Why Model Two Risk Groups Together?
A portfolio can contain distinct kinds of risks even when each claim is recorded using the same severity variable. For example, a hypothetical insurer might cover two classes of equipment: one class with relatively small, frequent repair losses and another with less frequent but larger losses. A single severity distribution for the combined portfolio must reflect both the severity patterns within each class and the proportion of claims coming from each class.
A two-component mixture gives a direct way to do this. It first selects a risk group according to its share of claims, then generates a loss from that group’s severity distribution. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
The group assignment is a latent label in the combined severity model: a claim comes from one group, but the overall model may not condition on observing that label. The weights must match the sampling basis. If the model describes a randomly selected claim, use each group’s share of claims, not automatically its share of policies or insured values.
Distribution and Moments of the Mixture
Write \(F_j(x)=P(X\leq x\mid Z=j)\) for the component distribution function, and \(S_j(x)=P(X>x\mid Z=j)\) for the component survival function. Conditioning on the group gives the overall distribution and survival function:
When the components have densities \(f_1\) and \(f_2\), the overall density is \(f_X(x)=\pi_1f_1(x)+\pi_2f_2(x)\). The weights apply to the entire component distribution, not just to a component’s mean or tail behavior.
Suppose the component means are \(\mu_1,\mu_2\) and the component variances are \(\sigma_1^2,\sigma_2^2\), all finite. The overall mean is the weighted component mean:
The overall variance contains two sources of variation: variation within each group and variation between the group means. In general,
The first two terms are the weighted average of the component variances. The last two terms measure how far the group means lie from the overall mean. Omitting these between-group terms understates the total variance whenever the component means differ.
Worked Calculations
For a threshold \(x\), the overall probability of exceeding it is obtained by adding the weighted component exceedance probabilities. A related but different quantity is the proportion of exceedances that come from a particular group. If \(S_X(x)>0\), conditioning on the event \(X>x\) gives
The numerator is the probability that a randomly selected claim both belongs to group \(j\) and exceeds \(x\). Dividing by the probability of any exceedance gives the group’s share among exceedances. The formula applies whenever the denominator is positive; it does not require a particular component family.
Worked Example: Exponential Groups and Large-Claim Composition
A hypothetical portfolio has equal claim weights in two groups. Conditional severity is exponential with mean 6 in Group 1 and mean 18 in Group 2. Find the overall mean and variance, the probability a claim exceeds 12, and the share of claims exceeding 12 that come from Group 2.
Method and component parameters. The weights are \(\pi_1=\pi_2=0.5\). For an exponential distribution with mean \(\theta\), the variance is \(\theta^2\) and the survival function is \(S(x)=e^{-x/\theta}\), for \(x\geq0\). Thus the component means are 6 and 18, and the component variances are 36 and 324.
Overall mean.
Overall variance. The within-group contribution is \(0.5(36)+0.5(324)=18+162=180\). The between-group contribution is \(0.5(6-12)^2+0.5(18-12)^2=0.5(36)+0.5(36)=36\). Hence
Exceedance probability. At \(x=12\), the component survival probabilities are \(e^{-12/6}=e^{-2}=0.1353353\) and \(e^{-12/18}=e^{-2/3}=0.5134171\). Their weighted contributions are \(0.5(0.1353353)=0.0676676\) and \(0.5(0.5134171)=0.2567086\). Thus \(S_X(12)=0.0676676+0.2567086=0.3243762\), or 0.3244.
Group 2 share among exceedances. Divide its weighted tail contribution by the total tail:
The mean is 12, the variance is 216, and about 79.14% of claims above 12 come from Group 2. The group share is not the Group 2 weight of 50%: Group 2 has the heavier tail at this threshold.
Common mistake and correction. Using only the weighted component variances gives 180 and misses between-group variation. The correction is to add the between-group contribution, 36, for an overall variance of 216.
Worked Example: Exponential and Gamma Components
A hypothetical claim portfolio has 70% of its claims in Group 1, whose severity is exponential with mean 10, and 30% in Group 2, whose severity is gamma with shape 2 and scale 10. Find the overall mean and variance, \(S_X(10)\), and the Group 2 share among claims exceeding 10. Severities are measured in thousands of dollars.
Method and component moments. For an exponential distribution with mean 10, the variance is \(10^2=100\), so its second moment is \(100+10^2=200\). For a gamma distribution with shape \(k=2\) and scale \(\theta=10\), the mean is \(k\theta=20\), the variance is \(k\theta^2=200\), and the second moment is \(200+20^2=600\).
Overall mean.
Overall variance. First calculate the second moment:
Then subtract the square of the overall mean:
Exceedance probability. At 10, the exponential component survival is \(e^{-10/10}=e^{-1}=0.3678794\). A gamma distribution with shape 2 and scale 10 has survival \(e^{-x/10}(1+x/10)\), so Group 2’s survival at 10 is \(e^{-1}(1+1)=2e^{-1}=0.7357589\). The weighted contributions are \(0.7(0.3678794)=0.2575156\) and \(0.3(0.7357589)=0.2207277\). Therefore,
which rounds to 0.4782.
Group 2 share. The Group 2 contribution is divided by the overall exceedance probability:
The overall mean is 13 thousand dollars, the variance is 151 thousand dollars squared, and about 46.15% of claims above 10 thousand dollars come from Group 2. Although Group 2 has the larger component mean, its exceedance share at this threshold is below one half because its claim weight is only 30%.
Common mistake avoided. Do not use the Group 2 survival probability, 0.7358, as the group’s share of all exceedances. It is a conditional probability within Group 2. The overall tail must first be formed using both weighted tails, and the Group 2 share is then its weighted tail divided by that total.
Worked Example: Two Gamma Groups with Different Tail Weights
A hypothetical portfolio has 60% of its claims in Group 1 and 40% in Group 2. Group 1 severity is gamma with shape 2 and scale 5; Group 2 severity is gamma with shape 3 and scale 5. Calculate the overall mean and variance, the probability a claim exceeds 10, and the share of those exceedances from Group 2. Severity is measured in thousands of dollars.
Component means and variances. For a gamma distribution with shape \(k\) and scale \(\theta\), the mean is \(k\theta\) and the variance is \(k\theta^2\). Group 1 therefore has mean \(2(5)=10\) and variance \(2(25)=50\). Group 2 has mean \(3(5)=15\) and variance \(3(25)=75\).
Overall mean and variance.
The component second moments are \(50+10^2=150\) and \(75+15^2=300\). Thus \(E[X^2]=0.6(150)+0.4(300)=90+120=210\), and
Exceedance probability. For an integer-shape gamma distribution with shape \(k\) and scale \(\theta\), the survival function is \(e^{-x/\theta}\sum_{r=0}^{k-1}(x/\theta)^r/r!\). At \(x=10\), \(x/\theta=2\). Group 1’s survival is \(e^{-2}(1+2)=3e^{-2}=0.4060058\). Group 2’s survival is \(e^{-2}(1+2+2)=5e^{-2}=0.6766764\). The weighted tails are \(0.6(0.4060058)=0.2436035\) and \(0.4(0.6766764)=0.2706706\). Hence
or 0.5143.
Group 2 share.
The component means are below and above the threshold, respectively, and the heavier Group 2 tail makes its share among exceedances larger than its original 40% claim weight.
Interpreting a Mixture in Practice
A mixture describes the distribution for a randomly selected claim from a specified population. Its weights are not arbitrary fitting multipliers: they represent the probabilities of selecting each group on that basis. Changing from claim weights to policy weights can change the model, particularly if the groups have different claim frequencies. Keep the population and weighting basis explicit when constructing or interpreting a mixture.
The overall survival probability and the group share answer different questions. \(S_X(x)\) asks what fraction of all claims exceed \(x\). In contrast, \(P(Z=j\mid X>x)\) asks how the exceedances are divided among the groups. A group can be a small share of all claims yet account for a large share of high-severity claims if its component tail is sufficiently heavy.
Identify what is sampled and set \(\pi_1+\pi_2=1\) using the corresponding group proportions.
State its distribution and parameters, then obtain its mean, variance, and survival function as needed.
Weight component means and tails; for variance, include both within-group and between-group variation.
Divide the selected group’s weighted survival probability by the overall survival probability at the same threshold.
The most common exam error is to confuse a component-level quantity with an overall mixture quantity. A component mean is not generally the overall mean; a component survival is not the overall survival; and the weighted average of component variances is not generally the overall variance. Check both the weights and the target probability or moment before finalizing a calculation.
Check Your Understanding
For each problem, use the stated claim weights and report probabilities to four decimal places.
- A hypothetical mixture has weights 0.25 and 0.75. The component distributions are exponential with means 8 and 12. Find the overall mean and variance. (Answer: Mean \(=11\); variance \(=127\).)
- A hypothetical portfolio has 40% of claims from an exponential group with mean 5 and 60% from an exponential group with mean 15. Find \(S_X(10)\) and the share of claims above 10 from the group with mean 15. (Answer: \(S_X(10)=0.3622\); Group 2 share \(=0.8505\).)
- Two equally weighted groups have gamma severities: Group 1 has shape 2 and scale 4, and Group 2 has shape 3 and scale 4. Find the overall mean, variance, \(S_X(8)\), and Group 2’s share among claims exceeding 8. (Answer: Mean \(=10\); variance \(=44\); \(S_X(8)=0.5413\); Group 2 share \(=0.6250\).)
- A hypothetical claim population has weights 0.6 and 0.4. Group 1 has mean 20 and variance 100; Group 2 has mean 50 and variance 900. Find the overall mean and variance, showing the within-group and between-group contributions. (Answer: Mean \(=32\); within-group contribution \(=420\); between-group contribution \(=216\); variance \(=636\).)