Continuous Severity Convolution And Numerical Integration

Why Continuous Convolution Matters

When claim severities are continuous, the aggregate loss conditional on a fixed number of claims has a density formed by convolution. If the claim count is random, the unconditional aggregate distribution combines those count-conditioned distributions using the probabilities of each count. In practice, the convolution and the final probability calculation may require numerical integration.

The key is to keep two operations distinct: first calculate the distribution of the sum for each fixed claim count; then weight those results by the count probabilities. This avoids treating a density as if it were a probability, or losing the probability mass at zero.

Definition: Let \(X_1,X_2,\ldots\) be independent, identically distributed, nonnegative continuous claim severities with density \(f_X\). For a fixed count \(n\geq 1\), define \(S_n=\sum_{j=1}^{n}X_j\). Its density is \(f_n\). If \(N\) is an independent claim count with \(P(N=n)=p_n\), the aggregate is \(S_N\). For \(s>0\), its density is \(f_{S_N}(s)=\sum_{n\geq 1}p_n f_n(s)\). If \(P(N=0)>0\), the aggregate also has a probability mass \(p_0\) at zero.

The assumptions matter. The convolution formulas below use independent severities with the same distribution, and the mixture formula uses independence between \(N\) and the severities. If severities depend on claim count, claim order, or one another, the relevant conditional joint distributions must be used instead. No interest or mortality assumption is needed here: the examples concern undiscounted claim aggregates, not life-contingent cash flows.

Convolution for a Fixed Number of Claims

For two independent continuous severities, the density of their sum is obtained by integrating over all ways the first severity could contribute to a total of \(s\). For nonnegative severities,

f_2(s)=\int_{0}^{s} f_X(x)f_X(s-x)\,dx.

If each severity has support \([a,b]\), the integration range must also respect that support. For \(n\) claims, a useful recursive form is

f_n(s)=\int_{\max(a,\,s-(n-1)b)}^{\min(b,\,s-(n-1)a)} f_X(x)f_{n-1}(s-x)\,dx,

where an integral is zero if its lower limit exceeds its upper limit. This formula assumes the severities have support \([a,b]\); for unbounded severities, the limits instead follow from the actual support and can be infinite.

Why the convolution formula holds: Conditional on \(X_1=x\), the remaining sum \(X_2+\cdots+X_n\) must equal \(s-x\) when the total is \(s\). By independence, the density contribution at that split is \(f_X(x)f_{n-1}(s-x)\). Integrating over all feasible \(x\) adds the contributions from every possible split of the total. The support limits exclude splits that would require an impossible severity or remaining sum.

These integrals are often evaluated numerically. Composite Simpson’s rule approximates \(\int_A^B g(x)\,dx\) using an even number \(m\) of equal subintervals of width \(h=(B-A)/m\):

\int_A^B g(x)\,dx\ \approx\ \frac{h}{3} \left[g(x_0)+g(x_m)+4\sum_{\substack{j=1\\j\text{ odd}}}^{m-1}g(x_j) +2\sum_{\substack{j=2\\j\text{ even}}}^{m-2}g(x_j)\right].

For accurate work, split the integration range at support boundaries and points where the integrand changes formula. Simpson’s rule is particularly convenient on smooth portions of the integrand, but a coarse grid across a kink can give a poor approximation. Increasing the number of panels or using a suitable numerical integration routine provides a check on stability.

Worked Example: Convolving Three Uniform Severities

A hypothetical small portfolio has exactly three claims. Each severity is independent and distributed uniformly on \([0,1]\) thousand dollars, so \(f_X(x)=1\) for \(0<x<1\). Find \(f_3(1.5)\) using numerical integration. There is no discounting or interest assumption, and no mortality model applies.

Step 1: Find the two-claim density. For \(0\leq s\leq1\), the convolution range is \(0\leq x\leq s\), and both uniform densities equal 1. Thus \(f_2(s)=s\). For \(1<s\leq2\), the feasible range is \(s-1\leq x\leq1\), giving \(f_2(s)=2-s\). In particular, \(f_2\) is triangular on \([0,2]\).

Step 2: Set up the three-claim convolution. At \(s=1.5\), the first severity \(x\) must lie in \([0,1]\), and \(f_X(x)=1\) there:

f_3(1.5)=\int_0^1 f_2(1.5-x)\,dx.

Step 3: Apply Simpson’s rule. Use four equal panels, so \(h=0.25\). At \(x=0,0.25,0.50,0.75,1\), the corresponding values of \(1.5-x\) are \(1.5,1.25,1,0.75,0.5\). The values of \(f_2(1.5-x)\) are \(0.5,0.75,1,0.75,0.5\), respectively. Therefore,

f_3(1.5)\approx \frac{0.25}{3} \left[0.5+0.5+4(0.75+0.75)+2(1)\right] =\frac{0.25}{3}(9)=0.75.

In this case the integrand is piecewise linear and the grid includes its kink, so the Simpson calculation gives the exact value. The density is \(0.75\) per thousand dollars; it is not a probability of \(0.75\) at exactly \(1.5\).

Mixing Over the Claim Count

Once the fixed-count densities are available, the aggregate density for positive loss is a probability-weighted mixture. The weights are the count probabilities, not the conditional densities themselves. The zero-claim case contributes an atom at zero rather than a continuous density.

P(S_N=0)=p_0,\qquad f_{S_N}(s)=\sum_{n=1}^{\infty}p_n f_n(s),\quad s>0,

The density mixture applies when severities are strictly positive with probability one. If a severity itself can be zero with positive probability, additional point masses may occur and must be handled explicitly. For a probability over a positive interval, integrate the mixture density over that interval. Equivalently, calculate each conditional probability and weight it by \(p_n\).

Calculate count probabilities.
Obtain \(p_n=P(N=n)\) for the counts that can contribute, including \(p_0\) if zero claims are possible.

Obtain fixed-count distributions.
Use convolution for each \(n\), respecting the severity support and any changes in the density formula.

Form the mixture.
Weight each \(f_n(s)\) by \(p_n\) for a density, or each conditional probability by \(p_n\) for a probability.

Integrate and check.
Split at discontinuities or kinks, perform the numerical integration, and verify that the continuous probability and zero atom are consistent with the total probability.

Worked Example: Integrating a Binomial Aggregate Density

A hypothetical product has \(N\sim\operatorname{Binomial}(2,0.5)\) claims. Conditional on a claim, severity is independent and uniformly distributed on \([0,1]\) thousand dollars. Find \(P(0.5<S_N<1.5)\) by integrating the aggregate density. There is no interest or mortality assumption.

The count probabilities are \(p_0=0.25\), \(p_1=0.50\), and \(p_2=0.25\). The zero-claim atom is at zero, outside the interval. For one claim, \(f_1(s)=1\) on \((0,1)\). For two claims, the triangular density is \(f_2(s)=s\) on \((0,1)\) and \(f_2(s)=2-s\) on \((1,2)\).

On \(0.5<s<1\), the aggregate density is \(0.50(1)+0.25s\). At the Simpson nodes \(0.5,0.75,1\), its limiting values from this interval are \(0.625,0.6875,0.75\). Simpson’s rule over width \(0.5\) gives

P(0.5<S_N<1\text{ from this interval}) \approx \frac{0.5}{6}\left[0.625+4(0.6875)+0.75\right] =0.34375.

On \(1<s<1.5\), the one-claim density is zero, so the aggregate density is \(0.25(2-s)\). At nodes \(1,1.25,1.5\), its limiting values from the right are \(0.25,0.1875,0.125\). Hence,

P(1<S_N<1.5)\approx\frac{0.5}{6} \left[0.25+4(0.1875)+0.125\right]=0.09375.

Adding the two interval probabilities gives \(0.34375+0.09375=0.43750\), or \(0.4375\) to four decimal places. This is exact here because each piece of the mixture density is linear. As a check, the one-claim contribution is \(0.50(0.5)=0.25\), and the two-claim contribution is \(0.25(0.75)=0.1875\); their sum is also \(0.4375\).

Combining Conditional Cumulative Probabilities

It is often simpler to compute an aggregate probability by mixing conditional cumulative probabilities rather than integrating a full mixture density. For a threshold \(c\), the identity is

P(S_N\leq c)=\sum_{n=0}^{\infty}p_nP(S_n\leq c),

where \(S_0=0\), so \(P(S_0\leq c)=1\) when \(c\geq0\). This is conditioning on the claim count. It is exact when the sum includes every possible count. If a count distribution has an infinite tail and only finitely many terms are evaluated, report a truncation bound or a convergence check; do not silently treat omitted counts as impossible.

Worked Example: An Aggregate Cumulative Probability

A hypothetical account has \(N\sim\operatorname{Binomial}(3,0.5)\) claims, with independent severities uniformly distributed on \([0,1]\) thousand dollars. Calculate \(P(S_N\leq1)\). There is no interest or mortality assumption.

The count probabilities for \(n=0,1,2,3\) are \(0.125,0.375,0.375,0.125\). For zero claims, the aggregate is zero, so the conditional probability is 1. For one claim, \(P(S_1\leq1)=1\). For two claims, the triangular density is \(f_2(s)=s\) from 0 to 1, giving

P(S_2\leq1)=\int_0^1s\,ds=0.5.

For three claims and \(0\leq s\leq1\), the convolution gives \(f_3(s)=s^2/2\): in the recursive integral the feasible range is \(0\leq x\leq s\), and \(f_2(s-x)=s-x\), so \(\int_0^s(s-x)\,dx=s^2/2\). Integrating this density gives

P(S_3\leq1)=\int_0^1\frac{s^2}{2}\,ds=\frac{1}{6}.

Weighting the four cases,

P(S_N\leq1)=0.125(1)+0.375(1)+0.375(0.5)+0.125\left(\frac16\right) =0.708333\ldots.

Thus the probability is \(0.7083\) to four decimal places. The terms sum to \(0.125+0.375+0.1875+0.020833\ldots=0.708333\ldots\), confirming the calculation.

Numerical Checks and Common Traps

Numerical convolution can be sensitive to support limits, quadrature grids, and count truncation. A useful check is that each fixed-count density integrates to 1 over its full support. The mixture’s continuous density integrates to \(1-p_0\) when severities are strictly positive and the count probabilities sum to 1. Adding the zero atom then restores total probability 1.

Using a density value as a probability. A density such as \(f_3(1.5)=0.75\) is not the probability of an exact aggregate value. A probability over an interval requires integration.
Ignoring support limits. Integrating outside the feasible range can include impossible severity splits. Derive the limits before applying a numerical rule.
Applying one quadrature rule across a kink or jump. Split the integral where the density changes formula. The one-claim component in the binomial example changes at \(s=1\), so integrating across that point without splitting can reduce accuracy.
Forgetting the zero atom. The density for positive aggregate values does not describe the probability \(p_0\) at zero. Include that mass when calculating probabilities that include zero or checking total probability.
Dropping count probabilities or tail terms. A mixture requires the weights \(p_n\). For a count distribution with infinitely many possible counts, any omitted tail must be controlled or acknowledged.

Key takeaway: Convolve continuous severities conditional on a fixed count, respect the feasible integration range, and only then mix over the count probabilities. Integrate the resulting density for interval probabilities, while accounting separately for any atom at zero.

Check Your Understanding

For each problem, state the fixed-count or count-mixture calculation before giving the result. Severities are independent of the claim count in every problem, and no interest or mortality assumption applies.

Claims have severity \(X\sim\operatorname{Uniform}(0,1)\) thousand dollars. Find the density of the sum of two claims at \(s=0.6\). (Answer: \(f_2(0.6)=0.6000\) per thousand dollars.)
A portfolio has \(N\sim\operatorname{Binomial}(2,0.4)\) claims and \(X\sim\operatorname{Uniform}(0,1)\) thousand dollars. Find both \(P(S_N=0)\) and the aggregate density at \(s=0.5\). (Answer: \(P(S_N=0)=0.3600\) and \(f_{S_N}(0.5)=0.5600\) per thousand dollars.)
Let \(N\sim\operatorname{Binomial}(2,0.5)\) and \(X\sim\operatorname{Uniform}(0,1)\) thousand dollars. Calculate \(P(0.25<S_N<0.75)\), integrating separately the one-claim and two-claim contributions. (Answer: \(0.3125\).)
Let \(N\sim\operatorname{Binomial}(3,0.5)\) and \(X\sim\operatorname{Uniform}(0,1)\) thousand dollars. Calculate \(P(S_N\leq0.5)\), including the zero-claim case. (Answer: \(0.3620\) to four decimal places.)
With exactly three independent severities \(X\sim\operatorname{Uniform}(0,1)\) thousand dollars, set up \(f_3(1.25)=\int_0^1 f_2(1.25-x)\,dx\). Split at the point where \(1.25-x=1\), and evaluate the integral. (Answer: \(f_3(1.25)=0.6875\) per thousand dollars.)

What You'll Learn