Variance and Standard Deviation of Claim Severity

Why Measure Severity Dispersion?

The expected value of a claim severity describes its average size, but it does not describe how far individual losses tend to vary around that average. Two hypothetical portfolios can have the same expected claim amount while one has nearly identical claims and the other has widely varying claims. Variance and standard deviation quantify this dispersion. They are useful only when interpreted alongside the mean: a large standard deviation indicates substantial spread, not necessarily a large expected loss.

Let $X$ denote a claim severity, measured on a specified basis and in a stated unit, such as dollars per claim. The severity may be discrete, continuous, or mixed. The formulas below require a finite second moment, $E[X^2] < \infty$. If this condition fails, the usual finite variance and standard deviation do not exist.

Definition: The variance of claim severity is the expected squared distance between the severity and its mean: $$ \operatorname{Var}(X)=E\left[(X-E[X])^2\right]. $$ The standard deviation is the nonnegative square root of the variance: $$ \operatorname{SD}(X)=\sqrt{\operatorname{Var}(X)}. $$ If $X$ is measured in dollars, variance is measured in dollars squared and standard deviation is measured in dollars.

Calculating Variance from Two Expectations

In most severity calculations, the efficient route is to find $E[X]$ and $E[X^2]$, then use the identity below. This works for discrete, continuous, and mixed distributions, provided the expectations are calculated using the appropriate probabilities, density, or mixture components.

Theorem: If $E[X^2]$ is finite, then $$ \operatorname{Var}(X)=E[X^2]-\left(E[X]\right)^2. $$ Consequently, $$ \operatorname{SD}(X)=\sqrt{E[X^2]-\left(E[X]\right)^2}. $$

For a discrete severity taking values $x_j$ with probabilities $p_j$, calculate $E[X]=\sum_j x_jp_j$ and $E[X^2]=\sum_j x_j^2p_j$. For a continuous severity with density $f_X(x)$, calculate the corresponding expectations by integration: $E[X]=\int x f_X(x)\,dx$ and $E[X^2]=\int x^2 f_X(x)\,dx$. For a mixed model, include both its probability masses and its continuous part. These are applications of expectation to functions of severity.

Why the Shortcut Formula Holds

The shortcut follows by expanding the squared deviation and using the fact that $E[X]$ is a constant. Write $\mu=E[X]$. Then

\begin{aligned} \operatorname{Var}(X) &=E[(X-\mu)^2]\\ &=E[X^2-2\mu X+\mu^2]\\ &=E[X^2]-2\mu E[X]+\mu^2\\ &=E[X^2]-2\mu^2+\mu^2\\ &=E[X^2]-\mu^2. \end{aligned}

The formula is not an approximation: it is exactly the expected squared deviation written in a computationally convenient form. The final subtraction is between quantities with the same units, since both $E[X^2]$ and $(E[X])^2$ are in squared loss units.

Worked Example: A Discrete Claim Severity

A hypothetical equipment policy has claim severity $X$, in dollars per claim, with the following distribution: $P(X=0)=0.20$, $P(X=1{,}000)=0.50$, and $P(X=3{,}000)=0.30$. Find the variance and standard deviation. No interest or mortality assumption is involved; the stated discrete severity distribution is the complete model.

Method. Calculate the first and second expectations from the probability table, then apply the variance identity. This method avoids confusing the average claim with the variability around that average.

First, check the probabilities: $0.20+0.50+0.30=1.00$.

Mean severity:

E[X]=(0)(0.20)+(1{,}000)(0.50)+(3{,}000)(0.30) =0+500+900 =\$1{,}400.

Second moment: square each possible loss before multiplying by its probability:

\begin{aligned} E[X^2] &=(0^2)(0.20)+(1{,}000^2)(0.50)+(3{,}000^2)(0.30)\\ &=0+500{,}000+2{,}700{,}000\\ &=3{,}200{,}000\text{ dollars}^2. \end{aligned}

Therefore,

\begin{aligned} \operatorname{Var}(X) &=3{,}200{,}000-(1{,}400)^2\\ &=3{,}200{,}000-1{,}960{,}000\\ &=1{,}240{,}000\text{ dollars}^2, \\[4pt] \operatorname{SD}(X) &=\sqrt{1{,}240{,}000}\\ &=\$1{,}113.55. \end{aligned}

The variance is reported in dollars squared; the standard deviation is in dollars. Rounding only at the final step, the standard deviation is $\$1{,}113.55$ to the nearest cent.

Check. Directly weight the squared deviations from $\$1{,}400$: $0.20(1{,}400^2)+0.50(400^2)+0.30(1{,}600^2)=392{,}000+80{,}000+768{,}000=1{,}240{,}000$ dollars squared, agreeing with the shortcut.

Continuous and Mixed Severity Models

The same identity applies when losses are not restricted to a finite set of values. For a continuous model, integrate to obtain the expectations; for a mixed model, account for each point mass as well as the continuous probability. In all cases, retain the units through the calculation and take a square root only after the variance has been found.

Worked Example: Uniform Continuous Severity

For a hypothetical repair policy, severity $X$ is uniformly distributed from $\$0$ to $\$6{,}000$, inclusive endpoints having probability zero. Find its variance and standard deviation. The model is $X\sim\operatorname{Uniform}(0,6{,}000)$; no interest or mortality assumption is involved.

Method. The density is $f_X(x)=1/6{,}000$ for $0\le x\le6{,}000$, and zero otherwise. Integrate $x f_X(x)$ and $x^2 f_X(x)$. This avoids treating a continuous distribution as though it assigned positive probability to individual dollar values.

Mean:

\begin{aligned} E[X] &=\int_0^{6{,}000}x\frac{1}{6{,}000}\,dx\\ &=\frac{1}{6{,}000}\left[\frac{x^2}{2}\right]_0^{6{,}000}\\ &=\frac{36{,}000{,}000}{12{,}000} =\$3{,}000. \end{aligned}

Second moment:

\begin{aligned} E[X^2] &=\int_0^{6{,}000}x^2\frac{1}{6{,}000}\,dx\\ &=\frac{1}{6{,}000}\left[\frac{x^3}{3}\right]_0^{6{,}000}\\ &=\frac{216{,}000{,}000{,}000}{18{,}000} =12{,}000{,}000\text{ dollars}^2. \end{aligned}

Thus,

\begin{aligned} \operatorname{Var}(X) &=12{,}000{,}000-(3{,}000)^2\\ &=12{,}000{,}000-9{,}000{,}000\\ &=3{,}000{,}000\text{ dollars}^2,\\ \operatorname{SD}(X)&=\sqrt{3{,}000{,}000}=\$1{,}732.05. \end{aligned}

The standard deviation is rounded to the nearest cent. As a check, the uniform interval has width $\$6{,}000$, and its variance is width squared divided by 12: $6{,}000^2/12=3{,}000{,}000$ dollars squared.

Worked Example: A Mixed Severity with a Zero-Loss Mass

A hypothetical coverage has $P(X=0)=0.40$. With probability $0.60$, a positive claim is observed, and its severity is uniform from $\$1{,}000$ to $\$5{,}000$. Find the unconditional variance and standard deviation of $X$. The model is a point mass at zero plus a uniform component; no interest or mortality assumption is involved.

Method. First calculate the conditional mean and second moment of the uniform component, then weight them by the probability $0.60$. The zero-loss component contributes zero to both expectations. The key is to calculate dispersion for the full mixture, not just for positive claims.

For the uniform component on $[1{,}000,5{,}000]$, its mean is $(1{,}000+5{,}000)/2=\$3{,}000$. Its conditional second moment is

\begin{aligned} E[X^2\mid X>0] &=\frac{1}{4{,}000}\int_{1{,}000}^{5{,}000}x^2\,dx\\ &=\frac{5{,}000^3-1{,}000^3}{12{,}000}\\ &=\frac{124{,}000{,}000{,}000}{12{,}000}\\ &=10{,}333{,}333.33\text{ dollars}^2. \end{aligned}

The unconditional expectations are therefore

\begin{aligned} E[X]&=(0.40)(0)+(0.60)(3{,}000)=\$1{,}800,\\ E[X^2]&=(0.40)(0)+(0.60)(10{,}333{,}333.33) =6{,}200{,}000\text{ dollars}^2. \end{aligned}

Then

\begin{aligned} \operatorname{Var}(X) &=6{,}200{,}000-(1{,}800)^2\\ &=6{,}200{,}000-3{,}240{,}000\\ &=2{,}960{,}000\text{ dollars}^2,\\ \operatorname{SD}(X)&=\sqrt{2{,}960{,}000}=\$1{,}720.47. \end{aligned}

The standard deviation is rounded to the nearest cent. The mean of $\$1{,}800$ is below the positive-claim mean of $\$3{,}000$, because the unconditional model includes the zero-loss probability. The calculation uses unconditional expectations throughout, so it measures dispersion across all insured exposures.

Interpreting the Measures and Avoiding Traps

The mean and standard deviation answer different questions. $E[X]$ describes the long-run average severity under the model. $\operatorname{SD}(X)$ describes a scale of deviation around that mean, in the same units as severity. It is not a guaranteed maximum deviation, nor does it imply that a specified proportion of claims lies within one standard deviation unless additional distributional information justifies such a statement.

A standard exam error is to stop at $E[X^2]$ and call it variance. In the uniform example, that mistake would give $12{,}000{,}000$ dollars squared as the variance, rather than subtracting the squared mean of $9{,}000{,}000$ dollars squared. It would also give an incorrect standard deviation of $\sqrt{12{,}000{,}000}=\$3{,}464.10$. The correct variance is $3{,}000{,}000$ dollars squared and the correct standard deviation is $\$1{,}732.05$.

A second trap is a units error: variance is not in dollars, but in dollars squared. The square root restores the loss unit. Keep full precision during the calculation, subtract $(E[X])^2$ rather than $E[X]$, and round only the final requested quantity.

Key takeaway: Find the mean and second moment on the same severity basis, calculate $\operatorname{Var}(X)=E[X^2]-(E[X])^2$, and take the nonnegative square root for standard deviation. The mean measures expected size; the standard deviation measures spread in loss units.

Check Your Understanding

For each hypothetical severity model, calculate the requested variance or standard deviation. Answers are rounded to the nearest cent when expressed in dollars.

A discrete severity has probabilities $0.50, 0.30,$ and $0.20$ at $\$0,\$2{,}000,$ and $\$6{,}000$, respectively. Find its mean, variance, and standard deviation. (Answer: mean $\$1{,}800$; variance $5{,}160{,}000$ dollars squared; standard deviation $\$2{,}271.56$)
A hypothetical severity is uniform from $\$200$ to $\$2{,}600$. Find its mean, variance, and standard deviation. (Answer: mean $\$1{,}400$; variance $480{,}000$ dollars squared; standard deviation $\$692.82$)
A severity is zero with probability $0.20$, and equals $\$4{,}000$ with probability $0.80$. Find its unconditional mean, variance, and standard deviation. (Answer: mean $\$3{,}200$; variance $2{,}560{,}000$ dollars squared; standard deviation $\$1{,}600.00$)
Portfolio A has a certain severity of $\$2{,}500$. Portfolio B has severity $\$0$ or $\$5{,}000$, each with probability $0.50$. Find each portfolio's mean and standard deviation, and identify which has greater dispersion. (Answer: both means are $\$2{,}500$; A has standard deviation $\$0.00$, B has standard deviation $\$2{,}500.00$, so B has greater dispersion)
A discrete severity is $\$1{,}000$ with probability $0.60$ and $\$4{,}000$ with probability $0.40$. Find $E[X]$, $E[X^2]$, variance, and standard deviation, keeping squared units for the second moment and variance. (Answer: $E[X]=\$2{,}200$; $E[X^2]=7{,}000{,}000$ dollars squared; variance $2{,}160{,}000$ dollars squared; standard deviation $\$1{,}469.69$)

What You'll Learn