Continuous Severity Convolution

Why Convolution Is Needed

When a portfolio has a fixed number of claims, its total loss is the sum of the individual claim severities. Even when each severity has a simple density, the sum generally has a different density. Convolution is the method for deriving that density. It accounts for every way the individual severities can combine to produce the same total.

This tutorial concerns the distribution of the sum conditional on a specified number of claims. It does not mix results across different claim counts. That distinction matters: a density conditional on two claims describes a different distribution from the unconditional aggregate loss, which can also include zero, one, or more claims.

Definition: Let $X$ and $Y$ be independent continuous random variables with densities $f_X$ and $f_Y$. The density of their sum $Z=X+Y$ is $$ f_Z(s)=\int_{-\infty}^{\infty} f_X(x)f_Y(s-x)\,dx. $$ The integrand is zero wherever either density is zero, so the effective integration limits are the values of $x$ for which both $x$ and $s-x$ are in their respective supports.

For nonnegative claim severities, the support of the sum begins at zero. If both severities can take any nonnegative value, the formula simplifies to

f_Z(s)=\int_0^s f_X(x)f_Y(s-x)\,dx,\qquad s\geq 0.

This simplified form assumes both variables are nonnegative and have densities on the relevant range. If either severity has a bounded support, the limits must be narrowed accordingly. The convolution formula also relies on independence; multiplying the marginal densities is not valid for dependent severities.

Why the Convolution Formula Holds

For a fixed total $s$, if the first severity is $x$, the second must be $s-x$. The density of the pair at those values is $f_X(x)f_Y(s-x)$, by independence. Adding that contribution over all feasible values of $x$ gives the density of the total. More formally, use the transformation $Z=X+Y$ and retain $X$ as the other coordinate. Its inverse is $x=x$, $y=z-x$, with Jacobian determinant of absolute value 1. The joint density in the new coordinates is therefore $f_X(x)f_Y(z-x)$; integrating out $x$ gives the stated density of $Z$.

For a fixed claim count $n$, let $S_n=X_1+\cdots+X_n$ denote the total severity, conditional on there being $n$ claims. If severities are independent and identically distributed with density $f$, the density can be built recursively:

f_{S_1}(s)=f(s),\qquad f_{S_n}(s)=\int_{-\infty}^{\infty} f_{S_{n-1}}(t)f(s-t)\,dt.

This recursive statement requires independent, identically distributed severities with a density. For independent but nonidentically distributed severities, convolve the relevant densities in turn, using the density of each claim at its step. In either case, the result is a density, not a point probability; for a continuous sum, the probability of one exact total is zero.

Worked Example: Two Uniform Claim Severities

A hypothetical contract has exactly two claims. Conditional on two claims, their severities $X$ and $Y$, in thousands of dollars, are independent and each has a uniform distribution on $[0,2]$. Find the density of the total $S=X+Y$, and calculate $P(1\leq S\leq 3)$.

Method. Convolve the two uniform densities, paying particular attention to the bounded supports. For each severity, the density is $1/2$ between 0 and 2 and zero elsewhere. For a given total $s$, feasible values of $x$ must satisfy both $0\leq x\leq2$ and $0\leq s-x\leq2$. Thus $x$ ranges from $\max(0,s-2)$ to $\min(2,s)$.

For $0\leq s\leq2$, the feasible interval is $0\leq x\leq s$. For $2<s\leq4$, it is $s-2\leq x\leq2$. Hence

f_S(s)= \begin{cases} \displaystyle\int_0^s \frac12\cdot\frac12\,dx=\frac{s}{4}, & 0\leq s\leq2,\\[6pt] \displaystyle\int_{s-2}^2 \frac12\cdot\frac12\,dx=\frac{4-s}{4}, & 2<s\leq4,\\[6pt] 0, & \text{otherwise}. \end{cases}

To find the requested probability, integrate the density over the interval. Split at $s=2$, where the density changes form:

P(1\leq S\leq3) =\int_1^2\frac{s}{4}\,ds+\int_2^3\frac{4-s}{4}\,ds =\frac{3}{8}+\frac{3}{8} =\frac{3}{4}.

The density is nonnegative and has total area $2\int_0^2(s/4)\,ds=1$, which checks the result. Common mistake avoided: using $0\leq x\leq s$ for every $s$. That interval would include impossible severities when $s>2$; the upper support limit $x\leq2$ must still apply. The calculation is conditional on exactly two claims, as specified.

Different Severity Distributions

The two severities need not have the same distribution. The convolution formula combines their separate densities, provided they are independent. This is useful when different claim types have different severity patterns, or when the model explicitly assigns different distributions to the first and second claims.

Worked Example: Convolving Two Exponential Severities

Conditional on exactly two independent claims, suppose their severities $X$ and $Y$, measured in thousands of dollars, have exponential distributions with rates 1 and 2 per thousand dollars, respectively. Their densities are $f_X(x)=e^{-x}$ for $x\geq0$ and $f_Y(y)=2e^{-2y}$ for $y\geq0$, and zero otherwise. Find the density of $S=X+Y$, its value at $s=1$, and $P(S>1)$.

Method. Apply the nonnegative-severity convolution integral, using each claim's own density:

\begin{aligned} f_S(s) &=\int_0^s e^{-x}\,2e^{-2(s-x)}\,dx\\ &=2e^{-2s}\int_0^s e^x\,dx\\ &=2e^{-2s}(e^s-1)\\ &=2(e^{-s}-e^{-2s}),\qquad s\geq0. \end{aligned}

For $s=1$, the density is

f_S(1)=2(e^{-1}-e^{-2})\approx 0.4651

in density units per thousand dollars. To calculate the probability above 1, integrate the density from 1 to infinity:

\begin{aligned} P(S>1) &=\int_1^\infty 2(e^{-s}-e^{-2s})\,ds\\ &=2e^{-1}-e^{-2}\\ &\approx 0.6004. \end{aligned}

As a check, $P(S\leq1)=1-2e^{-1}+e^{-2}\approx0.3996$, and the two probabilities sum to 1. Common mistake avoided: treating the severities as identically distributed exponentials. Their rates differ, so the integrand must use $e^{-x}$ for the first claim and $2e^{-2(s-x)}$ for the second.

Extending Convolution to Three Claims

For three independent severities, first find the density of the sum of two, then convolve that density with the third. The integration limits must reflect the supports of both the two-claim sum and the remaining severity. This support check is often the main step in a piecewise calculation.

Worked Example: The Conditional Total of Three Uniform Claims

A hypothetical benefit pays three independent claims, conditional on exactly three claims occurring. Each severity is uniform on $[0,1]$, in thousands of dollars. Find the density of the total $S=X_1+X_2+X_3$, calculate $f_S(1.5)$, and find $P(1\leq S\leq2)$.

Method. The sum of the first two claims has density $f_{S_2}(t)=t$ for $0\leq t\leq1$, $f_{S_2}(t)=2-t$ for $1<t\leq2$, and zero otherwise. This follows by convolving two uniform $[0,1]$ densities. Now convolve $S_2$ with the third uniform severity, whose density is 1 on $[0,1]$. For total $s$, the value $t$ must lie in $[0,2]$ and $s-t$ in $[0,1]$; consequently, the limits are $\max(0,s-1)$ and $\min(2,s)$.

For $0\leq s\leq1$, the integration range is $0\leq t\leq s$, where $f_{S_2}(t)=t$. For $1<s\leq2$, split the range at $t=1$. For $2<s\leq3$, the range lies between $s-1$ and 2, where $f_{S_2}(t)=2-t$. The resulting density is

f_S(s)= \begin{cases} \displaystyle\int_0^s t\,dt=\frac{s^2}{2}, & 0\leq s\leq1,\\[6pt] \displaystyle\int_{s-1}^1 t\,dt+\int_1^s(2-t)\,dt =-s^2+3s-\frac32, & 1<s\leq2,\\[6pt] \displaystyle\int_{s-1}^2(2-t)\,dt=\frac{(3-s)^2}{2}, & 2<s\leq3,\\[6pt] 0, & \text{otherwise}. \end{cases}

The density at $s=1.5$ uses the middle piece:

f_S(1.5)=-(1.5)^2+3(1.5)-\frac32=0.75.

For the probability from 1 to 2, integrate the middle piece:

\begin{aligned} P(1\leq S\leq2) &=\int_1^2\left(-s^2+3s-\frac32\right)\,ds\\ &=\left[-\frac{s^3}{3}+\frac{3s^2}{2}-\frac{3s}{2}\right]_1^2 =\frac23. \end{aligned}

As a check, the density is symmetric around $s=1.5$, as expected for the sum of three identically distributed uniform $[0,1]$ severities. Each outer tail, below 1 and above 2, has probability $1/6$, leaving $1-1/6-1/6=2/3$ between 1 and 2. Common mistake avoided: for $1<s\leq2$, integrating from 0 to $s$ would include values of $t$ for which the third severity $s-t$ exceeds 1. The lower limit $s-1$ is essential.

Checks and Exam Traps

A correct convolution should respect both supports and behave like a density. Use these checks before relying on a result:

Check the support. If severities lie in $[a,b]$ and $[c,d]$, their sum lies in $[a+c,b+d]$. A density outside that range must be zero.
Check the integration limits. For a proposed total $s$, include only values of the first severity for which the implied second severity $s-x$ is possible.
Check the total area. The density must integrate to 1 over its full support. For piecewise densities, integrate each piece and add the areas.
Keep density and probability distinct. $f_S(s)$ is a density value; a probability requires integrating the density over an interval.
Keep the conditioning explicit. The convolution of $n$ severity densities gives the distribution conditional on $n$ claims, not the unconditional aggregate distribution across claim counts.

Convolution is useful because it turns individual continuous severity models into distributions for conditional aggregate losses. The method is the same whether the component densities are identical or different; the work lies in using the correct densities, supports, and limits. These calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, state the relevant support and use convolution or its recursive form. Answers are provided so you can check the calculation.

Conditional on two independent claims, each severity is exponential with rate 3 per thousand dollars. Find the density of their total and evaluate it at $s=0.5$. (Answer: $f_S(s)=9se^{-3s}$ for $s\geq0$; $f_S(0.5)=4.5e^{-1.5}\approx1.0041$.)
Two independent severities are uniform on $[0,3]$. Derive the density of their sum and evaluate it at $s=4$. (Answer: $f_S(s)=s/9$ for $0\leq s\leq3$, $f_S(s)=(6-s)/9$ for $3<s\leq6$, and zero otherwise; $f_S(4)=2/9$.)
Two independent claim severities are exponential with rates 2 and 5 per thousand dollars. Find the density of their total, specifying its support. (Answer: $f_S(s)=\frac{10}{3}(e^{-2s}-e^{-5s})$ for $s\geq0$, and zero otherwise.)
Three independent severities are uniform on $[0,1]$. Find the probability that their total is between 0.5 and 1. (Answer: $F_S(1)-F_S(0.5)=\frac16-\frac1{48}=\frac7{48}$.)
Independent severities $X$ and $Y$ are uniform on $[0,1]$ and $[0,2]$, respectively. Find the density of $X+Y$ and evaluate it at 2.5. (Answer: $f(s)=s/2$ for $0\leq s\leq1$, $f(s)=1/2$ for $1<s\leq2$, $f(s)=(3-s)/2$ for $2<s\leq3$, and zero otherwise; $f(2.5)=0.25$.)

What You'll Learn