In general if $f$ is a scale family we have that if $X\sim f(x\mid\lambda)$ then $\frac{X}{\lambda}\sim f(x\mid 1)$.

However what if $f$ has the constraint that its scale parameter $\lambda \in (1, \infty ).$ So scale cannot be $1.$

Now what is the distribution of $\frac{X}{\lambda}$? We would think it would be $f(x|1)$ with scale $1,$ but as mentioned before the scale cannot equal 1.

(Updated to address the comments - I think my original Weibull example did not add much to the question, so just sticking with general $f$. However, I had originally made $f$ a Weibull and constrained the scale parameter as above, just for example. I include this just because it is mentioned in the accepted answer below.)

Why constrain the scale? I am not sure - I just found a problem where it asked to prove that $f(x|\lambda)$ is a distribution, but added the constraint above. Using a location scale theorem, we can prove this by just showing that $f(x|1)$ is a distribution, which is sometimes easier. However, can we still do this if the constraint is present?

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    $\begingroup$ Why would you constrain $\lambda > 1$? The point of the constraint in the definition of the distribution is that, for the range of values given, the distribution is a proper one, i.e., $\geq 0$ everywhere, integrates to 1. If you happen to know that $\lambda > 1$ a priori for something that you are working on, that's going to lead to a constraint on estimated values from a sample, but from the point of view of the distribution, it's still well-defined for $0 < \lambda \leq 1$. $\endgroup$
    – jbowman
    Commented Aug 16, 2019 at 1:23
  • $\begingroup$ Where do you find a definition of anything called $\mathrm{Weibull}'$ in the linked Wikipedia article or anything that says $\lambda>1$? Could you tell us what the definition there says? $\endgroup$ Commented Aug 16, 2019 at 5:12

1 Answer 1


The Weibull distribution $f(x \vert k,\lambda) = k x^{k-1} \lambda^{-k} e^{-(x/\lambda)^k}$ belongs to the scale family. Because when the space of parameters is $\lambda,k \in (0,\infty)$, then for any variable $X$ that is distributed as this Weibull distribution then some scaled version of this variable $c \cdot X$, for any value $c \in (0,\infty)$ will also be Weibull distributed (and more in particular the Weibull distribution of the scaled variable $c \cdot X$ will be just relating to a scaling of the parameter $\lambda$).

If for whatever reason you consider an altered, more restricted, version of this Weibull distribution where the scaling parameter $\lambda$ is not in the range $(0,\infty)$, then you will not have a distribution that is in the scale family anymore. That is because this restricted Weibull distribution is not anymore closed under scaling, which is a requirement for being distribution that is part of the scale family of probability distributions.

Say you consider $X \sim f(x \vert k=1, \lambda=2)$ then $X/2 \sim f(x \vert k=1, \lambda=1)$, but when you restrict the set of distributions such that $f(x \vert k=1, \lambda=1)$ aint in it then this means that this restricted set of distributions is not part of the scale family.


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