Raw Moments and Central Moments of Severity

Why Two Kinds of Moments?

A severity model describes the amount $X$ of a single loss. Its mean describes the average loss, while its variance describes how much losses vary around that mean. Both quantities are moments, but they are taken around different reference points. A raw moment is calculated using powers of $X$ itself; a central moment uses powers of the deviation from the mean. Keeping those reference points straight is essential when moving from a probability model to a variance calculation.

The focus here is the first and second moments. These are enough to find the mean and variance, and they are often supplied or calculated in different forms in an actuarial problem. A moment might be computed from a probability table, from a density, or from a distribution with both point masses and a continuous part.

Definition: For a severity $X$, the $r$th raw moment about zero is $\mu'_r=E[X^r]$, provided the expectation exists. The $r$th central moment about the mean is $\mu_r=E[(X-E[X])^r]$, provided that expectation exists. In particular, $\mu'_1=E[X]$, $\mu_1=0$, and $\mu_2=\operatorname{Var}(X)$.

For the first two moments, the key identities are

\mu'_1=E[X], \qquad \mu_2=\mu'_2-(\mu'_1)^2, \qquad \mu'_2=\mu_2+(\mu'_1)^2.

The second raw moment is $E[X^2]$, not the variance. The variance is the second central moment, which measures squared deviations from the mean. Both the raw second moment and the variance are expressed in squared units of severity.

Why the Conversion Formula Holds

Write $\mu=E[X]$, and assume $E[X^2]$ is finite so the variance exists. Expanding the squared deviation from the mean gives

\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]-\{E[X]\}^2. \end{aligned}

The third line uses linearity of expectation and the fact that $\mu$ is a constant. Since $E[X^2]=\mu'_2$ and $E[X]=\mu'_1$, this proves $\mu_2=\mu'_2-(\mu'_1)^2$. Rearranging gives the reverse conversion. The formula does not turn a nonexistent or infinite second moment into a finite variance: the required moments must exist for the calculation to apply.

For a discrete severity taking values $x_j$ with probabilities $p_j$, the raw moment is calculated as $\mu'_r=\sum_j p_jx_j^r$. For a continuous severity with density $f_X$, it is $\mu'_r=\int_{-\infty}^{\infty}x^rf_X(x)\,dx$, when the integral exists. A mixed model requires adding the contributions from its point masses and integrating its density part.

Worked Calculations

Worked Example: Discrete Severity and the Raw Second Moment

A hypothetical equipment policy has claim severity $X$, measured in thousands of dollars, with probabilities $P(X=0)=0.20$, $P(X=2)=0.50$, and $P(X=5)=0.30$. Find the first raw moment, second raw moment, and variance. Check the variance using deviations from the mean.

Method. Use the probability table to calculate each raw moment, then convert the raw second moment to the variance. The probabilities sum to $0.20+0.50+0.30=1.00$.

First raw moment.

\mu'_1=E[X] =(0.20)(0)+(0.50)(2)+(0.30)(5) =0+1.00+1.50 =2.50.

Thus the mean severity is $2.50$ thousand dollars, or $\$2{,}500$.

Second raw moment. Square each severity before weighting it:

\mu'_2=E[X^2] =(0.20)(0^2)+(0.50)(2^2)+(0.30)(5^2) =0+2.00+7.50 =9.50.

Variance from raw moments.

\operatorname{Var}(X) =\mu'_2-(\mu'_1)^2 =9.50-(2.50)^2 =9.50-6.25 =3.25.

The variance is $3.25$ (thousand dollars)$^2$. To check, the mean is $2.50$, so the squared deviations are $6.25$, $0.25$, and $6.25$. Their probability-weighted sum is $(0.20)(6.25)+(0.50)(0.25)+(0.30)(6.25)=1.25+0.125+1.875=3.25$, as required. The check confirms that the raw second moment $9.50$ is not the variance.

Worked Example: Continuous Severity from a Density

A hypothetical claim severity has density $f_X(x)=3x^2/8$ for $0<x<2$, and $f_X(x)=0$ otherwise. Find the mean, the raw second moment, and the variance.

Check the density. Its total probability is

\int_0^2 \frac{3x^2}{8}\,dx =\frac{3}{8}\left[\frac{x^3}{3}\right]_0^2 =\frac{3}{8}\cdot\frac{8}{3} =1.

Mean. Multiply the density by $x$, then integrate over its support:

\mu'_1=E[X] =\int_0^2 x\frac{3x^2}{8}\,dx =\frac{3}{8}\int_0^2x^3\,dx =\frac{3}{8}\left[\frac{x^4}{4}\right]_0^2 =\frac{3}{8}(4) =\frac{3}{2} =1.5.

Raw second moment. Multiply by $x^2$, not by $x$:

\mu'_2=E[X^2] =\int_0^2 x^2\frac{3x^2}{8}\,dx =\frac{3}{8}\int_0^2x^4\,dx =\frac{3}{8}\left[\frac{x^5}{5}\right]_0^2 =\frac{3}{8}\cdot\frac{32}{5} =\frac{12}{5} =2.4.

Variance.

\operatorname{Var}(X) =\mu'_2-(\mu'_1)^2 =2.4-(1.5)^2 =2.4-2.25 =0.15.

The results are $\mu'_1=1.5$, $\mu'_2=2.4$, and $\operatorname{Var}(X)=0.15$. The arithmetic can also be checked exactly: $12/5-(3/2)^2=12/5-9/4=(48-45)/20=3/20=0.15$. The density is supported on $(0,2)$, and the mean $1.5$ lies within that interval, a useful basic reasonableness check.

Worked Example: Combining Point Masses and a Density

A hypothetical warranty severity $X$, in thousands of dollars, has point masses $P(X=0)=0.20$ and $P(X=2)=0.30$, as well as density $f_X(x)=0.25$ for $0<x<2$, with density zero elsewhere. Find the mean, raw second moment, and variance.

Check the total probability. The density contributes $\int_0^2 0.25\,dx=0.50$. The point masses contribute $0.20+0.30=0.50$, so the total probability is $1$.

Mean. Add the weighted point values to the integral over the continuous part:

\begin{aligned} \mu'_1=E[X] &=(0.20)(0)+(0.30)(2)+\int_0^2x(0.25)\,dx\\ &=0+0.60+0.25\left[\frac{x^2}{2}\right]_0^2\\ &=0.60+0.25(2)=1.10. \end{aligned}

Raw second moment. Square the outcome in both the point-mass and density contributions:

\begin{aligned} \mu'_2=E[X^2] &=(0.20)(0^2)+(0.30)(2^2)+\int_0^2x^2(0.25)\,dx\\ &=0+1.20+0.25\left[\frac{x^3}{3}\right]_0^2\\ &=1.20+\frac{2}{3} =\frac{28}{15} \approx1.8667. \end{aligned}

Variance. Use the unrounded mean $1.10=11/10$:

\operatorname{Var}(X) =\frac{28}{15}-\left(\frac{11}{10}\right)^2 =\frac{560}{300}-\frac{363}{300} =\frac{197}{300} \approx0.6567.

Therefore the mean is $1.10$ thousand dollars, the raw second moment is $28/15$ (thousand dollars)$^2$, and the variance is $197/300$ (thousand dollars)$^2$. Decimal approximations are rounded to four places. As a check, the variance is positive and smaller than the raw second moment, consistent with subtracting the positive squared mean.

Choosing and Checking the Needed Moment

When a problem asks for an expected severity, the needed quantity is the first raw moment, $E[X]=\mu'_1$. When it asks for variance, either calculate the second central moment directly or calculate the first two raw moments and use $\mu_2=\mu'_2-(\mu'_1)^2$. If the mean and variance are already known, the reverse identity gives the raw second moment as $\mu'_2=\operatorname{Var}(X)+\{E[X]\}^2$.

Quantity requested	Moment to use	Interpretation
Mean severity	$\mu'_1$	Expected loss $E[X]$
Raw second moment	$\mu'_2$	Expected squared loss $E[X^2]$
Variance	$\mu_2$, or $\mu'_2-(\mu'_1)^2$	Expected squared deviation from the mean

Units help expose mistakes. If $X$ is measured in dollars, $\mu'_1$ is in dollars, while $\mu'_2$ and $\operatorname{Var}(X)$ are in dollars squared. A variance cannot be directly interpreted as a dollar amount; its square root is in the original severity units. Keep calculations in thousands if the problem defines $X$ in thousands, and convert only when the requested reporting unit requires it.

Common Exam Trap

Wrong calculation: treating the raw second moment as the variance, so reporting $\operatorname{Var}(X)=\mu'_2$. For the discrete example, that would give $9.50$ instead of the correct $3.25$ (thousand dollars)$^2$.

Correction: subtract the square of the mean: $\operatorname{Var}(X)=9.50-(2.50)^2=3.25$. A related integration error is to calculate the mean using $x^2f_X(x)$; that integral is the raw second moment. The mean uses $xf_X(x)$. Label each integral or sum by the moment it represents before evaluating it.

Key takeaway: The mean is the first raw moment, $\mu'_1$. The variance is the second central moment, $\mu_2$, and equals $\mu'_2-(\mu'_1)^2$ when the second moment exists. Always distinguish the moment's reference point, preserve the squared units, and check the result.

Check Your Understanding

For each problem, identify which moments are needed and retain sufficient precision until the final answer.

A hypothetical severity $X$, in thousands of dollars, equals $1$ with probability $0.40$ and $3$ with probability $0.60$. Find $\mu'_1$, $\mu'_2$, and $\operatorname{Var}(X)$. (Answer: $\mu'_1=2.2$, $\mu'_2=5.8$, $\operatorname{Var}(X)=0.96$, in the corresponding powers of thousands of dollars.)
A severity has density $f_X(x)=1/3$ for $0<x<3$ and zero otherwise. Find its mean, raw second moment, and variance. (Answer: $\mu'_1=1.5$, $\mu'_2=3$, $\operatorname{Var}(X)=0.75$.)
A model specifies $E[X]=4$ dollars and $\operatorname{Var}(X)=21$ dollars squared. Find the first raw moment, second raw moment, and second central moment. (Answer: $\mu'_1=4$ dollars, $\mu'_2=37$ dollars squared, $\mu_2=21$ dollars squared.)
A mixed severity has masses $P(X=0)=0.40$ and $P(X=2)=0.20$, plus density $f_X(x)=0.20$ for $0<x<2$ and zero elsewhere. Check total probability, then find its mean and variance. (Answer: total probability $1$, $\mu'_1=0.8$, $\mu'_2=4/3$, $\operatorname{Var}(X)=52/75$.)
A hypothetical portfolio model has $E[X]=3$ thousand dollars and $E[X^2]=13$ (thousand dollars)$^2$. Calculate its variance and state whether the raw second moment is the variance. (Answer: $\operatorname{Var}(X)=13-3^2=4$ (thousand dollars)$^2$; no, the raw second moment is $13$, not $4$.)

Quantity requested	Moment to use	Interpretation
Mean severity	\(\mu'_1\)	Expected loss \(E[X]\)
Raw second moment	\(\mu'_2\)	Expected squared loss \(E[X^2]\)
Variance	\(\mu_2\), or \(\mu'_2-(\mu'_1)^2\)	Expected squared deviation from the mean

What You'll Learn