Mixed Severity Distributions

Why Mixed Severity Models Matter

A claim severity model need not be entirely discrete or entirely continuous. For example, a hypothetical warranty may have a positive chance of no payment, a separate chance of a fixed payment, and otherwise a payment that varies continuously. Such a model has both point masses and a continuous component. Calculations must account for both: a density describes the continuous part but does not describe probability concentrated at individual values.

The key method is to handle the components separately and then add their contributions. This applies when constructing the cumulative distribution function, finding probabilities, and calculating the mean severity.

Definition: A severity random variable $X$ has a mixed distribution if its probability is divided between one or more point masses and a continuous component. If the point masses are $p_j=P(X=a_j)$, and the continuous component has density $f_X(x)$, then $$ \sum_j p_j+\int_{-\infty}^{\infty} f_X(x)\,dx=1. $$ The density represents only the continuous component; its integral can be less than one. The remaining probability is carried by the point masses.

Building the CDF and Calculating Probabilities

At a point mass $a_j$, the CDF $F_X(x)=P(X\leq x)$ jumps by $p_j$. The continuous component contributes the integral of its density up to $x$. Thus, for any real $x$,

F_X(x)=\sum_{j:a_j\leq x}p_j+\int_{-\infty}^{x}f_X(t)\,dt.

The sum includes an atom at $x$ itself because the CDF uses $X\leq x$. A continuous component assigns probability zero to any single point, so whether an interval includes an endpoint affects its probability only when that endpoint is an atom. This is why interval notation and jumps must be checked together.

Worked Example: Finding CDF Values and Interval Probabilities

A hypothetical repair plan has claim payment $X$, in dollars, with probability $0.20$ of a zero payment and probability $0.30$ of a payment of exactly $500. The remaining probability $0.50$ is spread uniformly over the interval from $0 to $1,000. Find $F_X(500)$, $P(0<X\leq 500)$, $P(X>500)$, and $P(X=500)$.

Method. Write the atomic probabilities separately from the continuous density. The continuous component has total probability $0.50$, so its density on $[0,1000]$ is $0.50/1000=0.0005$ per dollar. Its probability over an interval is the interval length multiplied by this density.

At $x=500$, the CDF includes both atoms at or below 500: the mass $0.20$ at zero and the mass $0.30$ at 500. The continuous contribution from 0 through 500 is $500(0.0005)=0.25$. Therefore, $$ F_X(500)=0.20+0.30+0.25=0.75. $$

For $0<X\leq 500$, the atom at zero is excluded, but the atom at $500 is included. The continuous contribution is again $0.25$, so $$ P(0<X\leq 500)=0.30+0.25=0.55. $$

For $X>500$, there is no atom above $500. The continuous interval from just above 500 to 1,000 has length 500, giving $$ P(X>500)=500(0.0005)=0.25. $$ Finally, $P(X=500)=0.30$: the continuous part has no probability at one exact value. As a check, $P(X\leq 500)+P(X>500)=0.75+0.25=1$. The interval probability also agrees with the CDF difference $F_X(500)-F_X(0)=0.75-0.20=0.55$.

Common mistake avoided. Treating the density as if it described the whole distribution would omit the 0.30 mass at $500 and the 0.20 mass at zero. The density integrates to only 0.50, not 1.

Expected Severity from Both Components

For a mixed severity distribution with finite mean, the expected value is the sum of the point-mass contributions and the integral contribution from the continuous component:

E[X]=\sum_j a_jp_j+\int_{-\infty}^{\infty}x f_X(x)\,dx.

This formula follows by partitioning the distribution into its atomic outcomes and its continuous part. An outcome equal to $a_j$ contributes severity $a_j$ with probability $p_j$, hence $a_jp_j$. For the continuous part, severity near $x$ has probability approximately $f_X(x)\,dx$, so its contribution to the mean is $x f_X(x)\,dx$; adding those contributions over the support gives the integral. The total is the expected severity because these components account for all outcomes without overlap in probability.

Equivalently, if a model is specified as a mixture of component distributions, multiply each component's mean by its mixture probability and add. The continuous component's mean must be weighted by the probability assigned to that component; it is not automatically the mean of the full distribution.

Worked Example: A Weighted Continuous Component and a Fixed Payment

A hypothetical equipment policy has payment $X$, in dollars, with probability $0.25$ of no payment, probability $0.25$ of an $800 payment, and probability $0.50$ of a payment uniformly distributed from $200 to $1,000. Find $P(300<X\leq 800)$ and $E[X]$.

Method. The density for the continuous component is its probability, $0.50$, divided by its interval length, $1,000-200=800$. Thus the unconditional density is $0.50/800=1/1600$ on that interval. It integrates to $800/1600=0.50$, as required.

For $300<X\leq 800$, the continuous interval has length $800-300=500$. Its probability is $500/1600=0.3125$. The atom at $800 is included, adding 0.25; the atom at zero is outside the interval. Therefore, $$ P(300<X\leq 800)=0.3125+0.25=0.5625. $$

For the mean, the zero payment contributes $0(0.25)=0$, and the fixed payment contributes $800(0.25)=200$. The continuous component is uniform from $200 to $1,000, so its conditional mean is $(200+1000)/2=600$. Weighting that mean by the component probability gives $0.50(600)=300$. Hence $$ E[X]=0+200+300=\$500.00. $$

The probability result is consistent with the component totals: the continuous probability is 0.50, and the two atoms total 0.50. Common mistake avoided. Adding the continuous component's conditional mean of $600 without multiplying by 0.50 would incorrectly treat that component as having probability one.

When the Continuous Component Is Not Uniform

A mixed distribution does not require a uniform continuous part. Its density may vary over the severity range. The same procedure still applies: integrate the density over the requested interval, add any included atoms, and weight the continuous density by severity when calculating the continuous contribution to $E[X]$.

Worked Example: Integrating a Nonuniform Continuous Component

A hypothetical small-claim model has $P(X=0)=0.10$, $P(X=100)=0.20$, and continuous density $$ f_X(x)=\frac{7x}{200000},\qquad 0<x<200, $$ with density zero elsewhere. Find $P(X\leq 100)$, $P(X>100)$, and $E[X]$. All severities are in dollars.

Method. First verify that the continuous component has the stated remaining probability. Its total probability is $$ \int_0^{200}\frac{7x}{200000}\,dx =\frac{7}{200000}\left[\frac{x^2}{2}\right]_0^{200} =\frac{7(20000)}{200000}=0.70. $$ Together with the two masses, $0.10+0.20+0.70=1$.

For $X\leq 100$, include the atom at zero, the atom at 100, and the continuous probability from 0 to 100: $$ P(X\leq 100)=0.10+0.20+\int_0^{100}\frac{7x}{200000}\,dx =0.30+\frac{7(5000)}{200000} =0.475. $$

For $X>100$, the atom at 100 is excluded, and only the continuous part from 100 to 200 contributes: $$ P(X>100)=\int_{100}^{200}\frac{7x}{200000}\,dx =\frac{7}{200000}\left[\frac{x^2}{2}\right]_{100}^{200} =\frac{7(15000)}{200000}=0.525. $$ These probabilities sum to one. In particular, $F_X(100)=0.475$, while the left limit at 100 is $0.475-0.20=0.275$, confirming the jump of 0.20 at that point.

The atoms contribute $0(0.10)+100(0.20)=20$ dollars to the mean. The continuous contribution is $$ \int_0^{200}x\frac{7x}{200000}\,dx =\frac{7}{200000}\left[\frac{x^3}{3}\right]_0^{200} =\frac{7(8000000)}{600000} =\frac{280}{3}. $$ Thus $E[X]=20+280/3=113.333\ldots$, which rounds to $113.33 to the nearest cent. The result is plausible: the mean exceeds $100 because much of the continuous probability lies above $100, but it is below the maximum possible severity of $200.

Common mistake avoided. Integrating $f_X$ gives probabilities, not expected severity. For the continuous contribution to the mean, the integrand is $x f_X(x)$, so each severity is weighted by its size.

Endpoint and Normalization Checks

A reliable solution to a mixed-model problem includes checks on both probability and arithmetic. The following process helps prevent missed jumps and incorrect component weights.

List the atoms.
Record every value $a_j$ with its mass $p_j$, including a possible mass at zero.

Identify the density component.
Check its support and integrate it over that support. The integral must equal the probability not assigned to atoms.

Handle interval endpoints.
Include or exclude an atom according to the inequality in the event. A continuous component assigns zero probability to a single endpoint.

Add component contributions.
For probabilities, add included masses and density integrals. For the mean, add $a_jp_j$ and the integral of $x f_X(x)$.

A frequent exam trap is to infer that a density must integrate to one. That is true for a purely continuous distribution, but not for a mixed one. Another is to use $F_X(b)-F_X(a)$ for an interval without checking which endpoints are included: $F_X(b)-F_X(a)=P(a<X\leq b)$, so it excludes an atom at $a$. If the event is $a\leq X\leq b$, the mass at $a$ must be added.

Key takeaway: A mixed severity model is a complete distribution only when its point masses and continuous density together carry probability one. Keep the two components separate in the calculation, then combine them with the event's endpoint rules or the expected-value weights.

Check Your Understanding

For each problem, separate the atoms from the continuous component before calculating.

A severity has mass 0.15 at zero, mass 0.25 at $400, and remaining probability uniformly distributed from $0 to $1,000. Find $P(X\leq 400)$ and $E[X]$. (Answer: $P(X\leq 400)=0.64$; $E[X]=\$400.00$)
A severity has mass 0.10 at zero, mass 0.20 at $600, and probability 0.70 uniformly distributed from $200 to $1,000. Find $P(300<X\leq 600)$ and $E[X]$. (Answer: $0.4625$; $E[X]=\$540.00$)
A severity has mass 0.10 at zero, mass 0.20 at $200, and density $f_X(x)=7x/450000$ for $0<x<300$. Find $P(X\leq 150)$ and $E[X]$. (Answer: $0.2750$; $E[X]=\$180.00$)
A severity has mass 0.20 at $100, mass 0.10 at $500, and probability 0.70 uniformly distributed from $0 to $700. Find $P(100\leq X<500)$ and $E[X]$. (Answer: $0.6000$; $E[X]=\$315.00$)

What You'll Learn