$\newcommand{\E}{\mathrm{E}}$ I have been going over the Hill et al. (1976) paper on Johnson distributions and struggling with some notation that is undefined but I believe might have been considered common knowledge at the time.

The authors define $\sqrt{\beta_1}=\frac{\mu_3}{\sigma^3}$ which leads me to believe that $\mu_n = \E[X^n - \E[X]]$. Therefore skewness is given by $\sqrt{\beta_1}$ and kurtosis by $\beta_2 = \frac{\mu_4}{\sigma^4}$.

In the paper, the authors use the notation $\mu_2$ which would be the variance ($\sigma^2$ seems more familiar).

Hill, I. D., Hill, R. and Holder, R. E., 1976, Fitting Johnson curves by moments. Applied Statistics, 25, 180–189.

(1) They go on to use the notation $\mu_1'$. Is this simply $\mu_1' = \E[X]$?
(2) Does the notation $\mu_n$ just mean $\mu_n = \E[X^n - \E[X]], \; n\ge 2\;$?

  • 1
    $\begingroup$ 1. See en.wikipedia.org/wiki/Central_moment, (& maybe en.wikipedia.org/wiki/…) ... 2. while it's clearly an inconsistent use of notation, the first raw moment, $\mu'_1$ is often written $\mu_1$ (or often just $\mu$). That is, while $'$ is used for raw moments and omitted for central moments, in the case of the first moment, an omitted dash refers to the same thing as having the dash. (the first central moment being 0 any time the mean is finite so it's not so surprising that for $\mu_1$ the $'$ is often dropped without confusion) $\endgroup$
    – Glen_b
    Aug 15, 2018 at 2:20
  • $\begingroup$ Thanks @Glen_b, these links are helpful (hadn't seen them) $\endgroup$ Aug 19, 2018 at 17:28
  • $\begingroup$ Further review of the paper revealed the author also used $\bar x$ which I take to be the sample mean. Elsewhere in the paper the term $\mu_1'$ is also used to designate the target mean (obtained by the sample mean). $\endgroup$ Aug 29, 2018 at 16:29
  • $\begingroup$ $\bar{x}$ is conventional for an observed sample mean (we'd refer to a collection of random variables $X_1, X_2, ... X_n$ and a collection of observed realizations on them - observations - $x_1, x_2, ..., x_n$, for which $\bar{x}=(x_1+x_2+...+x_n)/n$. This is very much conventional in statistics. $\endgroup$
    – Glen_b
    Aug 30, 2018 at 0:13
  • $\begingroup$ @Glen_b, thank you. I wrote that comment because in older statistics papers I've seen notation ($\mu_1',\bar x,$ etc) used without definition and for the same concepts within the same paper and wanted to note that for anyone else who struggles with an old school paper. Thanks again! $\endgroup$ Aug 30, 2018 at 13:57

1 Answer 1


My assumption would be that $\mu_n = \mathbb{E}[(X-\mathbb{E}(X))^n]$. For $n=2$ this is also referred to as the "second central moment" or the variance. Then the notation $\mu_n^\prime = \mathbb{E}(X^n)$ often referred to as the $n$th non-central moment.


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