Nonidentical Independent Claim Severities

Why Nonidentical Severities Need Separate Treatment

A portfolio can have a fixed number of claims while the claims follow different severity models. For example, the first claim may be subject to one coverage limit and the second to another. Even if the claim amounts are independent, it is not appropriate to treat them as identically distributed unless their probability models actually match. The aggregate distribution must reflect each severity model, while its mean and variance can be assembled from the individual moments.

This tutorial focuses on the aggregate loss conditional on a specified claim count. The frequency model determines how likely that count is; once the count is fixed, the calculation is a combination of the corresponding individual severities.

Definition: Given $N=n$, let $X_1,\ldots,X_n$ be the claim severities, assumed to be mutually independent but not necessarily identically distributed. The conditional aggregate loss is $$ S_n=\sum_{j=1}^{n}X_j. $$ Each $X_j$ has its own distribution, mean $\mu_j=E[X_j\mid N=n]$, and variance $\sigma_j^2=\operatorname{Var}(X_j\mid N=n)$. The conditional aggregate distribution is the distribution of $S_n$ under these assumptions.

The notation allows the severity model for claim $j$ to depend on its position or other known information. In a random-frequency model, the formulas below apply conditional on $N=n$ when the severities are independent under that condition and have the stated conditional distributions. Independence is essential for the variance sum; without it, covariance terms may be needed.

Convolving the Individual Distributions

For discrete severities, the aggregate probability at a possible total $s$ is obtained by adding the probabilities of every combination of individual losses that sums to $s$. For two independent severities, this is the discrete convolution. If $p_j(x)=P(X_j=x\mid N=n)$, then

P(S_n=s\mid N=n) = \sum_{x_1+\cdots+x_n=s}\prod_{j=1}^{n}p_j(x_j).

The product appears because the severities are independent. The sum includes every combination of values whose total is $s$. For continuous severities, the corresponding operation integrates the product of densities over combinations with the required total; the discrete examples here make the combination of probabilities visible.

When the severities are nonnegative integer-valued, probability generating functions provide a compact alternative. Define $G_j(z)=E[z^{X_j}\mid N=n]$. Independence gives

G_{S_n}(z)=\prod_{j=1}^{n}G_j(z).

The coefficient of $z^s$ in this product is $P(S_n=s\mid N=n)$. This method does not require the individual generating functions to be identical. It is often convenient to multiply them successively, just as one would carry out repeated convolutions.

Worked Example: Convolving Two Different Severity Models

Conditional on exactly two claims, suppose the independent severities have these distributions: $X_1$ is 1 or 3, each with probability 0.5; $X_2$ is 2 with probability 0.25 and 4 with probability 0.75. Find the conditional aggregate distribution and verify its mean and variance. All amounts are in thousands of dollars.

Method. List each possible pair, multiply its probabilities using independence, and combine pairs with the same total.

The four pairs give:

$1+2=3$, with probability $0.5(0.25)=0.125$.
$1+4=5$, with probability $0.5(0.75)=0.375$.
$3+2=5$, with probability $0.5(0.25)=0.125$.
$3+4=7$, with probability $0.5(0.75)=0.375$.

Combining the two ways to obtain 5 gives $P(S_2=5\mid N=2)=0.375+0.125=0.500$. Thus the distribution is 3 with probability 0.125, 5 with probability 0.500, and 7 with probability 0.375. The probabilities sum to $0.125+0.500+0.375=1.000$.

For the individual moments, $E[X_1\mid N=2]=0.5(1)+0.5(3)=2$, and $\operatorname{Var}(X_1\mid N=2)=0.5(1-2)^2+0.5(3-2)^2=1$. For the second severity, $E[X_2\mid N=2]=0.25(2)+0.75(4)=3.5$, and $\operatorname{Var}(X_2\mid N=2)=0.25(2-3.5)^2+0.75(4-3.5)^2=0.75$. Independence therefore gives aggregate mean $2+3.5=5.5$ and aggregate variance $1+0.75=1.75$.

Check. The distribution gives $E[S_2\mid N=2]=3(0.125)+5(0.500)+7(0.375)=5.5$. Its second moment is $3^2(0.125)+5^2(0.500)+7^2(0.375)=32$, so its variance is $32-(5.5)^2=1.75$. The distribution and moment calculations agree. A common mistake would be to use one severity variance twice; the two individual variances are not equal.

Conditional Aggregate Moments

Convolution is necessary when the full distribution is required, but the first two moments have simpler formulas. Under the independence assumption,

E[S_n\mid N=n]=\sum_{j=1}^{n}\mu_j, \qquad \operatorname{Var}(S_n\mid N=n)=\sum_{j=1}^{n}\sigma_j^2.

The mean formula follows from linearity of expectation, which does not require independence. To justify the variance formula, write the centered aggregate as a sum of centered severities and expand its square:

\begin{aligned} \operatorname{Var}(S_n\mid N=n) &=E\left[\left(\sum_{j=1}^{n}(X_j-\mu_j)\right)^2\middle|N=n\right]\\ &=\sum_{j=1}^{n}\operatorname{Var}(X_j\mid N=n) +2\sum_{j<k}\operatorname{Cov}(X_j,X_k\mid N=n). \end{aligned}

Mutual independence implies that each covariance in the second sum is zero. Only the individual variances remain, giving the stated result. If the severities are dependent, that covariance sum cannot simply be discarded.

The second moment can also be found from the mean and variance:

E[S_n^2\mid N=n] = \operatorname{Var}(S_n\mid N=n) +\left(E[S_n\mid N=n]\right)^2.

This is useful when checking a distribution computed by convolution. It also highlights why summing individual second moments is not the formula for the aggregate second moment: the square of the aggregate mean contributes cross-products.

Worked Example: Three Severities and a Full Lattice Distribution

Conditional on three claims, suppose the severities are mutually independent. $X_1$ is 1 or 2 with probability 0.5 each; $X_2$ is 2 with probability 0.75 or 4 with probability 0.25; $X_3$ is 1 or 3 with probability 0.5 each. Find the aggregate distribution and its conditional mean and variance.

Method. First convolve $X_1$ and $X_2$. Their possible totals are 3, 4, 5, and 6:

Total 3: $1+2$, probability $0.5(0.75)=0.375$.
Total 4: $2+2$, probability $0.5(0.75)=0.375$.
Total 5: $1+4$, probability $0.5(0.25)=0.125$.
Total 6: $2+4$, probability $0.5(0.25)=0.125$.

Now add $X_3$, which equals 1 or 3 with probability 0.5 each. The resulting probabilities are: total 4, $0.375(0.5)=0.1875$; total 5, $0.375(0.5)=0.1875$; total 6, $0.375(0.5)+0.125(0.5)=0.2500$; total 7, $0.375(0.5)+0.125(0.5)=0.2500$; total 8, $0.125(0.5)=0.0625$; and total 9, $0.125(0.5)=0.0625$. These probabilities sum to 1.

The individual means are $E[X_1]=1.5$, $E[X_2]=2(0.75)+4(0.25)=2.5$, and $E[X_3]=2$. Therefore $E[S_3\mid N=3]=1.5+2.5+2=6$.

The individual variances are $\operatorname{Var}(X_1)=0.25$, $\operatorname{Var}(X_2)=0.75$, and $\operatorname{Var}(X_3)=1$. Hence $\operatorname{Var}(S_3\mid N=3)=0.25+0.75+1=2$.

Check. The distribution gives mean $4(0.1875)+5(0.1875)+6(0.2500)+7(0.2500)+8(0.0625)+9(0.0625)=6$. Its second moment is $16(0.1875)+25(0.1875)+36(0.2500)+49(0.2500)+64(0.0625)+81(0.0625)=38$, so the variance is $38-6^2=2$. The check confirms both moments.

Worked Example: A Conditional Tail Probability and Moment Check

Given $N=2$, let independent severities $X_1$ and $X_2$, in thousands of dollars, have distributions $P(X_1=1)=0.6$, $P(X_1=4)=0.4$, $P(X_2=2)=0.25$, and $P(X_2=5)=0.75$. Find the conditional probability that the aggregate is at least 6, and calculate its mean and variance.

Method. Enumerate the four independent pairs and combine equal totals. The pair totals and probabilities are 3 with $0.6(0.25)=0.15$, 6 with $0.6(0.75)=0.45$, 6 with $0.4(0.25)=0.10$, and 9 with $0.4(0.75)=0.30$. Thus $P(S_2=3\mid N=2)=0.15$, $P(S_2=6\mid N=2)=0.55$, and $P(S_2=9\mid N=2)=0.30$.

The requested probability is $P(S_2\geq 6\mid N=2)=0.55+0.30=0.85$.

The individual means are $E[X_1]=0.6(1)+0.4(4)=2.2$ and $E[X_2]=0.25(2)+0.75(5)=4.25$, so $E[S_2\mid N=2]=6.45$. The individual variances are $\operatorname{Var}(X_1)=0.6(1-2.2)^2+0.4(4-2.2)^2=2.16$ and $\operatorname{Var}(X_2)=0.25(2-4.25)^2+0.75(5-4.25)^2=1.6875$. Therefore, $\operatorname{Var}(S_2\mid N=2)=2.16+1.6875=3.8475$.

Check and trap. The distribution gives mean $3(0.15)+6(0.55)+9(0.30)=6.45$. Its second moment is $9(0.15)+36(0.55)+81(0.30)=45.45$, and $45.45-(6.45)^2=3.8475$. A tempting but incorrect shortcut is to calculate $2(2.16)=4.32$, as though both claims had the first claim's variance. The second severity has variance 1.6875, not 2.16, so the correct variance is 3.8475.

Practical Checks and Common Traps

A sound calculation keeps the conditional setup explicit. The distribution being computed is conditional on the specified claim count, and each claim index has its own severity probabilities. A frequency probability such as $P(N=n)$ is not part of the convolution conditional on $N=n$. If an unconditional aggregate distribution is later required, the conditional distributions must be combined across possible counts using their frequency probabilities.

Before accepting a discrete convolution, check that its probabilities sum to 1 and that its support includes every possible sum. Then compare its mean and variance with the sums of the individual moments. These checks are especially helpful when several combinations lead to the same aggregate value.

Assuming identical severities without evidence. Do not multiply one variance by $n$ unless every severity has that variance; for nonidentical severities, add their actual variances.
Forgetting coincident totals. Different claim combinations can yield the same aggregate amount, and their probabilities must be added.
Mixing conditional and unconditional quantities. A distribution given $N=n$ is not yet the aggregate distribution across random claim counts.
Using independence where it is absent. If severities are dependent, the variance includes their covariances.

Key result: Conditional on a fixed claim count, independent nonidentical severities combine by convolution. Their aggregate mean is the sum of their means, and their aggregate variance is the sum of their variances. Verify a calculated distribution by checking its total probability and moments.

Check Your Understanding

For each problem, condition on the stated claim count and use the severity distributions given. Amounts are in thousands of dollars unless otherwise stated.

Two independent severities satisfy $P(X_1=2)=0.75$, $P(X_1=6)=0.25$, $P(X_2=1)=0.4$, and $P(X_2=3)=0.6$. Find the aggregate distribution and its mean. (Answer: $S_2=3,5,7,9$ with probabilities $0.30,0.45,0.10,0.15$, respectively; mean $5.2$.)
Three independent severities have $X_1=1$ or 3 with equal probabilities, $X_2=2$ or 5 with equal probabilities, and $X_3=2$ with certainty. Find the aggregate mean and variance. (Answer: mean $7.5$; variance $3.25$.)
Given two claims, independent severities have $P(X_1=2)=0.4$, $P(X_1=5)=0.6$, $P(X_2=1)=0.25$, and $P(X_2=4)=0.75$. Find $P(S_2\geq 9\mid N=2)$ and $\operatorname{Var}(S_2\mid N=2)$. (Answer: probability $0.45$; variance $3.8475$.)
Three independent severities satisfy $P(X_1=1)=P(X_1=2)=0.5$, $X_2=3$ with certainty, and $P(X_3=2)=0.25$, $P(X_3=4)=0.75$. Find the full aggregate distribution. (Answer: $S_3=6,7,8,9$ with probabilities $0.125,0.125,0.375,0.375$, respectively.)
A frequency model has $N\sim\operatorname{Binomial}(4,0.25)$. Conditional on $N=2$, the two independent severities are $X_1=1$ or 3 with equal probabilities and $X_2=2$ with probability 0.75 or 6 with probability 0.25. Find the conditional aggregate mean, variance, and probability of an aggregate of at least 7. (Answer: conditional mean $5$; conditional variance $4$; $P(S_2\geq 7\mid N=2)=0.25$.)

What You'll Learn