Scale parameter definition

Question

I have seen two definitions of scale parameter of a distribution: Let $X$ be random variable and $\{{P_{\theta, X }: \theta \in \Theta \subset \mathbb{R}\}}$ be a parametric family of distribution of $X$ indexed by $\theta$

$$1)~~ \theta \text{ is a scale parameter iff} \rightarrow \text{ the distribution of } cX \text{ is in } \{{P_{\theta, X }: \theta \in \Theta \subset \mathbb{R}\}} \text{ for every positive constant c}$$

For example Let $X$ ~ $LogNormal(\theta, 1)~, \theta \in (-\infty, \infty)$ then $cX$ ~ $LogNormal(\theta^*,1)$ where $\theta^* = \theta + ln(c) \in (-\infty, \infty)$. With this definition $\theta$ is a scale parameter.

For the second definition it is required that $\theta$ is positive:

$$2) ~ \theta > 0 \text{ is a scale parameter } iff \rightarrow \text{ the distribution of } X/\theta \text{ is independent of } \theta$$

If $X$ ~ $LogNormal(\theta, 1)~, \theta \in (0, \infty)$ (We restrict the parameter space) then $X/\theta $~ $LogNormal(\theta + ln(1/\theta), 1)$. With this definition $\theta$ is not a scale parameter.

I was wondering which of these two is the correct definition of a scale parameter? Or it depends on the book and the author?

Answer 1

I think both can be seen as expressions of a more general idea of a location-scale family of distributions:

A family of distributions $f(x;\mu,\theta),\; \mu \in \mathbb{R}, \theta > 0$ is a location-scale family iff $g(x):=\frac{1}{\theta}f((x-\mu)/ \theta)$ is free of $\mu, \theta$.

This implies both other definitions

1. Location-scale families are closed under multiplication
2. The location and scale parameters can be sent to 0, 1 (respectively) via subtraction and division (repectively)

So, $\theta$ is a scale parameter if it transforms like a the scale parameter in a location-scale family (after "centering" by subtracting mean).

Essentially, $\theta \propto \sigma$ (the standard deviation of $f(x;\eta, \theta)$

Answer 2

Your (1) and (2) are saying similar but different things. (1) is saying that changing $\theta$ affects the scale of the distribution in some unstated way but keeps the distribution in the same parametric family, while (2) is saying the scale of the distribution is directly proportional to $\theta$. Personally I would regard the version in (1) as potentially leading to confusion. As an illustration, with Gaussian distributions with mean $0$, the standard deviation is a scale parameter in both definitions but the variance does not meet the condition in (2) (physically it is likely to be in the wrong units) and so I would not describe it as a scale parameter.

In the conventional parametrisation of $X \sim \mathrm{LogNormal}(\mu,\sigma^2)$ with density $f(x) =\frac 1 {x\sigma\sqrt{2\pi}}\ \exp\left(-\frac{\left(\log_e( x)-\mu\right)^2}{2\sigma^2}\right)$, I do not think you can helpfully say $\mu$ is a scale parameter: doubling $\mu$ does not double the scale of the distribution of $X$. Similaraly with your $\theta$.

You could for example reparametrise, say to use $\alpha$ and $\beta$ calling this $X \sim \mathrm{LogNormal}^*(\alpha,\beta)$, and have a density $g(x)=\frac 1 {x\sqrt{2\pi\log_e(\beta)}}\ \exp\left(-\frac{\left(\log_e( x/\alpha)\right)^2}{2\log_e(\beta)}\right)$ where $\alpha$ is a scale parameter meeting both your conditions. As an illustration, $X$ has mean $\alpha \sqrt{\beta}$ and variance $\alpha^2\beta(\beta-1)$, so doubling $\alpha$ quadruples the variance, as you might hope for a scale parameter. In terms of the conventional parametrisation, $\alpha= e^\mu$ and $\beta = e^{\sigma^2}$, and so there is a bijection $\mathbb R \to \mathbb R^+$ between $\mu$ and $\alpha$, with one determining the other.

My conclusion from this is that I think you can say that the log-normal distribution family is a scalable family of distributions, since if you simply change the scale then you still have a log-normal distribution, and this is what your (1) is trying to say. But I would also conclude that $\mu$ and your $\theta$ are not scale parameters for the log-normal distribution, and your (2) demonstrates that.