Moment Transformations in Scalar Location-Scale Families

Why Track Moments Under Location and Scale?

A location-scale transformation changes the numerical values of a random variable in a simple way: \(X=\mu+\sigma Z\) shifts \(Z\) by \(\mu\) and multiplies it by \(\sigma\). The distribution’s center and spread therefore change, but the effect on higher moments is not always as obvious. The key is to distinguish raw moments, taken about zero, from central moments, taken about the mean.

We use the usual convention \(\sigma>0\) for a scale parameter. The reference variable \(Z\) need not have mean zero or variance one. Writing the formulas for a general \(Z\) makes clear which conclusions depend on the reference distribution and which follow just from shifting and scaling.

Definition: The \(r\)th raw moment of \(Z\) is \(E[Z^r]\), when it exists. If \(m=E[Z]\), the \(r\)th central moment is \(c_r=E[(Z-m)^r]\), when it exists. In particular, \(c_2=\operatorname{Var}(Z)\). The standardized skewness is \(c_3/c_2^{3/2}\), and the kurtosis is \(c_4/c_2^2\), provided the required moments exist and \(c_2>0\). Excess kurtosis is kurtosis minus \(3\).

Raw Moments: Expand About Zero

Let \(r\) be a nonnegative integer, and suppose \(E[|Z|^r]<\infty\). This condition ensures that the moments \(E[Z^k]\) exist for every integer \(k\) from \(0\) through \(r\). The binomial theorem gives a formula for the raw moment of \(X=\mu+\sigma Z\).

Theorem: If \(E[|Z|^r]<\infty\), then

E[X^r] =\sum_{k=0}^{r}\binom{r}{k}\mu^{r-k}\sigma^k E[Z^k].

To prove the formula, expand the power and take expectations. Every term is integrable under the stated moment condition, and the sum is finite, so expectation distributes over it:

X^r=(\mu+\sigma Z)^r =\sum_{k=0}^{r}\binom{r}{k}\mu^{r-k}\sigma^k Z^k, \qquad E[X^r] =\sum_{k=0}^{r}\binom{r}{k}\mu^{r-k}\sigma^k E[Z^k].

This formula is useful when raw moments of \(Z\) are available. For example, its first two cases give \(E[X]=\mu+\sigma E[Z]\) and \(E[X^2]=\mu^2+2\mu\sigma E[Z]+\sigma^2E[Z^2]\). A shift contributes to every raw moment; it does not simply disappear from them.

Worked Example: Raw Moments of a Shifted and Scaled Exponential Variable

Let \(Z\) have an exponential distribution with rate \(1\), so its density is \(e^{-z}\) for \(z>0\), and set \(X=10+3Z\). First, for each integer \(k\geq 0\),

E[Z^k]=\int_0^\infty z^k e^{-z}\,dz=k!.

The integral equals \(k!\) by repeated integration by parts, starting from \(\int_0^\infty e^{-z}\,dz=1\); the boundary terms vanish because \(z^k e^{-z}\) tends to zero as \(z\) tends to infinity. Thus \(E[Z]=1\), \(E[Z^2]=2\), and \(E[Z^3]=6\). Applying the raw-moment formula,

\begin{aligned} E[X] &= 10+3(1)=13,\\ E[X^2] &= 10^2+2(10)(3)(1)+3^2(2)=100+60+18=178,\\ E[X^3] &= 10^3+3(10^2)(3)(1)+3(10)(3^2)(2)+3^3(6)\\ &=1000+900+540+162=2602. \end{aligned}

As a check on the second raw moment, \(E[X^2]-E[X]^2=178-169=9\), the variance calculated directly from the scale below.

Central Moments: Center First, Then Scale

Raw moments include the origin in their definition, so a location shift affects them. Central moments instead subtract the variable’s own mean. If \(E[|Z|]<\infty\), then \(E[X]=\mu+\sigma m\), where \(m=E[Z]\). Consequently,

X-E[X] =(\mu+\sigma Z)-(\mu+\sigma m) =\sigma(Z-m).

This identity yields the central-moment transformation directly.

Theorem: Let \(r\geq 1\) be an integer, suppose \(E[|Z|^r]<\infty\), and write \(m=E[Z]\). For \(X=\mu+\sigma Z\) with \(\sigma>0\),

E[(X-E[X])^r]=\sigma^r E[(Z-m)^r].

Indeed, \(E[X]=\mu+\sigma m\), and hence \((X-E[X])^r=\sigma^r(Z-m)^r\). Taking expectations gives the stated result. The expectation is finite: \(|Z-m|^r\leq 2^{r-1}(|Z|^r+|m|^r)\) for \(r\geq1\), so the assumed absolute moment suffices. \(\square\)

For \(r=2\), \(3\), and \(4\), this gives

E[X]=\mu+\sigma m,\qquad \operatorname{Var}(X)=\sigma^2c_2,\qquad E[(X-E[X])^3]=\sigma^3c_3,\qquad E[(X-E[X])^4]=\sigma^4c_4.

The variance formula requires only a finite second moment. The third- and fourth-central-moment formulas require the corresponding absolute moments. To define skewness, require finite third absolute moment and positive variance; to define kurtosis, require finite fourth absolute moment and positive variance. These conditions prevent ratios from being used when their numerators or denominators are not well-defined.

Worked Example: Mean, Variance, and Shape for a Normal Variable

Let \(Z\sim N(0,1)\) and \(X=5+2Z\). The standard normal has mean \(0\), variance \(1\), third central moment \(0\) by symmetry, and fourth central moment \(3\). Therefore,

E[X]=5+2(0)=5,\qquad \operatorname{Var}(X)=2^2(1)=4,

E[(X-E[X])^3]=2^3(0)=0,\qquad E[(X-E[X])^4]=2^4(3)=48.

The transformed skewness is \(0/4^{3/2}=0\), and its kurtosis is \(48/4^2=3\). The location shift changes the mean, while the scale changes the variance and the unstandardized central moments.

Skewness and Kurtosis Under Transformation

Let \(v=c_2=\operatorname{Var}(Z)>0\), with \(c_3\) and \(c_4\) finite as needed. Since \(\sigma>0\), the variance of \(X\) is \(\sigma^2v\). Substituting the central-moment formulas into the definitions gives

\operatorname{Skewness}(X) =\frac{\sigma^3c_3}{(\sigma^2v)^{3/2}} =\frac{c_3}{v^{3/2}}, \qquad \operatorname{Kurtosis}(X) =\frac{\sigma^4c_4}{(\sigma^2v)^2} =\frac{c_4}{v^2}.

Thus a location shift changes neither skewness nor kurtosis, and a positive scale change cancels from both standardized ratios. Excess kurtosis also stays the same because it is kurtosis minus the same constant, \(3\). The cancellation depends on standardizing by the transformed variance; the central moments themselves do change with scale.

Worked Example: Shape of a Shifted and Scaled Exponential Variable

Continue with \(Z\) exponential with rate \(1\) and \(X=10+3Z\). The raw moments already calculated give \(E[Z]=1\) and \(\operatorname{Var}(Z)=2-1^2=1\). The central third moment of \(Z\) is

$$ E[(Z-1)^3] =E[Z^3]-3E[Z^2]+3E[Z]-1 =6-3(2)+3(1)-1=2. $$

The fourth central moment is

\begin{aligned} E[(Z-1)^4] &=E[Z^4]-4E[Z^3]+6E[Z^2]-4E[Z]+1\\ &=24-4(6)+6(2)-4(1)+1=9. \end{aligned}

For \(X\), the mean is \(13\), the variance is \(3^2(1)=9\), and the third and fourth central moments are \(3^3(2)=54\) and \(3^4(9)=729\). Therefore, its skewness is \(54/9^{3/2}=54/27=2\), and its kurtosis is \(729/9^2=729/81=9\). The location and positive scale have left both standardized shape measures unchanged.

Worked Example: A Transformed Bernoulli Variable

Let \(Z\) be Bernoulli with \(P(Z=1)=1/4\) and \(P(Z=0)=3/4\), and set \(X=2+4Z\). The values of \(X\) are \(2\) with probability \(3/4\) and \(6\) with probability \(1/4\), so

E[X]=\frac34(2)+\frac14(6)=3,\qquad \operatorname{Var}(X)=\frac34(2-3)^2+\frac14(6-3)^2=3.

For the reference variable, \(E[Z]=1/4\), and direct calculation gives

c_2=\frac{3}{16},\qquad c_3=\frac34\left(-\frac14\right)^3+\frac14\left(\frac34\right)^3=\frac{3}{32},

c_4=\frac34\left(-\frac14\right)^4+\frac14\left(\frac34\right)^4=\frac{21}{256}.

Scaling these central moments by \(4^2\), \(4^3\), and \(4^4\) gives \(X\)'s variance \(3\), third central moment \(6\), and fourth central moment \(21\). Its skewness is \(6/3^{3/2}=2/\sqrt{3}\), and its kurtosis is \(21/3^2=7/3\). Directly, the third central moment is \(\frac34(2-3)^3+\frac14(6-3)^3=6\), and the fourth is \(\frac34(2-3)^4+\frac14(6-3)^4=21\), confirming the scaled results.

Common Pitfall: Applying the Raw-Moment Rule to Central Moments

A tempting but incorrect argument is that shifting by \(\mu\) adds \(\mu\) to every moment. That statement confuses the raw-moment formula with central moments: the first raw moment is the mean, but a central moment is calculated after subtracting the transformed mean. In particular, variance does not equal \(\mu+\sigma\operatorname{Var}(Z)\). The correct calculation centers first:

X-E[X]=\sigma(Z-E[Z]), \qquad \operatorname{Var}(X)=\sigma^2\operatorname{Var}(Z).

A second common error is to say that scale leaves every moment unchanged. Scale multiplies the \(r\)th central moment by \(\sigma^r\); it is only the standardized skewness and kurtosis ratios that remain unchanged under a positive scale. If a nonzero multiplier \(a\) of either sign were allowed instead of the scale convention \(\sigma>0\), the skewness would be multiplied by \(\operatorname{sign}(a)\), while kurtosis would still be unchanged.

Key takeaway: Raw moments follow the binomial expansion of \(\mu+\sigma Z\). Central moments follow from \(X-E[X]=\sigma(Z-E[Z])\), so order \(r\) is multiplied by \(\sigma^r\). With positive scale and positive variance, standardized skewness and kurtosis are unchanged.

Check Your Understanding

Use the moment-existence conditions as well as the transformation formulas in your answers.

If \(E[Z]=2\), \(\operatorname{Var}(Z)=5\), and \(X=7+3Z\), find \(E[X]\) and \(\operatorname{Var}(X)\).
State the condition on \(Z\) that guarantees the raw-moment formula for \(E[X^4]\) applies.
If \(c_3=4\), \(c_2=9\), and \(X=\mu+2Z\), find the third central moment and skewness of \(X\).
Explain why a shift affects raw moments but not central moments.
If \(Z\) has kurtosis \(5\) and positive variance, what is the kurtosis of \(X=\mu+\sigma Z\) for \(\sigma>0\)?