Expected Severity For Discrete And Continuous Losses

Why Expected Severity Is a Weighted Average

A loss model describes uncertainty about the amount of a single loss $X$. Its expected severity, or mean loss, summarizes the model with one value in the same units as the loss. The calculation depends on how probability is represented: a discrete model assigns probability to individual loss amounts, while a continuous model describes probability across intervals using a density.

Expected severity is not necessarily an amount that will actually occur. It is a probability-weighted average across possible outcomes. In a portfolio calculation, it can be useful as a measure of average loss per claim under the specified model. Its interpretation depends on the model being appropriate and on the mean being finite.

Definition: If $X$ is a discrete loss with possible values $x_i$ and probabilities $p_i=P(X=x_i)$, its expected severity is $E[X]=\sum_i x_i p_i$, provided the sum is finite. If $X$ is continuous with probability density $f_X(x)$, its expected severity is $E[X]=\int_{-\infty}^{\infty}x f_X(x)\,dx$, provided the integral exists. For a nonnegative loss, the mean may be infinite; a finite expected severity requires the corresponding sum or integral to be finite.

For severity models, $X$ is usually nonnegative, so outcomes below zero have probability zero and can be omitted from the calculation. A discrete model may assign positive probability to a zero loss, such as a reported event that produces no payment. That probability still matters when checking the model totals, even though its contribution to the mean is $0$ times its probability.

The word probability is important in both formulas. In a discrete model, each $p_i$ is a probability. In a continuous model, $f_X(x)$ is a density, not the probability of the exact loss $x$. Probabilities over intervals come from integrating the density over those intervals. The density itself may be greater than 1, provided it is nonnegative and integrates to 1.

Discrete Losses: Sum Loss Amount Times Probability

For a discrete loss, list the possible values and their probabilities. Multiply each loss amount by its probability, then add the products. Before calculating, check that the probabilities are nonnegative and sum to 1. If one probability is missing, find it from this total before calculating the expected severity.

E[X]=\sum_i x_i P(X=x_i)

Worked Example: Finding a Discrete Mean With a Missing Probability

A hypothetical equipment policy has a gross loss $X$, measured in thousands of dollars, with possible values $0$, $0.4$, $1.2$, and $2.5$. The corresponding probabilities for the first three values are $0.12$, $0.38$, and $0.30$. Find the missing probability and the expected gross loss.

Method and model check. This is a discrete loss model, so use a probability-weighted sum. The probabilities must add to 1. The stated probabilities sum to $0.12+0.38+0.30=0.80$. Therefore, the probability of a loss of $2.5$ thousand dollars is $1-0.80=0.20$. All four probabilities are nonnegative and now sum to 1.

Calculate the mean. Multiply each loss, in thousands of dollars, by its probability: $0(0.12)=0$, $0.4(0.38)=0.152$, $1.2(0.30)=0.360$, and $2.5(0.20)=0.500$. Add the contributions: $0+0.152+0.360+0.500=1.012$ thousand dollars.

Thus, the expected gross loss is $1.012$ thousand dollars, or $1,012.00. No rounding is needed. As a check, the result lies between the smallest and largest possible losses, $0$ and $2.5$ thousand dollars.

Common mistake avoided: Treating the missing probability as zero would leave the listed probabilities summing to $0.80$, which is not a complete probability model. The missing probability must be recovered before interpreting the expected value.

Continuous Losses: Integrate Loss Times Density

For a continuous loss, multiply the loss value $x$ by its density $f_X(x)$, then integrate across the support of the loss. The support is the set of values where the density can be positive. A valid density must be nonnegative and have total integral 1. If a proposed density includes an unknown constant, determine that constant from the total-probability condition before using the mean formula.

\int_{-\infty}^{\infty} f_X(x)\,dx=1, \qquad E[X]=\int_{-\infty}^{\infty}x f_X(x)\,dx

The reason the formula uses $x f_X(x)$ is that probability is spread over intervals rather than assigned to individual points. On a short interval of width $\Delta x$ near $x$, the probability is approximately $f_X(x)\Delta x$. The contribution of losses in that interval to the average is approximately $x f_X(x)\Delta x$. Adding these contributions across the support gives an approximation to the mean. Taking the interval widths toward zero turns the sum into the integral.

\sum_j x_j P(X\text{ is in interval }j) \ \approx\ \sum_j x_j f_X(x_j)\Delta x \ \longrightarrow\ \int x f_X(x)\,dx

This is the continuous counterpart of the discrete weighted sum: probability is still the weight, but the density must be integrated over ranges of loss. In particular, substituting a density value directly as though it were a point probability does not calculate an expected loss.

Worked Example: Normalizing and Integrating a Piecewise Density

A hypothetical repair loss $X$, in thousands of dollars, has density $f_X(x)=c$ for $0\leq x<2$, $f_X(x)=2c$ for $2\leq x\leq3$, and $0$ otherwise. Find $c$ and $E[X]$.

Method and model check. First require the density to integrate to 1. The first interval has width 2 and density $c$, so its probability is $2c$. The second interval has width 1 and density $2c$, so its probability is also $2c$. Thus $2c+2c=1$, giving $c=1/4$. The density is nonnegative and has total probability $4(1/4)=1$.

Calculate the mean. Integrate $x$ times the density over each interval:

E[X] =\int_0^2 x\left(\frac14\right)\,dx +\int_2^3 x\left(\frac12\right)\,dx =\frac14\left[\frac{x^2}{2}\right]_0^2 +\frac12\left[\frac{x^2}{2}\right]_2^3

The first contribution is $\frac14(2-0)=0.50$. The second is $\frac12(\frac92-2)=\frac12(\frac52)=1.25$. Therefore, $E[X]=0.50+1.25=1.75$ thousand dollars, or $1,750.00. No rounding is needed. This is between the support endpoints, $0$ and $3$ thousand dollars, as expected.

Common mistake avoided: Integrating $x$ against the unnormalized expression without first finding $c$ would give a quantity that is not the mean of a probability distribution. The density must first have total integral 1.

Combining an Atom and a Continuous Component

Some loss distributions have both a point mass and a continuous part. For example, a loss may be exactly zero with positive probability, while positive losses vary continuously. In that case, the expected value includes both contributions: sum each atom’s value times its probability, and integrate $x$ against the density of the continuous part. The continuous density integrates to the probability assigned to the continuous part, which may be less than 1.

Result: If a nonnegative loss has atoms at values $a_j$, with probabilities $p_j$, and a continuous component with density $g(x)$, then $E[X]=\sum_j a_jp_j+\int xg(x)\,dx$, provided the contributions are finite. Here $g$ represents only the continuous component; its integral is the probability of being in that component, not necessarily 1.

Worked Example: Mean Loss With a Zero-Loss Atom

A hypothetical service claim has a $0.10$ probability of producing no loss. Otherwise, its loss $X$, measured in thousands of dollars, is described by a continuous component with density $g(x)=0.30$ for $0<x<3$ and $0$ elsewhere. Find the expected loss.

Check the model. The atom at zero has probability $0.10$. The continuous component has total probability $\int_0^3 0.30\,dx=0.30(3)=0.90$. The two components therefore have total probability $0.10+0.90=1$.

Calculate both contributions. The atom contributes $0(0.10)=0$. The continuous part contributes

\int_0^3 x(0.30)\,dx =0.30\left[\frac{x^2}{2}\right]_0^3 =0.30\left(\frac92\right) =1.35

The expected loss is $1.35$ thousand dollars, or $1,350.00. No rounding is needed. As a check, the continuous component has probability $0.90$ and, conditional on being in that component, the density is $0.30/0.90=1/3$ on $(0,3)$, whose mean is $1.5$ thousand dollars. The overall mean is $0.90(1.5)+0.10(0)=1.35$ thousand dollars, agreeing with the direct calculation.

Common mistake avoided: The continuous density integrates to $0.90$, not 1. Renormalizing it to integrate to 1 and then treating that conditional mean as the overall mean would omit the probability of having no loss.

Checks and Exam Traps

Expected severity is a straightforward calculation once the probability model is identified. A few checks help prevent errors that can otherwise produce plausible-looking answers.

Check the total probability. Discrete probabilities must sum to 1. A density for a purely continuous model must integrate to 1. In a mixed model, the atom probabilities and the continuous probability must sum to 1 together.
Do not confuse a density with a probability. For a continuous model, $P(X=x)=0$ at any single point. Probabilities are obtained by integrating the density over intervals.
Include every outcome appropriately. A zero loss contributes zero to the mean but still counts in checking the probability total.
Keep the units consistent. If losses are listed in thousands of dollars, the calculated mean is also in thousands of dollars. Convert to dollars only at the end.
Check the result against the support. For a loss confined between finite endpoints, its finite mean must also lie between those endpoints.

The discrete sum and continuous integral are two forms of the same principle: weight each possible loss by how much probability it carries. This skill is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus and is a foundation for later calculations involving more detailed features of severity distributions.

Check Your Understanding

For each problem, identify whether the loss is discrete, continuous, or mixed before calculating.

A discrete loss in thousands of dollars takes values $0$, $2$, and $5$, with probabilities $0.25$, $0.50$, and $0.25$. Find its expected loss. (Answer: $2.25$ thousand dollars.)
A discrete loss takes values $100$, $600$, and $1{,}000$ dollars with probabilities $0.20$, $0.50$, and an unspecified remaining probability. Find the missing probability and expected loss. (Answer: missing probability $0.30$; expected loss $$620.00$.)
A continuous loss in thousands of dollars has density $f_X(x)=3x^2/8$ for $0\leq x\leq2$, and zero otherwise. Verify the density integrates to 1 and find the mean. (Answer: total probability $1$; mean $1.5$ thousand dollars.)
A continuous loss has density $c$ on $0\leq x<1$, density $3c$ on $1\leq x\leq2$, and zero otherwise. Find $c$ and the expected loss. (Answer: $c=0.25$; expected loss $1.25$.)
A mixed loss in thousands of dollars has probability $0.10$ at zero and continuous-component density $0.30$ on $0<x<3$, with no other probability. Find the total continuous probability and the overall mean. (Answer: continuous probability $0.90$; mean $1.35$ thousand dollars.)

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