Finite Mixture Severity Models

Why Use a Finite Mixture?

A severity model may describe losses from several underlying groups with different risk characteristics. For example, a hypothetical insurer might cover two types of equipment: one type tends to produce smaller claims, while the other is less common but can produce larger ones. A single severity distribution may not represent both groups well. A finite mixture combines a finite number of component distributions, with a weight assigned to each one.

The key calculation rule is simple but important: calculate the requested quantity under each component, then take the weighted sum. The weights describe the probabilities of belonging to the components; they are not adjustments to the component distribution parameters.

Definition: Let $Z$ be an unobserved component label taking values $1,\ldots,k$, with $P(Z=j)=w_j$, where $w_j\geq 0$ and $\sum_{j=1}^{k}w_j=1$. A severity $X$ has a finite mixture distribution if, conditional on $Z=j$, it follows component distribution $j$, with density $f_j(x)$, survival function $S_j(x)$, and mean $\mu_j=E[X\mid Z=j]$, when these quantities exist. The mixture density, survival function, and mean are denoted by $f_X(x)$, $S_X(x)$, and $E[X]$.

This tutorial considers continuous, nonnegative severities with component densities. A component can have a restricted support, such as a uniform distribution on a bounded interval or a Pareto Type I distribution above its threshold. For the mixture mean formula, each component mean must exist and be finite. If a component has no finite mean and receives positive weight, the mixture mean is not finite.

Combining Component Densities and Survival Functions

The mixture density at a loss amount $x$ is the weighted sum of the component densities at that same amount. The mixture survival function, which gives the probability of a loss exceeding $x$, is the weighted sum of the component survival probabilities. These formulas apply when the component label is selected according to the stated weights and the severity then follows the selected component model.

f_X(x)=\sum_{j=1}^{k}w_j f_j(x), \qquad S_X(x)=P(X>x)=\sum_{j=1}^{k}w_j S_j(x).

For a continuous component with cumulative distribution function $F_j(x)$, its survival function is $S_j(x)=1-F_j(x)$. At points outside a component’s support, its density is zero, and its survival function is determined by that support. The weights must sum to one; otherwise they do not define probabilities for the component labels.

Worked Example: Combining an Exponential and a Uniform Component

A hypothetical portfolio has two types of claim, selected independently for each claim. With probability $0.70$, the severity is exponential with rate $\lambda=0.001$ per dollar. With probability $0.30$, the severity is uniform on $[0,2000]$ dollars. Calculate the mixture density, survival probability at $x=\$1{,}000$, and mean.

Method and assumptions. The component weights are $w_1=0.70$ and $w_2=0.30$, which sum to $1$. For an exponential distribution with rate $\lambda$, $f_1(x)=\lambda e^{-\lambda x}$, $S_1(x)=e^{-\lambda x}$, and $E[X\mid Z=1]=1/\lambda$. For a uniform distribution on $[0,b]$, the density within its support is $1/b$, its survival probability at $x\in[0,b]$ is $(b-x)/b$, and its mean is $b/2$.

Density. The exponential component density at $\$1{,}000$ is $0.001e^{-0.001(1000)}=0.001e^{-1}=0.000367879$ per dollar. The uniform component density is $1/2000=0.0005$ per dollar. Therefore,

f_X(1000) =0.70(0.000367879)+0.30(0.0005) =0.000257515+0.000150000 =0.000407515\text{ per dollar}.

Survival probability. The exponential component survival probability is $e^{-1}=0.367879$. The uniform component survival probability is $(2000-1000)/2000=0.5$. Thus,

$$ S_X(1000) =0.70(0.367879)+0.30(0.5) =0.257515+0.150000 =0.407515. $$

Mean. The exponential component mean is $1/0.001=\$1{,}000$. The uniform component mean is $2000/2=\$1{,}000$. Hence $E[X]=0.70(1000)+0.30(1000)=\$1{,}000$. Rounded as specified, $f_X(1000)=0.000408$ per dollar to six decimal places, $S_X(1000)=0.4075$ to four decimal places, and $E[X]=\$1{,}000.00$ to the cent.

Check. Each component calculation is evaluated at the same loss amount, and the two mixture weights sum to one. The component means agree, so their weighted average is also $\$1{,}000$.

The Mixture Mean and Why the Formulas Hold

The mean is the weighted average of the component means, provided the component means are finite. The same weights used to combine densities and survival probabilities must be used to combine means. In particular, the overall mean is not generally the unweighted average of the component means.

E[X]=\sum_{j=1}^{k} w_j\mu_j =\sum_{j=1}^{k}w_j E[X\mid Z=j].

The reason for the density formula is that an outcome near $x$ can arise through any component. For a small interval around $x$, the probability of falling in that interval is the sum over mutually exclusive component labels of the probability of choosing that component times the conditional probability of falling in the interval. Dividing by the interval width and taking the limit gives the weighted sum of component densities.

The survival formula follows directly by partitioning the event $X>x$ according to the component label. The events $\{X>x,Z=j\}$ are disjoint, so their probabilities add; within component $j$, $P(X>x,Z=j)=P(Z=j)P(X>x\mid Z=j)=w_jS_j(x)$. For the mean, the law of total expectation gives $E[X]=\sum_j P(Z=j)E[X\mid Z=j]$. These arguments also show why the same weights appear in all three formulas.

Verify the component weights.
Check that every weight is nonnegative and that their sum is one.

Calculate within each component.
Use the component model to find its density or survival probability at the requested loss, and its mean if required.

Weight and add.
Multiply each component result by its weight, then add the weighted contributions.

Check units and supports.
A density has units of inverse dollars when loss is measured in dollars; a survival probability is dimensionless; a mean has units of dollars.

Worked Calculations With Different Component Families

Mixtures can combine different severity families. The calculation method does not change: only the component formulas do. In the next example, use the lognormal convention $\ln X\sim N(\mu,\sigma^2)$, where $\sigma^2$ is the variance of the normal variable. The lognormal mean is $e^{\mu+\sigma^2/2}$.

Worked Example: A Pareto and Lognormal Mixture

A hypothetical loss has a $0.40$ probability of following a Pareto Type I distribution with threshold $\theta=\$1{,}000$ and shape parameter $\alpha=2$, and a $0.60$ probability of following a lognormal distribution with $\mu=\ln(1000)$ and $\sigma=1$. Calculate $f_X(1000)$, $S_X(1000)$, and $E[X]$.

Method and assumptions. For the Pareto component, $f_1(x)=\alpha\theta^\alpha/x^{\alpha+1}$ and $S_1(x)=(\theta/x)^\alpha$ for $x\geq\theta$; its mean is $\alpha\theta/(\alpha-1)$ when $\alpha>1$. For the lognormal component, $f_2(x)=\phi((\ln x-\mu)/\sigma)/(x\sigma)$, $S_2(x)=1-\Phi((\ln x-\mu)/\sigma)$, and its mean is $e^{\mu+\sigma^2/2}$. Here $\phi$ and $\Phi$ are the standard normal density and cumulative distribution functions.

Pareto component at $\$1{,}000$. Since $x=\theta=1000$, its survival probability is $(1000/1000)^2=1$, and its density is $2(1000)^2/(1000)^3=0.002$ per dollar. Its mean is $2(1000)/(2-1)=\$2{,}000$.

Lognormal component at $\$1{,}000$. The standardized log value is $(\ln(1000)-\ln(1000))/1=0$. Therefore its survival probability is $1-\Phi(0)=0.5$. Since $\phi(0)=0.398942$, its density is $0.398942/1000=0.000398942$ per dollar. Its mean is $e^{\ln(1000)+1/2}=1000e^{0.5}=\$1{,}648.72$, rounded to the cent.

Mixture density and survival. Applying the weights to the component results gives

f_X(1000) =0.40(0.002)+0.60(0.000398942) =0.000800000+0.000239365 =0.001039365\text{ per dollar}, $$ $$ S_X(1000) =0.40(1)+0.60(0.5) =0.40+0.30 =0.7000.

Mixture mean. The weighted mean is $0.40(\$2{,}000)+0.60(\$1{,}648.72)=\$800.00+\$989.23=\$1{,}789.23$, to the cent. Thus the density is $0.001039$ per dollar to six decimal places, the survival probability is $0.7000$ to four decimal places, and the mean is $\$1{,}789.23$.

Check. The Pareto component has survival $1$ at its threshold because all of its continuous probability is at or above that threshold. The resulting mixture survival $0.7000$ lies between the two component survival probabilities $1$ and $0.5$.

Worked Example: Three Components and a Threshold Trap

A hypothetical severity follows one of three models: with weight $0.20$, an exponential distribution with mean $\$1{,}000$; with weight $0.30$, a uniform distribution on $[0,\$2{,}000]$; or with weight $0.50$, a Pareto Type I distribution with threshold $\theta=\$1{,}000$ and shape $\alpha=2$. Find the density and survival probability at $\$1{,}000$, and the mixture mean.

Method and assumptions. The weights sum to $0.20+0.30+0.50=1$. The exponential rate is the reciprocal of its mean, $1/1000=0.001$ per dollar. The uniform and Pareto component formulas are as stated above.

Density contributions. The exponential density is $0.001e^{-1}=0.000367879$, so its weighted contribution is $0.20(0.000367879)=0.000073576$ per dollar. The uniform density is $1/2000=0.0005$, so its weighted contribution is $0.30(0.0005)=0.000150000$ per dollar. At the Pareto threshold, the density is $2/1000=0.002$, so its weighted contribution is $0.50(0.002)=0.001000000$ per dollar. Therefore,

f_X(1000) =0.000073576+0.000150000+0.001000000 =0.001223576\text{ per dollar}.

Survival contributions. The exponential survival is $e^{-1}=0.367879$, contributing $0.20(0.367879)=0.073576$. The uniform survival is $(2000-1000)/2000=0.5$, contributing $0.30(0.5)=0.150000$. The Pareto survival at its threshold is $(1000/1000)^2=1$, contributing $0.50(1)=0.500000$. Hence,

$$ S_X(1000) =0.073576+0.150000+0.500000 =0.723576. $$

Mean. The component means are $\$1{,}000$, $\$1{,}000$, and $\$2{,}000$, respectively. Thus $E[X]=0.20(1000)+0.30(1000)+0.50(2000)=200+300+1000=\$1{,}500$.

Common mistake and correction. Treating the Pareto component survival at its threshold as $0.7$, rather than $1$, would give the incorrect calculation $0.20e^{-1}+0.30(0.5)+0.50(0.7)=0.5736$. The value $0.7$ is not the conditional Pareto survival; the component survival is $(\theta/\theta)^\alpha=1$. The corrected survival is $0.7236$ to four decimal places. The density is $0.001224$ per dollar to six decimal places, and the mean is $\$1{,}500.00$. The density and mean also check against their component contributions above.

Common Mixture Errors

Mixture calculations are short, but several errors can substantially change the result. Before finalizing a calculation, check that the component formulas correspond to the correct model and that the mixture weights have been applied to the component results, not to the argument $x$ or to distribution parameters.

Using weights that do not sum to one. Component weights are probabilities of the mutually exclusive component labels. Check their sum before calculating.
Averaging parameters instead of distributions. A weighted average of component means gives the mixture mean, but averaging rate parameters does not generally produce the mixture distribution’s density or survival function.
Forgetting support restrictions. A uniform density is zero outside its interval; a Pareto Type I component has no probability below its threshold.
Confusing component and mixture survival. A component survival probability is conditional on that component. The mixture survival requires weighting and adding all component survival probabilities.
Mixing units. If $x$ is measured in dollars, a density is expressed per dollar, while a survival probability has no units and a mean is reported in dollars.

Key takeaway: For a finite mixture, first calculate the requested quantity separately under every component model, then take the weighted sum using the component probabilities. This gives the mixture density, survival function, or mean, as long as the model assumptions and component supports are handled correctly.

Check Your Understanding

For each problem, use the stated component models and weights. Report densities per dollar, probabilities to four decimal places, and means to the cent.

A severity is exponential with mean $\$500$ with probability $0.25$, and exponential with mean $\$1{,}500$ with probability $0.75$. Calculate $S_X(1000)$, $f_X(1000)$, and $E[X]$. (Answer: $S_X(1000)=0.4189$; $f_X(1000)=0.000324$ per dollar; $E[X]=\$1{,}250.00$.)
A severity is uniform on $[0,\$3{,}000]$ with weight $0.40$, or Pareto Type I with $\theta=\$1{,}000$, $\alpha=3$ with weight $0.60$. Find $S_X(1500)$, $f_X(1500)$, and $E[X]$. (Answer: $S_X(1500)=0.3778$; $f_X(1500)=0.000489$ per dollar; $E[X]=\$1{,}500.00$.)
A severity is exponential with mean $\$2{,}000$ with weight $0.25$, uniform on $[0,\$4{,}000]$ with weight $0.25$, or Pareto Type I with $\theta=\$2{,}000$, $\alpha=2$ with weight $0.50$. Calculate $S_X(2000)$, $f_X(2000)$, and $E[X]$. (Answer: $S_X(2000)=0.7170$; $f_X(2000)=0.000608$ per dollar; $E[X]=\$3{,}000.00$.)
A mixture has only two components: an exponential severity with mean $\$800$, and a Pareto Type I severity with $\theta=\$1{,}000$, $\alpha=2$. The mixture mean is $\$1{,}400$. What is the weight on the exponential component? (Answer: $0.5000$.)

What You'll Learn