Raw Moments And Variance Of Severity

Why Raw Moments Matter

A severity distribution describes the amount of a loss, but its full probability model is not always needed for every calculation. The first raw moment gives the expected loss; the second raw moment, together with the first, gives the variance. These summaries help describe the distribution’s average level and spread. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

The word raw distinguishes moments taken about zero from central moments taken about the mean. In particular, the second raw moment is \(E[X^2]\), while the variance is \(E[(X-E[X])^2]\). They are generally not equal.

Definition: For a loss random variable \(X\), the \(r\)th raw moment is \(E[X^r]\), when this expectation exists. The first raw moment is \(E[X]\), and the second raw moment is \(E[X^2]\). The variance is the second central moment, \(\operatorname{Var}(X)=E[(X-E[X])^2]\).

Calculating Raw Moments From a Distribution

For a discrete loss with possible values \(x_j\) and probabilities \(p_j=P(X=x_j)\), calculate a raw moment by raising each possible loss to the required power, weighting it by its probability, and summing. The probabilities must be nonnegative and sum to 1.

E[X^r]=\sum_j x_j^r p_j

For a continuous loss with density \(f_X(x)\), the corresponding calculation weights each value by the density and integrates over the support. A valid density is nonnegative and integrates to 1.

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

For a nonnegative loss, the integral can be restricted to its support. These formulas require the relevant expectation to exist: the sum or integral of the absolute value must be finite. A finite mean alone does not guarantee a finite second moment. If \(E[X^2]\) is infinite, the variance is not finite.

Variance From the First Two Raw Moments

The variance can be computed directly from deviations from the mean, but expanding the square gives a convenient formula in terms of raw moments. Let \(\mu=E[X]\), assumed finite, and suppose \(E[X^2]\) is finite.

Start with the definition.
\(\operatorname{Var}(X)=E[(X-\mu)^2]\).

Expand the square.
\((X-\mu)^2=X^2-2\mu X+\mu^2\).

Take expectations.
By linearity, \(E[X^2-2\mu X+\mu^2]=E[X^2]-2\mu E[X]+\mu^2\).

Substitute \(E[X]=\mu\).
\(E[X^2]-2\mu^2+\mu^2=E[X^2]-\mu^2\).

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

This identity works because variance measures squared deviations around the mean, and the cross-term in the expansion simplifies using \(E[X]=\mu\). It does not say that the second raw moment equals the square of the mean. Indeed, their difference is the variance.

Units provide a useful check. If \(X\) is measured in thousands of dollars, then \(E[X]\) is in thousands of dollars, while \(E[X^2]\) and \(\operatorname{Var}(X)\) are in squared thousands of dollars. The standard deviation, if needed, is the square root of variance and returns to the original loss units.

Worked Example: Discrete Severity in Thousands of Dollars

A hypothetical repair-cost model assigns a loss \(X\), measured in thousands of dollars, as follows: \(P(X=0)=0.20\), \(P(X=2)=0.50\), and \(P(X=5)=0.30\). Find the first two raw moments and the variance.

Method. Apply the discrete moment formula separately for powers 1 and 2, then use the raw-moment variance identity. First check the probabilities: \(0.20+0.50+0.30=1.00\).

The first raw moment is

$$ E[X]=(0)(0.20)+(2)(0.50)+(5)(0.30)=0+1.00+1.50=2.50. $$

Thus, the expected loss is \(2.50\) thousand dollars. The second raw moment is

$$ E[X^2]=(0^2)(0.20)+(2^2)(0.50)+(5^2)(0.30) =0+2.00+7.50=9.50. $$

Therefore,

\operatorname{Var}(X)=9.50-(2.50)^2=9.50-6.25=3.25

in squared thousands of dollars. As a check, calculate the probability-weighted squared deviations from the mean: \(0.20(0-2.50)^2+0.50(2-2.50)^2+0.30(5-2.50)^2 =1.25+0.125+1.875=3.25\). The direct calculation agrees.

Common mistake avoided. Replacing \(E[X^2]\) by \((E[X])^2\) would give \(6.25\) for both quantities and incorrectly produce zero variance. The square must be applied to each possible loss before probability-weighting: here, \(E[X^2]=9.50\), not \(6.25\).

Worked Example: A Continuous Density With a Normalizing Constant

A hypothetical severity \(X\), measured in thousands of dollars, has density \(f_X(x)=c x^2\) for \(0<x<4\), and zero elsewhere. Find \(c\), \(E[X]\), \(E[X^2]\), and \(\operatorname{Var}(X)\).

Method. First choose \(c\) so that the density integrates to 1. Then integrate \(x f_X(x)\) and \(x^2 f_X(x)\) over the support.

Normalization gives

1=\int_0^4 c x^2\,dx =c\left[\frac{x^3}{3}\right]_0^4 =c\frac{64}{3}, \qquad c=\frac{3}{64}.

The first raw moment is

E[X]=\int_0^4 x\frac{3x^2}{64}\,dx =\frac{3}{64}\left[\frac{x^4}{4}\right]_0^4 =\frac{3}{64}(64)=3.

The second raw moment is

E[X^2]=\int_0^4 x^2\frac{3x^2}{64}\,dx =\frac{3}{64}\left[\frac{x^5}{5}\right]_0^4 =\frac{3}{64}\frac{1024}{5} =\frac{48}{5}=9.6.

Consequently,

\operatorname{Var}(X)=\frac{48}{5}-3^2 =\frac{48}{5}-9 =\frac{3}{5}=0.6.

The variance is \(0.6\) squared thousands of dollars. To check it directly, integrate the squared deviation:

\int_0^4 (x-3)^2\frac{3x^2}{64}\,dx =\frac{3}{64}\int_0^4(x^4-6x^3+9x^2)\,dx =\frac{3}{64}\left(\frac{1024}{5}-384+192\right) =\frac{3}{64}\frac{64}{5} =\frac{3}{5}.

This confirms the variance identity. A frequent error is to integrate over the wrong interval or omit the density’s normalizing constant; either mistake changes both moments.

Worked Example: Moments and a Tail Probability From a Density

A hypothetical loss \(X\), in thousands of dollars, has density \(f_X(x)=c(3-x)\) for \(0<x<3\), and zero elsewhere. Determine the normalizing constant, the first two raw moments, the variance, and \(P(X>2)\).

Method. Normalize the density, then use the continuous moment formulas. For the final probability, integrate the density only over the requested interval.

The normalization condition is

1=c\int_0^3(3-x)\,dx =c\left[3x-\frac{x^2}{2}\right]_0^3 =c\left(9-\frac{9}{2}\right) =\frac{9c}{2}, \qquad c=\frac{2}{9}.

Now calculate the first moment:

E[X]=\frac{2}{9}\int_0^3x(3-x)\,dx =\frac{2}{9}\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3 =\frac{2}{9}\left(\frac{27}{2}-9\right) =1.

The second moment is

E[X^2]=\frac{2}{9}\int_0^3x^2(3-x)\,dx =\frac{2}{9}\left[x^3-\frac{x^4}{4}\right]_0^3 =\frac{2}{9}\left(27-\frac{81}{4}\right) =\frac{3}{2}.

Hence \(\operatorname{Var}(X)=\frac{3}{2}-1^2=\frac{1}{2}\), in squared thousands of dollars. The requested probability is

P(X>2)=\frac{2}{9}\int_2^3(3-x)\,dx =\frac{2}{9}\left[3x-\frac{x^2}{2}\right]_2^3 =\frac{2}{9}\left(\frac{9}{2}-4\right) =\frac{1}{9}=0.1111.

The result is internally consistent: \(c=2/9\) makes the total density 1, and the variance is nonnegative. The interval for the probability begins at 2, not 0; integrating over the full support would instead return 1.

Common Traps and Checks

Moment calculations are usually short, but a small notation or unit error can change the result substantially. Use the following checks before accepting a calculation:

Keep the order of operations clear. For a discrete distribution, calculate \(x_j^2\) first, then multiply by \(p_j\), then sum. In general, \(E[X^2]\ne(E[X])^2\).
Do not confuse raw and central moments. \(E[X^2]\) is measured about zero; variance is measured about the mean. The connection is \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\).
Check the support and normalization. A discrete probability mass function must sum to 1; a continuous density must integrate to 1. Use the stated support in every sum or integral.
Track squared units. Variance has the square of the loss unit. Do not subtract a mean in dollars from a second moment in squared dollars.
Check finiteness and sign. When the relevant moments are finite, variance cannot be negative. A negative computed value signals an arithmetic or modeling error.

Raw moments turn a probability model into useful numerical summaries while preserving a direct connection to the underlying distribution. The first moment records average severity; the second captures squared scale and, after subtracting the squared mean, yields variance. Always calculate the moments from the stated mass function or density rather than assuming that the mean alone determines the spread.

Check Your Understanding

For each problem, calculate the requested raw moments and variance where applicable. Loss amounts are expressed in the units stated.

A discrete loss \(X\), in thousands of dollars, is 1 with probability 0.75 and 4 with probability 0.25. Find \(E[X]\), \(E[X^2]\), and \(\operatorname{Var}(X)\). (Answer: \(E[X]=1.75\), \(E[X^2]=4.75\), and \(\operatorname{Var}(X)=1.6875\) squared thousands of dollars.)
A discrete severity \(X\), in thousands of dollars, takes values 0, 3, and 7 with probabilities 0.10, 0.60, and 0.30, respectively. Find its first two raw moments and variance. (Answer: \(E[X]=3.9\), \(E[X^2]=20.1\), and \(\operatorname{Var}(X)=4.89\) squared thousands of dollars.)
A continuous loss \(X\) has density \(f_X(x)=2x/9\) for \(0<x<3\), and zero elsewhere. Find \(E[X]\), \(E[X^2]\), and \(\operatorname{Var}(X)\). (Answer: \(E[X]=2\), \(E[X^2]=4.5\), and \(\operatorname{Var}(X)=0.5\).)
A continuous severity has density \(f_X(x)=c(4-x)\) for \(0<x<4\), and zero elsewhere. Find \(c\), the first two raw moments, and the variance. (Answer: \(c=1/8\), \(E[X]=4/3\), \(E[X^2]=8/3\), and \(\operatorname{Var}(X)=8/9\).)
A hypothetical loss has an atom with \(P(X=0)=0.40\) and density \(f_X(x)=0.20\) for \(0<x<3\), with no other probability. Find \(E[X]\), \(E[X^2]\), and \(\operatorname{Var}(X)\). (Answer: \(E[X]=0.9\), \(E[X^2]=1.8\), and \(\operatorname{Var}(X)=0.99\).)

What You'll Learn