Variance and Standard Deviation of Loss

Why Measure Loss Spread?

An expected loss describes the distribution’s average, but it does not show how concentrated losses are around that average. Two hypothetical portfolios can have the same expected annual loss while one has losses clustered near the mean and the other has much greater variation. Variance and standard deviation provide measures of this spread. They are useful summaries, but they do not, by themselves, describe the shape of a distribution or how likely a particular extreme loss is.

This tutorial concerns a random loss $X$ measured on a fixed basis, such as dollars of loss over one year. Unless stated otherwise, assume losses are nonnegative, the relevant expected values are finite, and all monetary amounts are on the same basis. No interest or mortality adjustment is used in the examples. The variance formulas also apply to signed random variables when their second moments are finite.

Definition: For a loss random variable $X$ with finite mean $\mu=E[X]$, its population variance is $\operatorname{Var}(X)=E[(X-\mu)^2]$. Its standard deviation is $\operatorname{SD}(X)=\sqrt{\operatorname{Var}(X)}$. Variance measures the expected squared distance from the mean; standard deviation expresses spread in the same units as $X$.

Calculating Variance From a Distribution

For a discrete loss distribution, multiply each squared distance from the mean by the probability of that loss, then add the results. The probabilities must sum to one. For a continuous loss with density $f_X(x)$, take the corresponding integral over the support of the distribution.

\operatorname{Var}(X)=\sum_x (x-\mu)^2P(X=x) \qquad\text{(discrete)}

\operatorname{Var}(X)=\int_{-\infty}^{\infty}(x-\mu)^2 f_X(x)\,dx \qquad\text{(continuous)}

These formulas require a finite second moment. If the squared deviations have an infinite expected value, the variance is not finite, and the standard deviation is not a finite spread measure.

Theorem: When $E[X^2]$ is finite, variance can also be calculated as $\operatorname{Var}(X)=E[X^2]-(E[X])^2$. This shortcut is valid for both discrete and continuous distributions.

To see why the shortcut holds, let $\mu=E[X]$. Expand the squared deviation and use linearity of expectation:

\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 direct squared-deviation method makes the interpretation visible and can be numerically stable. The second-moment method is often quicker, particularly when $E[X^2]$ is easy to calculate. Both must use the same distribution and the same units.

Worked Example: Discrete Annual Loss

A hypothetical small commercial portfolio has annual covered loss $X$ with the following distribution: $P(X=0)=0.25$, $P(X=1{,}000)=0.50$, and $P(X=2{,}000)=0.25$. Losses are in dollars. Calculate the expected loss, variance, and standard deviation. There is no interest or mortality adjustment; the distribution is the stated one-year loss model.

Method. First find the mean, then calculate the probability-weighted squared deviations. As a check, also use the second-moment formula. This avoids treating variance as an average absolute deviation or forgetting the probabilities.

The probabilities sum to $0.25+0.50+0.25=1.00$. The expected loss is

E[X]=0(0.25)+1{,}000(0.50)+2{,}000(0.25) =0+500+500=\$1{,}000.

The deviations from the mean are $-\$1{,}000$, $\$0$, and $\$1{,}000$. Squaring and weighting them gives

\begin{aligned} \operatorname{Var}(X) &=(1{,}000)^2(0.25)+(0)^2(0.50)+(1{,}000)^2(0.25)\\ &=250{,}000+0+250{,}000\\ &=500{,}000\ \text{dollars}^2. \end{aligned}

Therefore,

\operatorname{SD}(X)=\sqrt{500{,}000} =\$707.1068\ldots\approx \$707.11.

Check. The second moment is $E[X^2]=0.25(0)^2+0.50(1{,}000)^2+0.25(2{,}000)^2=0+500{,}000+1{,}000{,}000=1{,}500{,}000$ dollars squared. Subtracting $(E[X])^2=1{,}000{,}000$ gives $500{,}000$ dollars squared, as above. The standard deviation is rounded to cents only at the final step.

Variance From a Continuous Loss Density

For a continuous distribution, probabilities are represented by density rather than probability at an individual point. A continuous random variable has probability zero of taking any one exact value; expected values are calculated by integrating over intervals. In the next example, the uniform density places equal density across a bounded range.

Worked Example: Uniform Loss Amount

Suppose a hypothetical annual loss $X$, in dollars, is uniformly distributed from $\$0$ to $\$6{,}000$. Thus $X\sim\operatorname{Uniform}(0,6{,}000)$, with density $f_X(x)=1/6{,}000$ for $0\le x\le6{,}000$, and zero elsewhere. Find its mean, variance, and standard deviation. There is no interest or mortality adjustment.

Method. Integrate $x f_X(x)$ and $x^2 f_X(x)$, then apply the second-moment identity. This keeps the density and integration limits explicit.

\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}

Next calculate the 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}

The variance and standard deviation are

\begin{aligned} \operatorname{Var}(X) &=E[X^2]-(E[X])^2\\ &=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{,}732.0508\ldots\approx \$1{,}732.05. \end{aligned}

Check. The mean, $\$3{,}000$, is the midpoint of the uniform interval $[0,6{,}000]$, as expected. The calculated variance is positive, and the standard deviation is in dollars. It is rounded to cents at the final step.

Units and Interpretation

If $X$ is measured in dollars, then $X-\mu$ is measured in dollars and its square is measured in dollars squared. Thus $\operatorname{Var}(X)$ has squared monetary units, while $\operatorname{SD}(X)$ has monetary units. The variance is useful in calculations, but a value such as $500{,}000$ dollars squared is not itself a typical dollar loss. Taking the square root returns the measure to the original loss scale.

Standard deviation describes spread around the mean, not a guaranteed range for losses. Without additional assumptions about the distribution, it does not imply that a fixed proportion of losses lies within one or two standard deviations of the mean. Nor does a low standard deviation establish that the upper tail is harmless. Use the distribution and the question being asked when interpreting this summary.

A useful transformation rule follows directly from the definition. For constants $a$ and $b$, adding $b$ shifts both the loss and its mean by the same amount and does not change deviations; multiplying by $a$ multiplies deviations by $a$. Therefore $\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$ and $\operatorname{SD}(aX+b)=|a|\operatorname{SD}(X)$. A change from dollars to thousands of dollars divides the numerical standard deviation by $1{,}000$, and divides the numerical variance by $1{,}000{,}000$.

Population Variance and Sample Variance

A distributional variance is a population quantity: it describes the specified random variable or full population. A sample variance instead summarizes observed data and, under specified sampling assumptions, estimates the unknown variance of a population. The denominator depends on which quantity is being calculated.

Definition: For observed values $x_1,\ldots,x_n$, their sample mean is $\bar{x}=\frac{1}{n}\sum_{j=1}^n x_j$. The sample variance is $s^2=\frac{1}{n-1}\sum_{j=1}^n(x_j-\bar{x})^2$, and the sample standard deviation is $s=\sqrt{s^2}$. The $n-1$ denominator is used when these observations form a sample used to estimate an unknown population variance; it is not the formula for a complete finite population’s variance.

The reason for $n-1$ can be seen when $X_1,\ldots,X_n$ are independent, identically distributed observations with finite population mean $\mu$ and variance $\sigma^2$. Their sample mean is $\bar{X}$. The identity below separates deviations from the sample mean from deviations around the population mean:

\sum_{j=1}^n(X_j-\bar{X})^2 = \sum_{j=1}^n(X_j-\mu)^2-n(\bar{X}-\mu)^2.

Taking expectations, the first sum on the right has expectation $n\sigma^2$. Since $\operatorname{Var}(\bar{X})=\sigma^2/n$, the second term has expectation $n(\sigma^2/n)=\sigma^2$. Consequently,

E\left[\sum_{j=1}^n(X_j-\bar{X})^2\right] =n\sigma^2-\sigma^2=(n-1)\sigma^2, \qquad E[s^2]=\sigma^2.

Thus dividing by $n-1$ makes $s^2$ an unbiased estimator of the population variance under these assumptions. This does not mean that $s$, the square root of $s^2$, is generally an unbiased estimator of the population standard deviation.

Worked Example: Choosing the Sample or Population Denominator

A hypothetical analyst has four observed annual losses, in thousands of dollars: $0.5,\ 0.8,\ 1.2,\ 1.5$. Treat them first as a sample used to estimate an underlying loss variance, then calculate the variance and standard deviation of these four values as a complete empirical population. There is no interest or mortality adjustment.

Method. Find the common mean and sum of squared deviations. Use $n-1$ for the sample estimate and $n$ for the variance of the complete set. This avoids choosing a denominator without checking what the data represent.

The sample mean is $\bar{x}=(0.5+0.8+1.2+1.5)/4=4.0/4=1.0$ thousand dollars. The squared deviations, in squared thousands of dollars, are $(0.5-1.0)^2=0.25$, $(0.8-1.0)^2=0.04$, $(1.2-1.0)^2=0.04$, and $(1.5-1.0)^2=0.25$. Their sum is $0.25+0.04+0.04+0.25=0.58$.

As a sample, the variance estimate and standard deviation are

s^2=\frac{0.58}{4-1}=0.193333\ldots\ (\text{thousand dollars})^2, \qquad s=\sqrt{0.193333\ldots}=0.439696\ldots\ \text{thousand dollars}.

Rounded to cents in dollar units, the sample standard deviation is $\$439.70$. As the complete empirical population, divide by four instead: $0.58/4=0.145$ squared thousands of dollars, so the population standard deviation is $\sqrt{0.145}=0.380789\ldots$ thousand dollars, or $\$380.79$. The distinction is not a change in the observations; it is a change in the question. Each standard deviation is rounded to cents only at the final step.

Common Mistakes

Using the wrong denominator. Dividing a sample’s squared deviations by $n$ does not give the usual unbiased sample variance. Conversely, using $n-1$ for the variance of a complete population changes the defined quantity.
Forgetting to square deviations. Positive and negative deviations from the mean would cancel if simply added. Variance squares them so that both directions contribute to spread.
Subtracting the wrong quantity. The shortcut is $E[X^2]-(E[X])^2$, not $E[X^2]-E[X]$.
Reporting variance in dollars. Variance is in dollars squared; take its square root to express spread in dollars.
Calling standard deviation a loss bound. Standard deviation is a measure of spread, not a maximum loss or a distribution-free interval containing a specified probability.

Key takeaway: For a full loss distribution, calculate $\operatorname{Var}(X)=E[(X-E[X])^2]=E[X^2]-(E[X])^2$, then take the square root for standard deviation. For sample data used to estimate an unknown variance, use the sample mean and divide the squared deviations by $n-1$.

Check Your Understanding

For each problem, identify whether the calculation concerns a specified distribution, a sample estimate, or a complete empirical population. Give monetary standard deviations to cents when applicable.

A hypothetical one-year loss $X$ is $\$0$ with probability $0.60$, $\$1{,}000$ with probability $0.30$, and $\$4{,}000$ with probability $0.10$. Find $E[X]$, $\operatorname{Var}(X)$, and $\operatorname{SD}(X)$. (Answer: mean $\$700$; variance $1{,}410{,}000$ dollars squared; standard deviation $\$1{,}187.43$.)
A hypothetical continuous annual loss is uniform on $[\$200,\$1{,}400]$. Using the uniform density, find its mean, variance, and standard deviation. (Answer: mean $\$800$; variance $120{,}000$ dollars squared; standard deviation $\$346.41$.)
Four observed losses, in thousands of dollars, are $2,4,4,10$. Calculate the sample variance and sample standard deviation for estimating an underlying variance. (Answer: sample variance $12$ squared thousands of dollars; sample standard deviation $3.4641$ thousand dollars, or $\$3{,}464.10$.)
A hypothetical loss $X$ is $\$0$ with probability $0.50$, $\$2{,}000$ with probability $0.30$, and $\$5{,}000$ with probability $0.20$. Find the mean, variance, and standard deviation, and verify the variance with the second-moment identity. (Answer: mean $\$1{,}600$; variance $3{,}640{,}000$ dollars squared; standard deviation $\$1{,}907.88$; $E[X^2]=6{,}200{,}000$ dollars squared.)
A hypothetical discrete loss $X$ is $\$0$ or $\$1{,}000$, each with probability $0.50$. Define $Z=1.5X+\$200$. Find the mean and standard deviation of $Z$, using the transformation rule to obtain its variance and standard deviation. (Answer: mean $\$950$; variance $562{,}500$ dollars squared; standard deviation $\$750.00$.)

What You'll Learn