enter image description here

I have been struggling to convince myself that SD (standard deviation) is about variation or dispersion.

Standard deviation From Wikipedia, the free encyclopedia

In statistics, the standard deviation is a measure of the amount of variation or dispersion of a set of values.

Is it really an appropriate claim to say average distance from the mean is measure of variation or dispersion? It can be, with specific data sets, but is it really appropriate to say so in general?

As stated in Is standard deviation a proper measure of dispersion of data ?, it seems claiming SD as measure for variation or dispersion could be questionable without cautions in what situations it can be.

Standard deviation has become the most widely used measure of dispersion of data. A measure of dispersion can, in the true sense, be regarded as the proper measure of dispersion if the measure is based on the deviations between all pairs of data.

Standard deviation is, however, not based on the deviations between all pairs of data. Accordingly, the perfectness of standard deviation, as measure of dispersion, is questionable. Therefore, the question is " Is standard deviation a proper measure of dispersion of data ? "

I have not been able to understand how is standard deviation, which is not based on the deviations between all pairs of data, a proper measure of dispersion? The explanation is not as clear as to convince that standard deviation, is a proper measure of dispersion.

Is using the word standard appropriate as well? I suppose if each distance is divided by SD, it could be said as standardized, but what is SD itself is standardizing?

If I follow this Standardization explanation:

In statistics, standardization is the process of putting different variables on the same scale. This process allows you to compare scores between different types of variables.

If there are two data sets X and Y, and SD always makes X and Y comparable on the same meaningful scale/unit, then I believe it is. But is it so?


Pearson made up this term in 1894 paper "On the dissection of asymmetrical frequency-curves", here's the pdf.

  • 2
    $\begingroup$ Note that standard deviation is not actually "average distance from the mean". stats.stackexchange.com/questions/142154/… $\endgroup$
    – fblundun
    Dec 24 '20 at 9:34
  • 2
    $\begingroup$ Literally any function that estimates $|X-\mu|$ can be considered a measure of dispersion. Could you therefore elaborate on what you mean by "appropriate"? $\endgroup$
    – whuber
    Dec 24 '20 at 14:48
  • $\begingroup$ Hi @whuber, thanks for the follow up. If I said "X is as dispersed or varied as Y" based on the SD(X) == SD(Y), for instance in a paper to publish, I suppose a reviewer would say, "it is not an appropriate conclusion". I need more substances to say so. If a metrics/measure cannot be 90% sure, for instance one covid test T is only 50% precise, it would not be "appropriate" as I have no confidence in it. Then if I see a statement "T is a test for covid", it is so but the test itself would be of no use. Hence I expect more substances to the statement "SD is a measure for dispersion/variation". $\endgroup$
    – mon
    Dec 24 '20 at 16:42
  • 2
    $\begingroup$ When I write about "dispersion" in a technical paper I stipulate how I measure dispersion, and I cannot see any objection a reviewer possibly could advance in that regard. $\endgroup$
    – whuber
    Dec 24 '20 at 18:23
  • 3
    $\begingroup$ Definitions should be written like dispersion =: signalling that what is on the left is just a name and what is on the right is the machinery that does the work. Many tangles can arise if you take "dispersion" to be a pre-defined property that can only be made precise in one way. As Mosteller and Tukey comment more generally (Data Analysis and Regression) we have many vague concepts in statistics that can be made precise in different ways. $\endgroup$
    – Nick Cox
    Jan 28 at 20:06

I'm not aware of any widely understood set of formal criteria that a purported measure of dispersion has to meet—perhaps it's more that dispersion is a vague, pre-quantitative, notion that has motivated & can apply to various precisely defined measures. The assertion that "a measure of dispersion can, in the true sense, be regarded as the proper measure of dispersion if the measure is based on the deviations between all pairs of data" is intriguing; but its author doesn't provide any rationale (or any reference). This criterion would not in fact rule out the standard deviation: for a sample ${x_1, \ldots, x_n}$, the sum of squared deviations from the mean is equal to the mean of squared pairwise differences,

$$\sum_{i=1}^n{\left(x_i- \frac{\sum_{i =1}^n x_i}{n}\right)^2} = \frac{\sum_{i=1}^n\sum_{j = i + 1}^n(x_i - x_j)^2}{n}$$

(see @whuber's answer to "Why isn't variance defined as the difference between every value following each other?").

† @NickCox made me aware in short order that Bickel & Lehmann (1976), propose such a set of criteria. A measure of dispersion must (1) be location-invariant, (2) be scale-equivariant, & (3), given distributions $F_X$ symmetric about $\mu_X$ & $F_Y$ symmetric about $\mu_Y$ where $|X-\mu_X|$ stochastically dominates $|Y-\mu_Y|$, be greater (or at least equal) for $F_X$ than for $F_Y$.

That's quite intuitive, but to propound criteria for a measure of dispersion for an asymmetric distribution, you'd have to privilege one particular measure of central tendency—conjuring the spectre of mean-dispersion, median-dispersion, &c. Bickel & Lehmann (1979) take a different tack: noting that for symmetric distributions $X$ is more dispersed than $Y$ when

$$ \begin{align} F_X^{-1}(v) - F_X^{-1}\left(\frac{1}{2}\right) &\leq F_Y^{-1}(v) - F_Y^{-1}\left(\frac{1}{2}\right)\quad \forall\, v \leq \frac{1}{2}\\ F_X^{-1}(v) - F_X^{-1}\left(\frac{1}{2}\right) &\geq F_Y^{-1}(v) - F_Y^{-1}\left(\frac{1}{2}\right)\quad \forall\, v \geq \frac{1}{2} \end{align} $$

, they decide that for asymmetric distributions $X$ is more spread out than Y when

$$ F_X^{-1}(v) - F_X^{-1}(u) \geq F_Y^{-1}(v) - F_Y^{-1}(u)\quad \forall\, u < v $$ So any two quantiles of $X$ are farther apart than the corresponding quantiles of $Y$; the same probability mass is spread more thinly.

See the papers for proofs that standard deviation is both a measure of dispersion (for symmetric distributions) & of spread (for both symmetric & asymmetric distributions).

‡ I see the rationale now: it's equivalent to saying that a dispersion measure must be invariant to translation (& to reflection if the deviation is unsigned). From the set of pairwise distances you can reconstruct the original sample (or finite population) $(x_1, \ldots, x_n)$ only with unknown location & orientation: $\pm(x_{(1)}+k, \ldots, x_{(n)} + k)$.


Bickel & Lehmann (1976), Ann. Statist., 4, 6, "Descriptive statistics for nonparametric models. III. Dispersion".

Bickel & Lehmann (1979), in Contributions to Statistics: Jaroslav Hajek Memorial Volume, "Descriptive statistics for nonparametric models, IV. Spread".

  • 3
    $\begingroup$ Bickel, P. J. and Lehmann, E. L. 1976. Descriptive statistics for nonparametric models. III. Dispersion. Annals of Statistics 4: 1139-1158 seems of some relevance at the start. projecteuclid.org/download/pdf_1/euclid.aos/1176343648 should be quite widely accessible (meaning that it can be seen, not that it is an easy read). $\endgroup$
    – Nick Cox
    Jan 28 at 18:30
  • 1
    $\begingroup$ @NickCox: Thank you! I suspected my unawareness might not last long. I'll read this (& "IV. Spread") & update my answer accordingly. $\endgroup$ Jan 28 at 19:35
  • 1
    $\begingroup$ Isn't criterion (2) supposed to be scale covariant? Dispersion had better increase as the scale is increased! $\endgroup$
    – whuber
    Jan 28 at 22:44
  • $\begingroup$ @whuber: Typo, but was meant to be "equivariant". Is "covariant" the same? $\endgroup$ Jan 28 at 22:46
  • $\begingroup$ I wouldn't insist on any standard meaning in statistics so I hesitate to say yes; but among physicists, both "covariant" and "contravariant" have well-established meanings that would translate nicely to stats. $\endgroup$
    – whuber
    Jan 28 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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