Quantiles Of Finite Mixture Severities

Why Mixture Quantiles Need the Mixture Distribution

A severity model may represent losses from several underlying groups, such as different hypothetical types of insured risks. A percentile of the overall portfolio is a percentile of the combined loss distribution, not an average of the percentiles calculated separately for each group. The distinction matters because a quantile is obtained by inverting a cumulative distribution function, and inversion does not generally preserve weighted averages.

The earlier material on finite mixture severity models represents the group membership by a component label. Let $Z$ denote that label, with $P(Z=j)=\pi_j$, where each $\pi_j$ is nonnegative and $\sum_{j=1}^m \pi_j=1$. Conditional on $Z=j$, let the loss $X$ have CDF $F_j(x)=P(X\le x\mid Z=j)$. There is no interest or mortality assumption in these severity-quantile calculations: the component distributions and their parameters specify the model.

Definition: The CDF of a finite mixture severity is $$ F_X(x)=P(X\le x)=\sum_{j=1}^m \pi_j F_j(x). $$ For $0<p<1$, the $p$-quantile is the generalized inverse $$ Q_X(p)=\inf\{x:F_X(x)\ge p\}. $$ If $F_X$ is continuous and strictly increasing around the quantile, $Q_X(p)$ is the unique solution to $F_X(x)=p$. If the CDF has a jump, use the infimum definition; the quantile can be a point at which $F_X(x)>p$.

The weights in the mixture are probabilities, not quantities to apply after finding component quantiles. For a two-component model, for instance, the equation to solve is $\pi_1F_1(x)+\pi_2F_2(x)=p$. Each component CDF must be evaluated at the same candidate loss value $x$. Depending on the component supports, some terms may be zero or one in a particular interval.

Why the Weighted CDF Equation Holds

The mixture equation follows by conditioning on the component label. The events $Z=1,\ldots,Z=m$ are disjoint and cover all possibilities. Therefore, for any loss value $x$,

\begin{aligned} F_X(x) &=P(X\le x)\\ &=\sum_{j=1}^m P(X\le x,\ Z=j)\\ &=\sum_{j=1}^m P(X\le x\mid Z=j)P(Z=j)\\ &=\sum_{j=1}^m \pi_j F_j(x). \end{aligned}

This identity gives the calculation method: construct the weighted CDF at a candidate $x$, then find the smallest $x$ at which the CDF reaches the target probability. When the CDF is continuous and strictly increasing over the relevant interval, solving equality is sufficient. When a component support begins or ends, first identify the interval containing the quantile; the formula for the weighted CDF can change at those boundaries.

Write the mixture CDF.
List the component weights and the component CDFs, then form $F_X(x)=\sum_j\pi_jF_j(x)$.

Locate the relevant interval.
Check component support boundaries and evaluate the mixture CDF there. This identifies which component CDF formulas apply.

Solve and verify.
Set the applicable weighted CDF equal to $p$, solve for $x$, and check that the result lies in the interval used and that the CDF reaches $p$.

Worked Quantiles for Continuous Mixtures

Worked Example: A Quantile Beyond the First Component’s Support

A hypothetical portfolio has severity $X$ drawn from a two-component mixture. With probability $0.70$, $X$ is Uniform on $[0,100]$ dollars; with probability $0.30$, it is Uniform on $[100,300]$ dollars. Find the 80th percentile. There is no interest or mortality assumption. Use the stated uniform distributions and weights, and round the loss to the nearest cent.

Method. Solve the weighted CDF equation after identifying which support interval contains the 80th percentile. The common mistake to avoid is to average the two component 80th percentiles.

For $100\le x\le300$, the first component CDF is already one, while the second component CDF is $(x-100)/200$. Thus

F_X(x)=0.70+0.30\left(\frac{x-100}{200}\right).

At $x=100$, this CDF is $0.70$. Since the target $0.80$ is greater than $0.70$, the quantile is beyond 100 dollars, so the interval formula is appropriate. Set the CDF equal to 0.80:

\begin{aligned} 0.70+0.30\left(\frac{x-100}{200}\right)&=0.80,\\ 0.30\left(\frac{x-100}{200}\right)&=0.10,\\ \frac{x-100}{200}&=\frac{1}{3},\\ x&=166.666\ldots. \end{aligned}

The 80th percentile is $166.67. Check: at this value, the first component contributes $0.70(1)=0.70$ and the second contributes $0.30(1/3)=0.10$, giving $F_X(x)=0.80$.

The component 80th percentiles are $80$ and $260$ dollars. Their weighted average is $0.70(80)+0.30(260)=56+78=134$ dollars, not the mixture quantile. In fact, $F_X(134)=0.70+0.30(34/200)=0.751$, which is below 0.80.

Worked Example: An Overlapping-Support Mixture

A second hypothetical portfolio has $X\sim\text{Uniform}(0,200)$ dollars with probability $0.60$ and $X\sim\text{Uniform}(100,300)$ dollars with probability $0.40$. Find the 90th percentile. There is no interest or mortality assumption. Round the final loss to the nearest cent.

Method. The target is above the CDF at 200 dollars, so use the interval $200\le x\le300$. The common mistake is to average component percentiles even though the combined CDF must reach 0.90.

At $x=200$, the first component CDF is one. For $200\le x\le300$, the second component CDF is $(x-100)/200$. Therefore,

F_X(x)=0.60+0.40\left(\frac{x-100}{200}\right).

The CDF at 200 is $0.60+0.40(100/200)=0.80$, confirming that the 90th percentile is above 200. Solve:

\begin{aligned} 0.60+0.40\left(\frac{x-100}{200}\right)&=0.90,\\ 0.40\left(\frac{x-100}{200}\right)&=0.30,\\ \frac{x-100}{200}&=0.75,\\ x&=250. \end{aligned}

The 90th percentile is $250.00. Verification gives $F_X(250)=0.60+0.40(150/200)=0.60+0.30=0.90$.

The component 90th percentiles are $180$ and $280$ dollars. Their weighted average is $0.60(180)+0.40(280)=108+112=220$ dollars. But $F_X(220)=0.60+0.40(120/200)=0.84$, not 0.90. This directly checks that the average is not the requested percentile.

Mixtures with an Atom

A mixture need not have a continuous CDF. For example, one component can put all its probability at a fixed loss amount, producing an atom in the overall distribution. The generalized inverse definition handles this without change. At an atom, it is possible for $F_X(Q_X(p))$ to be strictly greater than $p$; the quantile is still the smallest value whose CDF is at least $p$.

Worked Example: A Point-Mass Component

A hypothetical severity model selects a fixed loss of $50 with probability $0.25$. With probability $0.75$, it selects a Uniform distribution on $[100,200]$ dollars. Find the 30th percentile and compare it with the weighted average of the component 30th percentiles. There is no interest or mortality assumption. Round losses to the nearest cent.

Method. Use the weighted CDF and the generalized inverse definition. The common mistake is to assume that a mixture with an atom must satisfy $F_X(x)=p$ exactly at its quantile.

The CDF is zero below 50 dollars, equals 0.25 from 50 up to 100 dollars, and for $100\le x\le200$ is

F_X(x)=0.25+0.75\left(\frac{x-100}{100}\right).

The CDF is only 0.25 below 100 dollars, so the 30th percentile lies in $[100,200]$. Solve:

\begin{aligned} 0.25+0.75\left(\frac{x-100}{100}\right)&=0.30,\\ 0.75\left(\frac{x-100}{100}\right)&=0.05,\\ x-100&=100\left(\frac{0.05}{0.75}\right)=6.666\ldots,\\ x&=106.666\ldots. \end{aligned}

The 30th percentile is $106.67. The mixture CDF there is 0.30, subject to rounding. The component quantiles are $50 and $100+0.30(100)=$130. Their weighted average is \(0.25(50)+0.75(130)=12.50+97.50=$110.00$. At $110, the mixture CDF is $0.25+0.75(10/100)=0.325$, so $110 is not the 30th percentile.

Why Component Quantiles Cannot Usually Be Averaged

If $Q_j(p)$ denotes the $p$-quantile of component $j$, then $F_j(Q_j(p))$ is usually $p$ for continuous strictly increasing component CDFs. This fact does not imply that $F_X(\sum_j\pi_jQ_j(p))=p$. The mixture CDF at one loss $x$ averages the component CDF values at that same $x$, whereas the weighted average of component quantiles combines different loss values before evaluating any CDF.

A weighted average of component quantiles can coincide with a mixture quantile in special cases, but it is not a general quantile formula. It is particularly unreliable when supports differ, when the mixture quantile lies in a region where some components have already reached CDF one, or when the mixture has an atom. Always return to the weighted CDF equation and verify the proposed value against the target probability.

Common exam trap: Do not calculate $\sum_j\pi_jQ_j(p)$ and label it the mixture $p$-quantile. For example, in the overlapping-uniform example, that calculation gives $0.60(180)+0.40(280)=220$, but the actual mixture CDF at 220 is 0.84, not 0.90. The correction is to calculate $\sum_j\pi_jF_j(x)$ at a common $x$, set it equal to $p$ where appropriate, and check the result.

Check Your Understanding

For each hypothetical severity mixture, find the requested quantile by using the weighted CDF. Round dollar answers to the nearest cent where needed.

A loss is Uniform on $[0,100]$ dollars with probability $0.40$, and Uniform on $[0,200]$ dollars with probability $0.60$. Find the median. (Answer: $71.43)
A loss is Uniform on $[0,80]$ dollars with probability $0.25$, and Uniform on $[80,200]$ dollars with probability $0.75$. Find the 50th percentile. (Answer: $120.00)
A loss is Uniform on $[0,100]$ dollars with probability $0.70$, and Uniform on $[100,400]$ dollars with probability $0.30$. Find the 90th percentile. (Answer: $300.00)
A loss is $40 with probability $0.20$, and otherwise is Uniform on $[100,250]$ dollars. Find the 30th percentile. (Answer: $118.75)
A loss is Uniform on $[0,200]$ dollars with probability $0.60$, and Uniform on $[100,300]$ dollars with probability $0.40$. Find the 80th percentile and state whether it is in the first component’s support endpoint or beyond it. (Answer: $200.00; it is exactly at the first component’s upper endpoint.)

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