For a given random variable $X$ with mean $\mu_{\mathrm{old}}$ and standard deviation $\sigma_{\mathrm{old}}$ I would like to perform a transformation $g$ to obtain a new random variable $Y := g(X)$ which has a desired mean and standard deviation $\mu_{\mathrm{new}}$ and $\sigma_{\mathrm{new}}$ respectively without (if possible) changing other characteristics of the distribution (higher moments). A simple calculation shows that the transformation $g$ has the form

$$Y = g(X) := \frac{\sigma_{\mathrm{new}}}{\sigma_{\mathrm{old}}}X-\frac{\sigma_{\mathrm{new}}}{\sigma_{\mathrm{old}}}\mu_{\mathrm{old}} + \mu_{\mathrm{new}}$$

This fulfills the condition above since

$$E[Y] = \frac{\sigma_{\mathrm{new}}}{\sigma_{\mathrm{old}}}E[X]-\frac{\sigma_{\mathrm{new}}}{\sigma_{\mathrm{old}}}\mu_{\mathrm{old}} + \mu_{\mathrm{new}}=\frac{\sigma_{\mathrm{new}}}{\sigma_{\mathrm{old}}}\mu_{\mathrm{old}}-\frac{\mu_{\mathrm{new}}}{\sigma_{\mathrm{old}}}\mu_{\mathrm{old}} + \mu_{\mathrm{new}}=\mu_{\mathrm{new}}$$


$$\operatorname{Var}(Y) = \frac{\sigma^2_{\mathrm{new}}}{\sigma^2_{\mathrm{old}}}\operatorname{Var}(X)=\sigma^2_{\mathrm{new}}.$$

Does this normalization and scaling change other moments? I did some simulation in R and it seems no to be the case, at least for kurtosis and skewness. What about higher moments? Are there any reference for this?


1 Answer 1


Skewness and kurtosis are not actually moments of the random variable itself (either raw or central). They're central moments divided by the corresponding power of $\sigma$, to make them unit-free. Equivalently they're moments of a standardized variable.

e.g. The third raw moment $\mu_3' = E(X^3)$ is affected by both changes of location and scale, the third central moment is $\mu_3=E[(X-\mu)^3]$ is not affected by changes of location (since that's removed when $\mu$ is subtracted) but is affected by change of scale. The skewness, by contrast, is $\mu_3/\sigma^3$; as suggested above it's effectively the third moment of a standardized version of the original random variable.

Let $\mu_k$ be the $k$-th central moment of $X$ and let $Y=c(X-\mu_X)$. Then $E(Y^k)=E([c(X-\mu_X)]^k) = E(c^k[(X-\mu_X)]^k) = c^kE([(X-\mu_X)]^k)=c^k\mu_k$

So while $\mu_3$ is affected by change of scale, $\mu_3/\sigma^3$ (third-moment skewness) is not.

Similar comments apply to raw fourth moments, central fourth moments and kurtosis and the same ideas extend to higher order moments - raw moments are affected by both kinds of change, central moments by scale changes only, and standardized central moments by neither.

  • 2
    $\begingroup$ +1. In terms often used, skewness and kurtosis measure aspects of distribution shape but shape is invariant under changes of location or scale. $\endgroup$
    – Nick Cox
    Commented Jul 6, 2016 at 13:18
  • $\begingroup$ @Glen_b or Nick, do you have any references on this analysis of moments. Most stats texts I have seen never do a deep analysis of moments beyond 2. I am working on a problem where a bit more precision is needed in understanding how scaling affects the skewness of a distribution. $\endgroup$
    – krishnab
    Commented Mar 6, 2019 at 4:18
  • 1
    $\begingroup$ The line of algebra in the answer ($E(Y^k)=E([c(X-\mu_X)]^k) = E(c^k[(X-\mu_X)]^k)$ $= c^kE([X-\mu_X]^k)=c^k\mu_k$) is pretty much the whole shebang; It follows from the fact that $(ab)^n = a^n b^n$, and the linearity of expectation. Some elementary books may have exercises on it but presumably nobody's going to devote more than a few lines to stating it. The skewness will have a factor of the cube of the scale in both numerator and denominator, which cancel, so scale doesn't affect skewness at all; it is unitless. $\endgroup$
    – Glen_b
    Commented Mar 6, 2019 at 5:43
  • $\begingroup$ What if instead of scaling, we take the logarithm of $X$ ? How does skewness and kurtosis change then ? $\endgroup$
    – Our
    Commented Jul 19, 2019 at 19:55
  • 1
    $\begingroup$ That's a whole new question... $\endgroup$
    – Glen_b
    Commented Jul 20, 2019 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.