How do you show that the mean of a Normal Distribution is a scale parameter?

Question

Suppose I have $$X_1,X_2,...,X_n $$ that are iid random variables following a $N(\mu ,\mu^2 )$ distribution, with $\mu>0$

I want to go about proving that $\mu$ is a scale parameter.

Is the only way to do this to use the density function $f(x|\mu)$, and why does $\mu$ have to be greater than $0$?

Answer 1

$\mu$ is a scale parameter if the distribution of $\frac{X}{\sigma} $ does not depend on $\sigma$. In other words: The standard deviation of $\frac{X}{\sigma} $ has to be a constant.

$P(Z<z)=P\left( \frac{X}{\mu}<z \right)=P(X<z\cdot \mu)$

$$\int_{-\infty}^{ z\mu} \frac{1}{\sqrt{2\pi}\mu}e^{-\frac{1}{2}\left(\frac{x-\mu}{\mu}\right)^2}\,dx$$

Now we can substitute

$u\mu=x$ differentiating gives $\mu \cdot du= dx$

$$\int_{-\infty}^{ z} \frac{1}{\sqrt{2\pi}\mu}e^{-\frac{1}{2}\left(u-1\right)^2}\cdot \mu \, du$$

Cancelling $\mu$.

Therefore $Z\sim \mathcal N(1,1)$ and $\sigma$ is $1$ (constant).