-x$ is not a possible value of $Y$. For independent losses $X_1,\ldots,X_n$, the aggregate is $S=\sum_{j=1}^n X_j$, and its mass at $s$ is the sum of the products of component probabilities over all combinations whose losses total $s$.

Deriving the Convolution Formula

For a fixed total $s$, the event $S=s$ can occur through any combination of component outcomes satisfying $X=x$ and $Y=s-x$. These events are mutually exclusive as $x$ ranges over the possible values of $X$. Therefore, their probabilities add. Independence gives the probability of each combination as the product of its marginal probabilities:

P(S=s)=\sum_x P(X=x,Y=s-x) =\sum_x P(X=x)P(Y=s-x) =\sum_x p_X(x)p_Y(s-x).

This argument explains both why the probabilities are multiplied within each combination and why the contributions from different combinations are then added. It also shows why the formula requires independence. Without independence, the joint probability $P(X=x,Y=s-x)$ cannot generally be replaced by a product.

For several losses, convolution can be applied in stages. First calculate the distribution of $X_1+X_2$, then convolve that aggregate with $X_3$, and continue. Because addition is associative, this produces the same distribution as enumerating all combinations at once, while often making the arithmetic easier to organize.

List each component support.
Record every possible loss amount and its probability, including zero-loss outcomes.

Combine outcomes.
For each combination of component losses, add the losses and multiply their probabilities, using independence.

Group equal totals.
Add the probabilities of all combinations that produce the same aggregate loss.

Check the result.
Confirm that all aggregate probabilities are nonnegative and sum to one, then accumulate them in increasing loss order to find quantiles.

Worked Examples

Worked Example: Three Independent Discrete Losses

A hypothetical insurer models three independent annual losses, in thousands of dollars. $X$ is 0 with probability 0.60 and 2 with probability 0.40. $Y$ is 0 with probability 0.50 and 3 with probability 0.50. $Z$ is 0 with probability 0.80 and 1 with probability 0.20. Find the aggregate probability mass function and $\operatorname{VaR}_{0.80}$, using the discrete quantile definition.

Method. Enumerate all $2^3=8$ combinations, multiply the three probabilities in each combination, and group by total loss. The probabilities for the eight combinations are:

\begin{aligned} (0,0,0)&:\ 0.60(0.50)(0.80)=0.24, & (0,0,1)&:\ 0.60(0.50)(0.20)=0.06,\\ (2,0,0)&:\ 0.40(0.50)(0.80)=0.16, & (2,0,1)&:\ 0.40(0.50)(0.20)=0.04,\\ (0,3,0)&:\ 0.60(0.50)(0.80)=0.24, & (0,3,1)&:\ 0.60(0.50)(0.20)=0.06,\\ (2,3,0)&:\ 0.40(0.50)(0.80)=0.16, & (2,3,1)&:\ 0.40(0.50)(0.20)=0.04. \end{aligned}

The combinations $(X,Y,Z)$ yield totals 0, 1, 2, 3, 4, 5, and 6. In particular, total 3 arises from both $(0,3,0)$ and $(2,0,1)$; total 4 arises from $(0,3,1)$. Grouping every combination gives:

\begin{aligned} P(S=0)&=0.24, & P(S=1)&=0.06, & P(S=2)&=0.16,\\ P(S=3)&=0.24+0.04=0.28, & P(S=4)&=0.06, & P(S=5)&=0.16, & P(S=6)&=0.04. \end{aligned}

The probabilities sum to $0.24+0.06+0.16+0.28+0.06+0.16+0.04=1.00$, as required. The cumulative probabilities at totals 0 through 6 are 0.24, 0.30, 0.46, 0.74, 0.80, 0.96, and 1.00. By definition, $\operatorname{VaR}_{0.80}$ is the smallest loss whose cumulative probability is at least 0.80. Thus it is 4 thousand dollars. The calculation avoids the common mistake of missing the $(2,0,1)$ combination at total 3 or assigning the probability of $(0,3,1)$ incorrectly.

Worked Example: Convolution of Two Uneven Supports

A hypothetical portfolio has two independent annual losses, in thousands of dollars. Loss $A$ is 0, 4, or 10 with probabilities 0.50, 0.30, and 0.20. Loss $B$ is 0 with probability 0.70 and 6 with probability 0.30. Calculate the aggregate distribution and $\operatorname{VaR}_{0.90}$.

Method. Pair each value of $A$ with each value of $B$. The six outcomes and their probabilities are:

\begin{aligned} (A,B)=(0,0)&:\ 0.50(0.70)=0.35, & S&=0,\\ (0,6)&:\ 0.50(0.30)=0.15, & S&=6,\\ (4,0)&:\ 0.30(0.70)=0.21, & S&=4,\\ (4,6)&:\ 0.30(0.30)=0.09, & S&=10,\\ (10,0)&:\ 0.20(0.70)=0.14, & S&=10,\\ (10,6)&:\ 0.20(0.30)=0.06, & S&=16. \end{aligned}

Totals 10 occur in two distinct ways, so their probabilities must be added. The aggregate masses, in increasing order of loss, are $P(S=0)=0.35$, $P(S=4)=0.21$, $P(S=6)=0.15$, $P(S=10)=0.09+0.14=0.23$, and $P(S=16)=0.06$. They sum to $0.35+0.21+0.15+0.23+0.06=1.00$.

The cumulative probability is 0.35 at 0, 0.56 at 4, 0.71 at 6, 0.94 at 10, and 1.00 at 16. The first cumulative probability at least 0.90 occurs at 10, so $\operatorname{VaR}_{0.90}=10$ thousand dollars. A common mistake is to use only one of the two ways to obtain total 10; doing so understates its probability and gives the wrong cumulative distribution.

Worked Example: Iterative Convolution With a Third Loss

Three independent hypothetical annual losses, in thousands of dollars, have the following distributions. $U$ is 0 with probability 0.60 or 2 with probability 0.40; $V$ is 0 with probability 0.50 or 3 with probability 0.50; and $W$ is 0 with probability 0.75 or 4 with probability 0.25. Find the aggregate distribution and $\operatorname{VaR}_{0.80}$, using iterative convolution.

First convolution. For $T=U+V$, the four combinations give:

\begin{aligned} P(T=0)&=0.60(0.50)=0.30, & P(T=2)&=0.40(0.50)=0.20,\\ P(T=3)&=0.60(0.50)=0.30, & P(T=5)&=0.40(0.50)=0.20. \end{aligned}

These masses sum to 1.00. Next convolve $T$ with $W$. Each value of $T$ contributes its probability times 0.75 to the same total and times 0.25 to a total four units higher:

\begin{aligned} P(S=0)&=0.30(0.75)=0.225, & P(S=2)&=0.20(0.75)=0.150,\\ P(S=3)&=0.30(0.75)=0.225, & P(S=5)&=0.20(0.75)=0.150,\\ P(S=4)&=0.30(0.25)=0.075, & P(S=6)&=0.20(0.25)=0.050,\\ P(S=7)&=0.30(0.25)=0.075, & P(S=9)&=0.20(0.25)=0.050. \end{aligned}

In increasing order, the aggregate support and probabilities are $0:0.225,\ 2:0.150,\ 3:0.225,\ 4:0.075,\ 5:0.150,\ 6:0.050,\ 7:0.075,\ 9:0.050$. Their sum is 1.000. The cumulative probability through 4 is $0.225+0.150+0.225+0.075=0.675$; through 5 it is $0.675+0.150=0.825$. Therefore $\operatorname{VaR}_{0.80}=5$ thousand dollars. The staged method avoids listing all eight three-component combinations separately, while still accounting for every combination exactly.

Quantiles and Checks

For a discrete aggregate loss $S$, define its cumulative distribution function by $F_S(s)=P(S\leq s)$. For $0<\alpha<1$, the lower $\alpha$-quantile, also called $\operatorname{VaR}_{\alpha}$ under this convention, is

\operatorname{VaR}_{\alpha}(S)=\inf\{s:F_S(s)\geq\alpha\}.

The quantile is a loss amount, not a probability. Because a discrete distribution has jumps, the cumulative probability at the selected loss can exceed $\alpha$; it need not equal it. Always order the aggregate support before accumulating probabilities, especially when the component losses have different increments or when several combinations produce the same total.

A useful independent check is that the aggregate probabilities are nonnegative and sum to one. If the component losses have finite means, another check follows from linearity of expectation: $E[S]=\sum_j E[X_j]$, whether or not the losses are independent. This mean check can reveal an arithmetic or grouping error, but it does not replace checking the full probability mass function. Exact convolution can require many combinations when supports are large; in those cases the computational method may change, but the probability logic remains the same.

Common Exam Traps

Adding instead of multiplying within a combination. Under independence, the probability of a particular joint outcome is the product of component probabilities. Add only after identifying distinct combinations that give the same aggregate total.
Assuming independence without it being stated. Marginal distributions alone do not determine the aggregate distribution when losses may be dependent.
Omitting an outcome or zero-loss case. List each component's full support before calculating. A missing combination often causes the final masses not to sum to one.
Reading a quantile as an exact equality. For a discrete distribution, select the smallest loss with cumulative probability at least the target, even if the cumulative probability jumps past that target.

Key takeaway: Exact convolution consists of multiplying probabilities for independent component outcomes, adding the losses in each combination, and summing probabilities for combinations with equal totals. Verify the resulting mass function before reading a quantile from its cumulative probabilities.

Check Your Understanding

For each problem, assume the stated losses are independent and are measured in thousands of dollars. Find the requested aggregate probabilities or quantile.

A portfolio has losses $A$ and $B$. $A$ is 0 with probability 0.80 or 5 with probability 0.20; $B$ is 0 with probability 0.60 or 2 with probability 0.40. Give the aggregate mass at each possible total and $\operatorname{VaR}_{0.80}$. (Answer: masses at 0, 2, 5, and 7 are 0.48, 0.32, 0.12, and 0.08; $\operatorname{VaR}_{0.80}=2$.)
Loss $C$ is 0, 3, or 8 with probabilities 0.50, 0.25, and 0.25; loss $D$ is 0 or 4 with probabilities 0.75 and 0.25. Find the aggregate mass function and $\operatorname{VaR}_{0.90}$. (Answer: masses at 0, 3, 4, 7, 8, and 12 are 0.3750, 0.1875, 0.1250, 0.0625, 0.1875, and 0.0625; $\operatorname{VaR}_{0.90}=8$.)
Three losses take values $0$ or $2$, $0$ or $5$, and $0$ or $1$. The respective probabilities of the positive values are 0.25, 0.20, and 0.50. Find the aggregate mass at each possible total and $\operatorname{VaR}_{0.90}$. (Answer: masses at 0, 1, 2, 3, 5, 6, 7, and 8 are 0.300, 0.300, 0.100, 0.100, 0.075, 0.075, 0.025, and 0.025; $\operatorname{VaR}_{0.90}=6$.)
Three independent losses have distributions $A: 0$ or $2$ with probabilities 0.70 and 0.30; $B: 1$ or $4$ with probabilities 0.40 and 0.60; and $C: 0$ or $3$ with probabilities 0.50 and 0.50. Find $P(A+B+C=6)$ and $\operatorname{VaR}_{0.75}$. (Answer: $P(A+B+C=6)=0.15$; $\operatorname{VaR}_{0.75}=7$.)

What You'll Learn