Mixture Models For Heterogeneous Severity

Why Use a Severity Mixture?

A portfolio can contain losses from different kinds of exposure. For example, a hypothetical insurer might cover both small household equipment and specialized commercial machinery. Even if each type has a simple severity distribution, the portfolio’s overall loss distribution reflects both types and their relative proportions. A mixture model represents this heterogeneity by first selecting a component according to specified weights and then drawing a loss from that component.

The weights describe the probability that a randomly selected loss belongs to each component. They are not multipliers applied to an individual loss, and a mixture is not the sum of independent component losses. These distinctions matter when calculating probabilities and moments.

Definition: A severity $X$ has a mixture distribution with component distributions $F_1,\ldots,F_k$ and weights $\pi_1,\ldots,\pi_k$ if $\pi_i\geq 0$, $\sum_{i=1}^k\pi_i=1$, and $$ F_X(x)=P(X\leq x)=\sum_{i=1}^k\pi_iF_i(x). $$ Equivalently, a component $i$ is selected with probability $\pi_i$, and the loss conditional on that selection has distribution $F_i$. The component weights and distributions together specify the full model.

The component distributions need not all be of the same kind. A mixture can combine continuous distributions, discrete distributions, or both. If component $i$ has density $f_i$, the mixture’s continuous density is $\sum_i\pi_i f_i(x)$ wherever the component densities exist. Any component point masses must also be retained: a mixture can have jumps in its CDF even when some of its components are continuous.

For a loss variable, define the survival function by $S_X(x)=P(X>x)=1-F_X(x)$. Because the weights sum to one, the mixture survival function is $$ S_X(x)=\sum_{i=1}^k\pi_iS_i(x). $$ This formula is useful for exceedance probabilities. At a point mass, the strict inequality in $P(X>x)$ matters: it excludes a loss equal to $x$.

Means and Variances of Mixtures

Let $\mu_i=E[X\mid\text{component }i]$ and $\sigma_i^2=\operatorname{Var}(X\mid\text{component }i)$. If the component means exist, the overall mean is the weighted average of those means. If the component variances also exist, the overall variance has two parts: variation within components and variation among their means.

Theorem: For a mixture with weights $\pi_i$, $$ E[X]=\sum_{i=1}^k\pi_i\mu_i, \qquad \operatorname{Var}(X)=\sum_{i=1}^k\pi_i\left(\sigma_i^2+(\mu_i-\mu)^2\right), $$ where $\mu=E[X]$. Equivalently, $$ \operatorname{Var}(X)=\sum_{i=1}^k\pi_i\sigma_i^2+ \sum_{i=1}^k\pi_i(\mu_i-\mu)^2. $$ These formulas require the corresponding moments to be finite. A component with positive weight and an infinite mean makes the mixture mean infinite; an infinite component variance with positive weight precludes a finite mixture variance.

The variance formula follows by conditioning on the component. Write $Z$ for the selected component and expand the squared deviation from the overall mean:

\begin{aligned} E[(X-\mu)^2] &=\sum_i\pi_i E[(X-\mu)^2\mid Z=i]\\ &=\sum_i\pi_i E[((X-\mu_i)+(\mu_i-\mu))^2\mid Z=i]\\ &=\sum_i\pi_i\left(\sigma_i^2+(\mu_i-\mu)^2\right). \end{aligned}

In the expansion, the cross term has conditional expectation zero because $E[X-\mu_i\mid Z=i]=0$. The second part of the variance therefore measures how far component means lie from the overall mean. Omitting it understates the variability whenever the component means differ.

Worked Example: Two Exponential Loss Components

A hypothetical repair portfolio has weight $0.60$ on exponential losses with mean $\$500$, and weight $0.40$ on exponential losses with mean $\$1{,}500$. Find the overall mean, variance, and probability that a loss exceeds $\$1{,}000$. For an exponential distribution with mean $\theta$, use $S(x)=e^{-x/\theta}$ and variance $\theta^2$, for $x\geq0$.

Method. Apply the mixture formulas to each quantity. The two weights are nonnegative and sum to $0.60+0.40=1$.

Mean.

E[X]=0.60(500)+0.40(1{,}500)=300+600=\$900.

Variance. The within-component contribution is $0.60(500^2)+0.40(1{,}500^2)=150{,}000+900{,}000=1{,}050{,}000$ dollars squared. The between-component contribution is $0.60(500-900)^2+0.40(1{,}500-900)^2=96{,}000+144{,}000=240{,}000$ dollars squared. Thus

\operatorname{Var}(X)=1{,}050{,}000+240{,}000 =1{,}290{,}000\text{ dollars}^2.

Exceedance probability. Weight each component’s survival probability at $\$1{,}000$:

\begin{aligned} P(X>1{,}000) &=0.60e^{-1{,}000/500}+0.40e^{-1{,}000/1{,}500}\\ &=0.60e^{-2}+0.40e^{-2/3}\\ &\approx 0.60(0.1353353)+0.40(0.5134171)\\ &\approx 0.0812012+0.2053668\\ &\approx 0.2866. \end{aligned}

The probability is rounded to four decimal places. The calculation avoids treating the mixture as a single exponential with the overall mean; that substitution would give $e^{-1{,}000/900}$, which is not the mixture’s survival probability. As a check, the component contributions add to $0.286568$, and the overall mean is between the two component means.

Mixtures Can Include Point Masses

A component may assign positive probability to one exact loss amount. In that case, the mixture CDF has a jump at that amount. Evaluating the weighted CDFs directly is a reliable way to handle such boundaries.

Worked Example: A Continuous Component and a Fixed Loss

A hypothetical portfolio has probability $0.70$ of a loss uniformly distributed from $\$0$ to $\$1{,}000$, inclusive, and probability $0.30$ of a fixed loss of $\$1{,}500$. Find $F_X(800)$, $P(X>1{,}200)$, the mean, and the variance.

Method. Combine the component CDFs and moments using their weights. For a uniform loss on $[0,1{,}000]$, the CDF is $x/1{,}000$ on that interval, its mean is $\$500$, and its variance is $1{,}000^2/12$. The fixed-loss component has mean $\$1{,}500$ and variance zero.

For $0\leq x<1{,}000$, only the uniform component contributes to the CDF, so $F_X(x)=0.70x/1{,}000$. For $1{,}000\leq x<1{,}500$, it is $0.70$; at and above $\$1{,}500$, it is $1$. Therefore,

F_X(800)=0.70\left(\frac{800}{1{,}000}\right)=0.56.

At $\$1{,}200$, the uniform component cannot exceed the threshold, while the fixed loss does. Hence $P(X>1{,}200)=0.30$. The strict inequality causes no ambiguity here because neither component has a loss at $\$1{,}200$.

The mean is

E[X]=0.70(500)+0.30(1{,}500)=350+450=\$800.

For the variance, the within-component contribution is $0.70(1{,}000^2/12)+0.30(0)=58{,}333.333$ dollars squared. The between-component contribution is $0.70(500-800)^2+0.30(1{,}500-800)^2=63{,}000+147{,}000=210{,}000$ dollars squared. Thus

\operatorname{Var}(X)=58{,}333.333+210{,}000 =268{,}333.333\text{ dollars}^2.

The component weights sum to one, $F_X(800)$ lies between zero and one, and the variance includes both the uniform component’s spread and the separation between the component means. The mixture CDF jumps by $0.30$ at $\$1{,}500$, reflecting the fixed-loss component.

Using Component Survival Functions

When component distributions have familiar survival functions, an exceedance probability is often simpler to calculate from those functions than from a piecewise CDF. The same weights must be used for every component, and the threshold must be evaluated under each component’s own parameters.

Worked Example: Exponential and Erlang Components

A hypothetical loss is exponential with mean $2$ with probability $0.50$, and has an Erlang distribution with shape $2$ and scale $2$ with probability $0.50$. Find the mean, variance, $F_X(2)$, and $P(X>2)$. For the Erlang component, $S(x)=e^{-x/2}(1+x/2)$, mean $4$, and variance $8$; the exponential component has $S(x)=e^{-x/2}$, mean $2$, and variance $4$.

Method. First combine component moments, then weight the component CDFs and survival functions at the specified threshold.

The mean is

$$ E[X]=0.50(2)+0.50(4)=3. $$

The within-component variance is $0.50(4)+0.50(8)=6$. The between-component contribution is $0.50(2-3)^2+0.50(4-3)^2=1$. Therefore $\operatorname{Var}(X)=6+1=7$.

At $x=2$, the exponential survival probability is $e^{-1}$, and the Erlang survival probability is $e^{-1}(1+1)=2e^{-1}$. Thus

\begin{aligned} P(X>2)&=0.50e^{-1}+0.50(2e^{-1})=1.5e^{-1}\approx0.5518,\\ F_X(2)&=1-P(X>2)=1-1.5e^{-1}\approx0.4482. \end{aligned}

Both probabilities are rounded to four decimal places. They sum to one, as required. Using only the exponential component’s tail, or using only the Erlang component’s tail, would omit half of the mixture.

Common Mistakes and Checks

Using the average component mean to select a component distribution. The overall mean alone does not determine the mixture. In general, the mixture of exponentials is not an exponential distribution with the weighted-average mean.
Forgetting between-component variance. The weighted average of the component variances captures only within-component variation. Add the weighted squared deviations of component means from the overall mean.
Confusing a mixture with a sum. In a mixture, one component is selected for a loss. A sum of component losses is a different model and does not use the mixture CDF formula.
Ignoring a component point mass. A point-mass component creates a jump in the CDF. Check whether the threshold includes or excludes that exact value.
Using inconsistent weights. The same component probabilities apply when combining the CDF, survival function, mean, and variance. Confirm that all weights are nonnegative and add to one.

Key takeaway: A mixture combines component probabilities and expectations using the component weights. Its variance includes both the variability inside each component and the variability caused by differences between component means.

Check Your Understanding

For each hypothetical model, identify the component weights and use the mixture formulas. Probabilities are rounded to four decimal places where needed.

A loss is exponential with mean $\$100$ with probability $0.25$, and exponential with mean $\$300$ with probability $0.75$. Find its mean and $P(X>\$200)$. (Answer: mean $\$250$; $P(X>\$200)=0.4189$)
A loss is uniform on $[0,10]$ with probability $0.40$, and equals $20$ with probability $0.60$. Find $F_X(5)$, $P(X>10)$, and the mean. (Answer: $F_X(5)=0.20$; $P(X>10)=0.60$; mean $14$)
A loss is uniform on $[0,6]$ with probability $1/3$, and uniform on $[6,12]$ with probability $2/3$. Find $F_X(6)$, $P(X>9)$, the mean, and the variance. (Answer: $F_X(6)=1/3$; $P(X>9)=1/3$; mean $7$; variance $11$)
A loss is exponential with mean $1$ with probability $0.60$, and Erlang with shape $2$ and scale $1$ with probability $0.40$. Find its mean, variance, and $P(X>2)$. (Answer: mean $1.4$; variance $1.64$; $P(X>2)=1.8e^{-2}\approx0.2436$)
A loss is exponential with mean $2$ with probability $0.20$, and exponential with mean $5$ with probability $0.80$. Find its mean, variance, and $P(X>5)$. (Answer: mean $4.4$; variance $22.24$; $P(X>5)=0.2e^{-2.5}+0.8e^{-1}\approx0.3107$)

What You'll Learn