Why Measure Deviations from the Mean?
The mean describes a distribution’s center, but it does not describe how observations are arranged around that center. To study that arrangement, take each possible value’s deviation from the mean and average a power of those deviations. The resulting central moments give measures of spread and asymmetry.
The second and third central moments have especially useful interpretations. The second is the variance: it measures spread using squared deviations. The third preserves the signs of deviations when cubed, so it records whether larger positive or negative deviations have greater weighted influence. Neither moment alone gives a complete picture of a distribution, but both provide precise summaries.
For an odd moment such as the third, the expectation must be well-defined; a standard sufficient condition is \(E[|X-\mu|^3]<\infty\). The second central moment must also be finite to define a finite variance. The centering value \(\mu\) matters: the central moment uses deviations from the mean, not powers of the original values.
The Second and Third Central Moments
The second central moment is always nonnegative because every squared deviation is nonnegative. It is zero only when \(X=\mu\) with probability 1; otherwise, at least some probability is assigned to nonzero deviations, and the variance is positive. Because deviations are squared, observations above and below the mean contribute positively to spread.
The third central moment instead retains the sign of each deviation: values above the mean contribute positively and values below it contribute negatively. Cubing also gives relatively greater weight to deviations farther from the mean. Consequently, the sign of \(\mu_3\) describes the net balance of these signed, cubed deviations. A positive value indicates that positive deviations dominate this balance; a negative value indicates that negative deviations dominate it.
The units also differ. If \(X\) is measured in minutes, then \(\mu_2\) is measured in minutes squared and \(\mu_3\) in minutes cubed. For comparisons across scales, a common dimensionless summary is the standardized skewness, defined when the variance is finite and positive and the third absolute central moment is finite.
Why Symmetry Makes the Third Central Moment Zero
A distribution symmetric about its mean has matching probability on either side of the mean. Each deviation \(d\) is balanced by a deviation \(-d\) with the same probability or density. Their contributions to the third central moment cancel, because \(d^3+(-d)^3=0\).
More formally, write \(Z=X-\mu\). Symmetry about \(\mu\) means that \(Z\) and \(-Z\) have the same distribution. If the third moment exists, \(E[Z^3]=E[(-Z)^3]=-E[Z^3]\). Therefore \(E[Z^3]=0\), which is \(\mu_3=0\). This result has a one-way interpretation: symmetry implies a zero third central moment, but a zero third central moment does not prove symmetry. Different positive and negative deviations can cancel even when a distribution is not symmetric.
Worked Example: Central Moments of a Discrete Delay Count
In a hypothetical system, let \(X\) be the number of processing delays recorded for a randomly selected file. Suppose \(X\) has the discrete distribution \(P(X=0)=0.50\), \(P(X=1)=0.25\), and \(P(X=3)=0.25\). Find the mean, second and third central moments, and standardized skewness. This is a probability model, not an insurance valuation; no interest or mortality assumption is involved.
Method. First calculate the mean, then apply the definition of each central moment to the deviations from that mean. This direct method avoids mistaking a raw moment for a central moment.
The mean is \( \mu=E[X]=0(0.50)+1(0.25)+3(0.25)=0+0.25+0.75=1. \) Thus the deviations at \(X=0,1,3\) are \(-1,0,2\).
The second central moment is \( \mu_2=(-1)^2(0.50)+0^2(0.25)+2^2(0.25)=0.50+0+1.00=1.50. \) The variance is \(1.50\) delay-count-squared units.
The third central moment is \( \mu_3=(-1)^3(0.50)+0^3(0.25)+2^3(0.25)=-0.50+0+2.00=1.50. \) The positive value means the positive deviations have the greater net contribution after cubing and weighting.
Since \(\sigma^2=\mu_2=1.50\), the standardized skewness is \( \gamma_1=1.50/(1.50)^{3/2}=1/\sqrt{1.50}\approx0.8165. \) The final value is rounded to four decimal places. As a check, the probabilities sum to \(0.50+0.25+0.25=1\), and the weighted deviations sum to \(-1(0.50)+0(0.25)+2(0.25)=0\), as they must when centered at the mean.
A Continuous Distribution Example
For a continuous model, the probability-weighted average becomes an integral against the density. The same sequence applies: find the mean, calculate deviations from it, and integrate the required power of each deviation.
Worked Example: Moments of an Increasing Density
Suppose a hypothetical measurement \(X\) has density \(f_X(x)=x/2\) for \(0\leq x\leq2\), and density 0 elsewhere. Find its mean, second and third central moments, and standardized skewness. This is a continuous probability model; no interest or mortality assumption is involved.
Method. Use the density integral for each expectation. The density is valid because it is nonnegative on its support and \( \int_0^2 (x/2)\,dx=[x^2/4]_0^2=1. \)
First calculate the mean: \( \mu=\int_0^2 x(x/2)\,dx=\frac12[x^3/3]_0^2=\frac12\cdot\frac83=\frac43. \)
For the second central moment, integrate the squared deviation directly: \( \mu_2=\frac12\int_0^2(x-\frac43)^2x\,dx. \) Expanding the integrand gives \( (x-\frac43)^2x=x^3-\frac83x^2+\frac{16}{9}x. \) Therefore \( \mu_2=\frac12\left([x^4/4]_0^2-\frac83[x^3/3]_0^2+\frac{16}{9}[x^2/2]_0^2\right) =\frac12(4-\frac{64}{9}+\frac{32}{9})=\frac12\cdot\frac49=\frac29. \)
For the third central moment, \( \mu_3=\frac12\int_0^2(x-\frac43)^3x\,dx. \) Expanding gives \( (x-\frac43)^3x=x^4-4x^3+\frac{16}{3}x^2-\frac{64}{27}x. \) Thus \( \mu_3=\frac12\left([x^5/5]_0^2-4[x^4/4]_0^2+\frac{16}{3}[x^3/3]_0^2-\frac{64}{27}[x^2/2]_0^2\right) =\frac12(\frac{32}{5}-16+\frac{128}{9}-\frac{128}{27})=\frac12(-\frac{16}{135})=-\frac{8}{135}. \)
The standardized skewness is \( \gamma_1=\frac{-8/135}{(2/9)^{3/2}}=-\frac{4}{5\sqrt{2}}\approx-0.5657. \) The final value is rounded to four decimal places. Its negative sign agrees with the result for \(\mu_3\): the signed cubic deviations have a negative net balance. The density increases toward 2, but that observation alone is not a calculation of skewness; the integral supplies the precise value.
Central Moments Are Not Raw Moments
A raw moment of order \(r\) is \(E[X^r]\); a central moment of order \(r\) is \(E[(X-\mu)^r]\). The distinction is especially important for the third moment. A raw third moment averages cubes of the original values, while the third central moment averages cubes of deviations from the mean. They answer different questions and generally have different values.
Worked Example: Avoiding a Raw-Moment Substitution
A hypothetical service records a workload \(X\), measured in units, with \(P(X=10)=0.50\), \(P(X=20)=0.30\), and \(P(X=40)=0.20\). Find the mean and the second and third central moments. This discrete distribution is fully specified by the three probabilities; no interest or mortality assumption is involved.
Method. Center the outcomes at the mean before squaring or cubing. The common mistake is to report \(E[X^3]\) as the third central moment.
The probabilities sum to \(0.50+0.30+0.20=1\). The mean is \( \mu=10(0.50)+20(0.30)+40(0.20)=5+6+8=19. \) The deviations are \(10-19=-9\), \(20-19=1\), and \(40-19=21\).
The second central moment is \( \mu_2=(-9)^2(0.50)+1^2(0.30)+21^2(0.20)=40.5+0.3+88.2=129. \)
The third central moment is \( \mu_3=(-9)^3(0.50)+1^3(0.30)+21^3(0.20)=-364.5+0.3+1852.2=1488. \) The weighted deviations check the centering: \((-9)(0.50)+(1)(0.30)+(21)(0.20)=-4.5+0.3+4.2=0\).
For comparison, the raw third moment is \( E[X^3]=10^3(0.50)+20^3(0.30)+40^3(0.20)=500+2400+12800=15700, \) not \(1488\). The requested third central moment is \(1488\) workload-units-cubed because its calculation uses deviations from 19, not the original workload values.
Interpreting the Results with Care
The variance gives a measure of spread around the mean, but it is expressed in squared units. The third central moment adds information about the signed balance of deviations, but its size depends on the units of measurement. Standardized skewness removes those units, making comparisons more interpretable when the underlying variables have different scales.
The sign of the third central moment should not be translated into an overly broad claim. A positive value does not establish that every observation above the mean is far from it, nor does it determine the median or the most likely value. It states only that the probability-weighted cubic deviations have a positive net balance. Likewise, a zero third central moment means that positive and negative cubic contributions balance; it does not prove the distribution is symmetric.
Check Your Understanding
For each problem, calculate from the stated distribution and center the values at the mean.
- A hypothetical count \(X\) equals 0 with probability 0.75 and 2 with probability 0.25. Find \(\mu\), \(\mu_2\), and \(\mu_3\). (Answer: \(\mu=0.5,\ \mu_2=0.75,\ \mu_3=0.75\))
- A random variable is uniform on \([-2,2]\). Find its mean, second central moment, and third central moment. (Answer: \(\mu=0,\ \mu_2=4/3,\ \mu_3=0\))
- A hypothetical measurement has \(P(X=1)=0.20\), \(P(X=2)=0.50\), and \(P(X=4)=0.30\). Find \(\mu\), \(\mu_2\), and \(\mu_3\). (Answer: \(\mu=2.4,\ \mu_2=1.24,\ \mu_3=0.648\))
- A variable takes values 0 and 2 with probability 0.50 each. Find its raw third moment and its third central moment. (Answer: \(E[X^3]=4,\ \mu_3=0\))
- A continuous variable has density \(f_X(x)=3x^2\) on \(0\leq x\leq1\) and 0 elsewhere. Find its mean and variance by integrating the centered square. (Answer: \(\mu=3/4,\ \mu_2=3/80\))