Why the distribution $Y/\sigma$ does not depend on $\sigma$?

Question

This book describes the scale parameter as below:

Suppose $\sigma$ is a scale parameter, in the sense that $p(y|\sigma)=\sigma^{-1}f(y/\sigma)$ for some function $f$, so that the distribution of $Y/\sigma$ does not depend on $\sigma$.

My question is:

Why "not dependent"? I think from the equation $p(y|\sigma)=\sigma^{-1}f(y/\sigma)$, we can know nothing about the relationship between $Y/\sigma$ and $\sigma$.

A normal distribution is tried for $Y$:

$p(y|\sigma)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(y-\mu)^{2}}{2\sigma^{2}}}$

To scale it, we get:

$p(y|\sigma)=\sigma^{-1}f(y/\sigma)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(y/\sigma-\mu/\sigma)^{2}}{2}}$

So, $f(y/\sigma)$ is: $\frac{1}{\sqrt{2\pi}}e^{-\frac{(y/\sigma-\mu/\sigma)^{2}}{2}}$

Denote $y/\sigma$ as z, then $$p(z|\sigma)=\sigma^{-1}\times\frac{1}{\sqrt{2\pi}}e^{-\frac{(z-\mu/\sigma)^{2}}{2}}\tag{1}$$

I think the distribution of $Z$ (i.e., $Y/\sigma$) still depend on $\sigma$, because in Equation (1) there is $\sigma^{-1}$?

Answer 1

The confusion stems from not processing the change of variable correctly. If one sets$$Z=\sigma^{-1}Y$$the density $f_Z$ of $Z$ writes as $$f_Z(z)=f_Y(\underbrace{\sigma z}_{y(z)}) \times \underbrace{\left|\frac{\text{d}y(z)}{\text{d}z}\right|}_\text{Jacobian}=\sigma^{-1}f(\sigma z / \sigma) \times \sigma = f(z)$$ and is therefore not dependent $Y\sim\mathcal{N}(0,\sigma^2)$ then $Z\sim\mathcal{N}(0,1)$. The example $Y\sim\mathcal{N}(\mu,\sigma^2)$ does not work because the density of $Y$ is then indexed or parameterised by both $\mu$ and $\sigma$.

Answer 2

If f(y/σ) was such that σ only appeared in f(y/σ) as a factor multiplying the whole thing, i.e. if f(y/σ) could be written as f(y/σ) = σ g(y), then f(y/σ)/σ wouldn't depend on σ.