Why $(n-1)S^2=σ^2(z_1^2+z_2^2+...+z_{n-1}^2)$?

Question

This question is about the Wishart distribution:

I can do the

$(n-1)\dfrac{Σ_{i=1}^n(x_i-x̄)^2}{n-1}=Σ_{i=1}^n(x_i-x̄)^2$

$=\dfrac{Σ_{i=1}^N(x_i-x̄)^2}{N}((\dfrac{x_1-x̄}{A})^2+(\dfrac{x_2-x̄}{A})^2+...+(\dfrac{x_{n-1}-x̄}{A})^2)$

But I do not know what the A is.

Please tell me how to prove $(n-1)S^2=σ^2(z_1^2+z_2^2+...+z_{n-1}^2)$

Thank you very much.

Answer 1

In a gaussian model it is well known that

$\frac{(n-1)S^2}{\sigma^2}\sim \chi_{(n-1)}^2 $

So

$(n-1)S^2\sim \sigma^2 \chi_{(n-1)}^2 $

That's all because the $\chi_{(n-1)}^2$ is defined as the sum of $(n-1)$ independent squared standard gaussian

It is trivial that, if $Z$ is standard gaussian, $\sigma Z \sim N(0;\sigma^2)$