A Worked Severity Distribution And Moments Problem

A Systematic Approach to a Severity Problem

A complete severity calculation involves more than finding a mean. An actuary may need to identify which loss values the model allows, compute probabilities over intervals or beyond thresholds, find a percentile, and summarize the distribution with moments. Those tasks depend on consistent definitions: endpoint conventions matter for discrete losses, while units matter whenever moments are converted into dollars.

The examples use $X$ for the individual loss amount. A model may be continuous, discrete, or a mixture. The CDF and survival function apply in all three cases, but integration is used for continuous densities and summation for discrete probabilities.

Definition: The support of a distribution is the set of points for which every open neighborhood has positive probability. Equivalently, it is the smallest closed set with probability 1. A support endpoint need not itself have positive probability. The CDF is $F_X(x)=P(X\le x)$, and the survival function is $S_X(x)=P(X>x)=1-F_X(x)$. For a distribution with an atom at $x$, $P(X\ge x)=1-F_X(x-)$, where $F_X(x-)$ is the limit of the CDF from the left.

For a continuous distribution with density $f_X$, $F_X(x)$ is the integral of the density up to $x$. For a discrete distribution, $F_X(x)$ is the sum of the probabilities at values no greater than $x$. A continuous distribution has no probability at an individual point, so strict and non-strict inequalities at a fixed endpoint give the same probability. That is not generally true for a discrete distribution.

Definition: For $0<p<1$, the $p$-quantile is $Q_X(p)=\inf\{x:F_X(x)\ge p\}$. For a continuous, strictly increasing CDF, it is the value solving $F_X(x)=p$. For a discrete distribution, it is the first value at which the CDF reaches or exceeds $p$.

From a Density to Probabilities and Moments

Suppose $X$ is nonnegative and has density $f_X$. The probability assigned to an interval is the area under the density over that interval:

P(a<X\le b)=\int_a^b f_X(x)\,dx.

Taking $a$ below the entire support gives $F_X(x)=\int_{-\infty}^{x} f_X(t)\,dt$. The $r$th raw moment follows from the expectation of the function $X^r$: $E[X^r]=\int_0^\infty x^r f_X(x)\,dx$, when the integral exists. In a discrete model with probabilities $p_j=P(X=x_j)$, the corresponding formula is $E[X^r]=\sum_j x_j^r p_j$.

The variance is the second central moment. Expanding its definition gives a useful calculation formula:

\operatorname{Var}(X)=E[(X-E[X])^2] =E[X^2]-2E[X]E[X]+(E[X])^2 =E[X^2]-(E[X])^2.

Thus, a reliable solution computes the first and second raw moments before subtracting the squared mean. If $X$ is measured in thousands of dollars, $E[X]$ is in thousands of dollars and $\operatorname{Var}(X)$ is in squared thousands of dollars.

Worked Example: A Triangular Density and Its Endpoints

A hypothetical insurer models an individual repair loss $X$, measured in thousands of dollars, with density $f_X(x)=x/8$ for $0\le x\le4$, and density zero otherwise. Find the support, CDF, $P(1<X\le3)$, $S_X(3)$, the 90th percentile, the mean, and the variance. State the moment units.

Method and model check. This is an absolutely continuous model. Its total probability is $\int_0^4 x/8\,dx=[x^2/16]_0^4=1$, so the proposed density is normalized.

Support. The density is positive for $0<x\le4$, and every neighborhood of 0 also contains an interval immediately to its right with positive probability. Every neighborhood of 4 contains an interval immediately to its left with positive probability. The support is therefore the closed interval $[0,4]$, even though $P(X=0)=P(X=4)=0$.

CDF and interval probability. Integrating the density, with the lower and upper regions included, gives $F_X(x)=0$ for $x<0$, $F_X(x)=x^2/16$ for $0\le x\le4$, and $F_X(x)=1$ for $x>4$. Therefore, $P(1<X\le3)=F_X(3)-F_X(1)=9/16-1/16=8/16=0.5000$.

Survival probability. $S_X(3)=1-F_X(3)=1-9/16=7/16=0.4375$. Because the model is continuous, using $X\ge3$ instead of $X>3$ would not change this probability.

90th percentile. On the support, solve $x^2/16=0.90$. Since $x\ge0$, $Q_X(0.90)=4\sqrt{0.90}=3.794733$ thousand dollars. This is $\$3,794.73$, rounded to the nearest cent.

Moments. The first raw moment is $E[X]=\int_0^4 x(x/8)\,dx=[x^3/24]_0^4=64/24=8/3$ thousand dollars. The second raw moment is $E[X^2]=\int_0^4 x^2(x/8)\,dx=[x^4/32]_0^4=256/32=8$ squared thousands of dollars. Hence $\operatorname{Var}(X)=8-(8/3)^2=8-64/9=8/9$ squared thousands of dollars.

Check. The mean $8/3$ lies within the support, and the variance is positive. In dollar units, the mean is $\$2,666.67$, and the variance is $(8/9)(1,000)^2=\$888,888.89$ squared dollars. The probability and survival results sum to $0.5000+0.4375$ only for their stated, different events; they are not complementary events.

A Piecewise Density: Respecting the CDF at a Boundary

A density can change abruptly without creating a probability atom. The CDF is still obtained by integrating all density accumulated to the left of the requested point. At a boundary between density regions, the accumulated probability from the earlier region must be carried into the next expression.

Worked Example: A Piecewise-Uniform Severity Model

For a hypothetical equipment claim, let $X$ be the loss in thousands of dollars, with density $1/6$ on $0\le x\le2$, density $1/3$ on $2<x\le4$, and zero otherwise. Find the support, the CDF and survival function, $P(X>3)$, the 20th and 80th percentiles, the mean, and the variance.

Method and model check. Verify the total mass first: $\int_0^2(1/6)\,dx+\int_2^4(1/3)\,dx=2/6+2/3=1$. The support is $[0,4]$: every neighborhood of either endpoint has positive probability, even though neither endpoint has positive probability by itself.

CDF. For $0\le x\le2$, integrate the first density to obtain $F_X(x)=x/6$. For $2<x\le4$, include the probability $1/3$ accumulated by $x=2$, then integrate the second density: $F_X(x)=1/3+(x-2)/3=x/3-1/3$. Including the exterior regions, $F_X(x)=0$ for $x<0$, $x/6$ for $0\le x\le2$, $x/3-1/3$ for $2<x\le4$, and $1$ for $x>4$. At the join, the second expression gives $2/3-1/3=1/3$, matching the first expression at $x=2$.

Survival probability. At $x=3$, $F_X(3)=3/3-1/3=2/3$, so $P(X>3)=S_X(3)=1-2/3=1/3=0.3333$.

Quantiles. For $p=0.20$, the quantile is in the first region because $0.20<1/3$. Solving $x/6=0.20$ gives $Q_X(0.20)=1.2$ thousand dollars. For $p=0.80$, use the second region: $x/3-1/3=0.80$, so $x/3=1.133333$ and $Q_X(0.80)=3.4$ thousand dollars. Each answer lies in the region whose CDF expression was used.

Moments. Split each integral at 2: $E[X]=(1/6)\int_0^2x\,dx+(1/3)\int_2^4x\,dx=(1/6)(2)+(1/3)(6)=1/3+2=7/3$ thousand dollars. For the second raw moment, $E[X^2]=(1/6)\int_0^2x^2\,dx+(1/3)\int_2^4x^2\,dx=(1/6)(8/3)+(1/3)(56/3)=4/9+56/9=20/3$ squared thousands of dollars. Therefore, $\operatorname{Var}(X)=20/3-(7/3)^2=60/9-49/9=11/9$ squared thousands of dollars.

Check. The mean is $7/3$, inside $[0,4]$, and the variance $11/9$ is positive. In dollars, the mean is $\$2,333.33$ and the variance is $\$1,222,222.22$ squared dollars. The CDF equals 1 at the upper endpoint, as required.

Discrete Severity: Quantiles and Endpoint Conventions

For discrete losses, probability is concentrated at individual values, so the CDF has jumps. The generalized-inverse definition of a quantile is essential: a percentile may be a loss value at which the CDF jumps past the requested probability. Also distinguish $P(X>x)$ from $P(X\ge x)$ whenever $X$ can equal the threshold.

Worked Example: A Three-Point Loss Distribution

A hypothetical portable-device policy has individual claim severity $X$, in dollars, equal to $\$100$, $\$300$, or $\$700$, with respective probabilities $0.20$, $0.50$, and $0.30$. Find the support, CDF, $P(X>300)$, $P(X\ge300)$, the 75th percentile, and the mean and variance.

Model check and support. The probabilities sum to $0.20+0.50+0.30=1$. Each of the three values has positive probability, and any point outside these values has a sufficiently small neighborhood with probability zero. The support is $\{100,300,700\}$.

CDF. Summing probability at values no greater than $x$, $F_X(x)=0$ for $x<100$, $0.20$ for $100\le x<300$, $0.70$ for $300\le x<700$, and $1$ for $x\ge700$.

Threshold probabilities. $P(X>300)=P(X=700)=0.30$. In contrast, $P(X\ge300)=P(X=300)+P(X=700)=0.50+0.30=0.80$. The difference, $0.50$, is the probability of an exact $\$300$ loss.

75th percentile. The CDF is $0.70$ at $\$300$, which is below $0.75$, and it reaches 1 at $\$700$. Thus the first value where $F_X(x)\ge0.75$ is $\$700$, so $Q_X(0.75)=\$700$.

Moments. $E[X]=100(0.20)+300(0.50)+700(0.30)=20+150+210=\$380$. The second raw moment is $E[X^2]=100^2(0.20)+300^2(0.50)+700^2(0.30)=2,000+45,000+147,000=\$194,000$ squared dollars. Therefore, $\operatorname{Var}(X)=194,000-380^2=194,000-144,400=\$49,600$ squared dollars.

Check. The mean $\$380$ lies between the smallest and largest possible losses, and the variance is positive. The CDF jump at $\$300$ is $0.70-0.20=0.50$, equal to $P(X=300)$.

Common Mistakes and a Reliable Check

A frequent error is to define support as the set of points that individually have positive probability or positive density. That rule fails for continuous distributions: every exact point has probability zero, and the value of a density at a single point does not determine support. The neighborhood definition avoids both problems.

For the first example, the incorrect argument would be: “$f_X(0)=0$ and $P(X=0)=0$, so 0 is not in the support.” The correction is to examine neighborhoods. For any $0<\varepsilon\le4$, $P(0\le X<\varepsilon)=\int_0^\varepsilon x/8\,dx=\varepsilon^2/16>0$. Similarly, $P(4-\varepsilon<X\le4)=1-(4-\varepsilon)^2/16=\varepsilon/2-\varepsilon^2/16>0$. Every neighborhood of each endpoint therefore has positive probability, so both endpoints belong to the support.

Another common error is to use a survival formula without checking the event definition. By convention here $S_X(x)=P(X>x)$; with an atom at the threshold, this differs from $P(X\ge x)$. Finally, moments inherit powers of the measurement unit: if losses are recorded in thousands of dollars, a variance calculated from those values is in squared thousands, not dollars. Multiplying the variance by $1,000$ rather than $1,000^2$ is a unit-conversion error.

Check the model.
Verify that probabilities sum or that the density integrates to 1, and record the loss unit.

Identify the support.
Use the values with positive probability or, for continuous models, the neighborhood definition; do not discard endpoints just because they have zero point probability.

Build the distribution functions.
Accumulate density or probability for the CDF, then use $S_X(x)=1-F_X(x)$ for the strict upper tail.

Calculate quantiles and moments.
Use the generalized inverse for quantiles and integrate or sum for raw moments; check the CDF region and units before reporting.

The practical value of this sequence is that it separates model interpretation from arithmetic. A correct numerical result still needs the correct event, quantile convention, support, and unit to answer an actuarial severity question.

Check Your Understanding

For each problem, identify the relevant convention before calculating. Probabilities are reported to four decimals; money is reported to the nearest cent where applicable.

A continuous loss in thousands of dollars has density $f_X(x)=1/5$ for $0\le x\le5$, and zero otherwise. Find the support, $P(X>4)$, the median, the mean, and the variance. (Answer: support $[0,5]$; $P(X>4)=0.2000$; median $2.5$; mean $2.5$ thousand dollars; variance $25/12$ squared thousands of dollars.)
A discrete severity is $\$50$ with probability $0.25$, $\$150$ with probability $0.50$, and $\$350$ with probability $0.25$. Find the mean, variance, 60th percentile, and $P(X\ge150)$. (Answer: mean $\$175$; variance $\$11,875$ squared dollars; 60th percentile $\$150$; $P(X\ge150)=0.7500$.)
A loss $X$, in thousands of dollars, has density $f_X(x)=x/18$ on $0\le x\le6$, and zero otherwise. Find $F_X(x)$ on the support, $P(2<X\le4)$, the 25th percentile, the mean, and the variance. (Answer: $F_X(x)=x^2/36$ on $[0,6]$; probability $1/3=0.3333$; 25th percentile $3$ thousand dollars; mean $4$ thousand dollars; variance $2$ squared thousands of dollars.)
A discrete loss in dollars is $\$200$, $\$500$, or $\$900$, with probabilities $0.10$, $0.60$, and $0.30$. Find $P(X>500)$, $P(X\ge500)$, and the 80th percentile. (Answer: $P(X>500)=0.3000$; $P(X\ge500)=0.9000$; 80th percentile $\$900$.)
A continuous severity has support $[2,8]$ in hundreds of dollars and a CDF that is continuous and strictly increasing over that interval. Explain why its support includes 2 and 8 even if neither exact endpoint has positive probability, and state the equation used to find its 40th percentile. (Answer: every neighborhood of either endpoint intersects the support in a set with positive probability; solve $F_X(x)=0.40$ for $x\in[2,8]$.)

What You'll Learn