Standardized Moments, Skewness, and Kurtosis

Why Standardize Moments?

Central moments describe variation around a distribution's mean, but their numerical values depend on the units of measurement. For example, measuring a variable in meters rather than centimeters multiplies its third central moment by \(100^3\). Standardized moments remove this scale dependence, allowing shape summaries such as skewness and kurtosis to be compared across distributions measured in different units.

These summaries are meaningful only when the corresponding moments exist. In particular, a distribution may have a finite mean and variance while having no skewness or kurtosis. The moment conditions are therefore part of the definitions, not technical details that can be ignored.

Standardized Moments and Their Existence

Definition: Suppose \(X\) has finite mean \(\mu=E[X]\) and finite, positive variance \(\sigma^2=\operatorname{Var}(X)\). For an integer \(r\geq 1\), the \(r\)th standardized moment is \[ \eta_r=E\left[\left(\frac{X-\mu}{\sigma}\right)^r\right], \] provided \(E[|X-\mu|^r]<\infty\). The standardized variable \(Z=(X-\mu)/\sigma\) has mean \(0\) and variance \(1\).

The absolute-moment condition ensures that the expectation defining \(\eta_r\) is finite. For any positive integer \(r\), if \(E[|X-\mu|^r]<\infty\), then every lower absolute central moment is finite as well. The standard summaries of interest are:

Definition: If \(E[|X-\mu|^3]<\infty\), the skewness is \[ \gamma_1=\eta_3=\frac{E[(X-\mu)^3]}{\sigma^3}. \] If \(E[|X-\mu|^4]<\infty\), the kurtosis is \[ \beta_2=\eta_4=\frac{E[(X-\mu)^4]}{\sigma^4}. \] The excess kurtosis is \(\gamma_2=\beta_2-3\).

A finite positive variance is needed to standardize at all. Skewness additionally requires a finite absolute third central moment; kurtosis requires a finite absolute fourth central moment. Because a finite fourth absolute moment implies a finite third absolute moment, a distribution with defined kurtosis also has defined skewness. By contrast, a distribution can have finite variance but lack both summaries.

How Standardization Affects the Summaries

Standardized moments are unchanged by shifting or rescaling the variable, apart from the sign of skewness under a reversal of direction. If \(Y=a+bX\), where \(a\) is real and \(b\neq 0\), then \[ E[Y]=a+b\mu,\qquad \operatorname{SD}(Y)=|b|\sigma. \] Consequently, \[ \frac{Y-E[Y]}{\operatorname{SD}(Y)} =\frac{b}{|b|}\frac{X-\mu}{\sigma}. \] Its \(r\)th standardized moment is \(\eta_r\) if \(b>0\), and \((-1)^r\eta_r\) if \(b<0\). Thus skewness changes sign when the measurement direction is reversed, while kurtosis and excess kurtosis do not change. This is why these summaries describe shape without depending on the choice of units or origin.

Worked Example: Normal Distribution

Let \(X\sim N(\mu,\sigma^2)\), with \(\sigma>0\), and set \(Z=(X-\mu)/\sigma\). Then \(Z\sim N(0,1)\). Its density is symmetric about zero, so the integrand \(z^3\) times the density is odd and \(E[Z^3]=0\).

For the fourth moment, write \(\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}\), so \(\phi'(z)=-z\phi(z)\). Integration by parts gives \[ E[Z^4]=\int_{-\infty}^{\infty}z^4\phi(z)\,dz =-\int_{-\infty}^{\infty}z^3\phi'(z)\,dz =3\int_{-\infty}^{\infty}z^2\phi(z)\,dz=3. \] The boundary term vanishes because \(z^3\phi(z)\) tends to zero at both ends, and \(E[Z^2]=1\). Therefore the normal distribution has skewness \(0\), kurtosis \(3\), and excess kurtosis \(0\).

Symmetry, Skewness, and Kurtosis

If a distribution is symmetric about its finite mean, and its absolute third central moment is finite, then its skewness is zero. Indeed, symmetry makes the centered variable \(X-\mu\) have the same distribution as \(-(X-\mu)\). The expectation of its odd function \((X-\mu)^3\) must therefore equal the expectation of \(-(X-\mu)^3\), so it is zero.

The converse is not true: zero skewness alone does not establish symmetry. Skewness records one particular signed moment, so different parts of an asymmetric distribution can balance in that moment. Kurtosis, being based on a fourth power, is not directional; it does not distinguish a long right tail from a long left tail.

Worked Example: Uniform Distribution on a Symmetric Interval

Let \(X\) be uniform on \([-\sqrt{3},\sqrt{3}]\). Symmetry gives \(\mu=0\), and \[ E[X^2]=\frac{1}{2\sqrt{3}}\int_{-\sqrt{3}}^{\sqrt{3}}x^2\,dx =\frac{1}{2\sqrt{3}}\left(\frac{2(\sqrt{3})^3}{3}\right)=1. \] Thus \(\sigma=1\). The third moment is zero by symmetry.

The fourth moment is \[ E[X^4]=\frac{1}{2\sqrt{3}}\int_{-\sqrt{3}}^{\sqrt{3}}x^4\,dx =\frac{1}{2\sqrt{3}}\left(\frac{2(\sqrt{3})^5}{5}\right)=\frac{9}{5}. \] Since \(\sigma=1\), the skewness is \(0\), the kurtosis is \(9/5\), and the excess kurtosis is \(9/5-3=-6/5\). This symmetric distribution has lower kurtosis than the normal distribution, despite having no skewness.

An Inequality Linking Skewness and Kurtosis

When both quantities exist, skewness and kurtosis cannot take arbitrary values. The following bound is a useful consistency check on calculations.

Theorem: If \(E[|X-\mu|^4]<\infty\) and \(0<\sigma^2<\infty\), then \[ \beta_2\geq 1+\gamma_1^2. \]

Proof. Let \(Z=(X-\mu)/\sigma\), so \(E[Z]=0\) and \(E[Z^2]=1\). For every real number \(a\), a square has nonnegative expectation: \[ 0\leq E[(Z^2-aZ-1)^2]. \] Expanding and using \(E[Z]=0\) and \(E[Z^2]=1\) gives \[ 0\leq E[Z^4]-2aE[Z^3]+a^2E[Z^2]-2E[Z^2]+2aE[Z]+1 =\beta_2-2a\gamma_1+a^2-1. \] Choose \(a=\gamma_1\). Then \(0\leq\beta_2-\gamma_1^2-1\), which proves the inequality. \(\square\)

For example, this bound says that a large absolute skewness requires sufficiently large kurtosis. It is a necessary condition, not a complete description of which pairs of values can occur.

Worked Example: Exponential Distribution

Let \(X\) have an exponential distribution with rate \(\lambda>0\). Set \(Y=\lambda X\), which has the rate-one exponential density \(e^{-y}\) for \(y>0\). Rescaling by a positive constant does not alter standardized moments, so it is enough to calculate them for \(Y\).

Integration by parts gives \(E[Y^k]=\int_0^\infty y^ke^{-y}\,dy=k!\) for nonnegative integers \(k\). In particular, \(E[Y]=1\), \(E[Y^2]=2\), \(E[Y^3]=6\), and \(E[Y^4]=24\). Hence \(\operatorname{Var}(Y)=2-1^2=1\), so the standard deviation is \(1\).

Expanding the third central moment, \[ E[(Y-1)^3]=E[Y^3]-3E[Y^2]+3E[Y]-1=6-6+3-1=2. \] Expanding the fourth, \[ E[(Y-1)^4]=E[Y^4]-4E[Y^3]+6E[Y^2]-4E[Y]+1 =24-24+12-4+1=9. \] Thus the exponential distribution has skewness \(2\), kurtosis \(9\), and excess kurtosis \(6\). Its positive skewness reflects its longer right side, while the fourth moment summarizes a different aspect of shape and does not by itself specify tail probabilities.

When the Summaries Do Not Exist

Finite variance does not guarantee finite skewness or kurtosis. For a concrete example, a Student \(t\) distribution with \(\nu\) degrees of freedom has density proportional in the tails to \(|x|^{-(\nu+1)}\). Its absolute \(r\)th moment has tail behavior governed by the integral \[ \int_1^\infty x^r x^{-(\nu+1)}\,dx =\int_1^\infty x^{r-\nu-1}\,dx. \] This integral is finite exactly when \(r<\nu\). In particular, for \(2<\nu\leq 3\), the variance is finite but the third absolute moment is infinite, so skewness and kurtosis are not defined. For \(3<\nu\leq 4\), skewness exists but kurtosis does not. The boundary values are excluded because the corresponding integral diverges logarithmically.

Common Mistakes and Interpretation

Ignoring existence conditions. A formula involving a third or fourth central moment does not make that moment finite. Check the relevant absolute moment before assigning skewness or kurtosis.
Calling zero skewness proof of symmetry. Symmetry with a finite third absolute moment implies zero skewness, but the reverse implication does not hold in general.
Confusing kurtosis with a direct tail probability. Kurtosis is a standardized fourth moment. It gives more weight to observations far from the mean, but it is not itself the probability of an extreme observation.
Mixing kurtosis conventions. The normal distribution has kurtosis \(3\) and excess kurtosis \(0\). Always check whether a reported value is \(\beta_2\) or \(\beta_2-3\).

Key takeaway: Skewness and kurtosis are standardized central moments: skewness uses the third moment and kurtosis the fourth. Their definitions require finite absolute moments of the corresponding orders, and their interpretations should be kept distinct from claims about symmetry or tail probabilities.

Check Your Understanding

Use the definitions and conditions above to answer the following.

What moment condition is needed for skewness to exist? What additional condition is needed for kurtosis?
A symmetric distribution has a finite third absolute central moment. What is its skewness, and which part of the argument depends on finiteness?
For a distribution with kurtosis \(4\), what is its excess kurtosis? Does that value alone determine its skewness?
A Student \(t\) distribution has \(\nu=3.5\) degrees of freedom. Which of variance, skewness, and kurtosis exist?
Why does changing measurement units leave kurtosis unchanged but reversing the direction of measurement reverse the sign of skewness?