I am trying to understand the exact definition of the location / scale / shape parameters (e.g. $a$ is called the shape parameter and $c$ is scale parameter in Pareto Type I). But the books I referred to (The Cambridge Dictionary of Statistics, HMC's Introduction to Mathematical Statistics, Feller's An Introduction to Probability Theory and its Applications, etc) only (seemingly) provided descriptive definitions for these parameters (location parameter are called centering parameter in Feller's). Wikipedia provided definitions in terms of cdf and pdf but without any sources given.

Based on the concepts in non-parametric statistics (say Ch.10 of HMC) I suspect the location / scale / shape parameters can be defined as the following:

Let $X$ be a random variable with cdf $F_X$. A parameter $\theta=T(F_X)$, where $T$ is a functional, is a location parameter if \begin{align*}T(F_{X+a})&=T(F_X)+a,&&\forall a\in\mathbb{R},\\ T(F_{aX})&=aT(F_X),&&\forall a\neq0;\end{align*} and it is a scale parameter if \begin{align*}T(F_{aX})&=aT(F_X),&&\forall a>0,\\ T(F_{X+b})&=T(F_X),&&\forall b\in\mathbb{R},\\ T(F_{-X})&=T(F_X);\end{align*} and it is a shape parameter if it is neither location nor scale.

Am I correct? Or did I confused some unrelated concepts?

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    $\begingroup$ My intuition is that the existence of such a functional is unique up to a parameterization of the parameters. I am not sure if such functional always exists, and they are not necessarily linear. Also, I think that the second property for the location parameter is not needed. $\endgroup$
    – Gumeo
    Oct 1, 2015 at 7:03
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    $\begingroup$ @Guðmundur If the second property of a location parameter is not assumed and $T$ is a functional satisfying the first property, then for all real $b$, $T + b$ (defined as $(T+b)(F) = T(F)+b$ for all distributions $F$) also satisfies the first property, whence $T$ would not be unique. $\endgroup$
    – whuber
    Oct 1, 2015 at 12:29
  • $\begingroup$ @whuber I missed that... I agree with you. $\endgroup$
    – Gumeo
    Oct 1, 2015 at 12:54
  • $\begingroup$ @whuber But should $T$ be unique? E.g. for symmetric distribution, $\mu$ is mean and median, which are distinct functionals. $\endgroup$
    – Francis
    Oct 1, 2015 at 20:02
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    $\begingroup$ @Francis On the set of symmetric distributions for which both the mean and median are defined, they agree, so they can be considered the same functional. Nevertheless I think you are right to challenge the implication that location parameters should be unique--they need not be. $\endgroup$
    – whuber
    Oct 1, 2015 at 20:13

1 Answer 1


It is often true that these correspond to (some function of) the first, second and third moment as noted by @GuðmundurEinarsson. However, there are exceptions: For example for a Cauchy distribution Evans, Hastings, and Peacock (2000) call the first parameter a location parameter, but it represents the median instead of the mean. The mean is not even defined for a Cauchy distribution.

A more encompasing but less precise description would be:

  • the location parameter shifts the entire distribution left or right
  • The scale parameter compresses or stretches the entire distribution
  • the shape parameter changes the shape of the distribution in some other way.

Merran Evans, Nicholas Hastings, and Brian Peacock (2000) Statistical Distributions, third edition. Wiley.

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    $\begingroup$ In the case of median we can define $T(F_X)=F_X^{-1}(1/2)$ (given inverse exists), which is a location functional since $$1/2=P(X+a<F_{X+a}^{-1}(1/2))=P(X+a<F_{X}^{-1}(1/2)+a)=1/2$$ and for $a>0$ (easy to check for $a<0$): $$1/2=P(aX<F_{aX}^{-1}(1/2))=P(aX<aF_{X}^{-1}(1/2))=1/2.$$ I suspect this is how location parameter being defined for Cauchy distribution (that's what you mean by Gauchy, right?). $\endgroup$
    – Francis
    Oct 1, 2015 at 8:54
  • $\begingroup$ My mistake: Gauchy should indeed have been Cauchy. I have edited the answer to fix it. $\endgroup$ Oct 1, 2015 at 9:28
  • $\begingroup$ @Maarten you might want to change your answer, I deleted mine to not invoke further confusion. $\endgroup$
    – Gumeo
    Oct 1, 2015 at 13:16

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