# If a distribution’s scale parameter cannot equal 1, is it part of a scale family?

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 However, I had originally made $$f$$ a Weibull and constrained the scale parameter as above, just for example.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?

• 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 $0 < \lambda \leq 1$. Commented Aug 16, 2019 at 1:23
• 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? Commented Aug 16, 2019 at 5:12

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.