Why do we have $(\frac{\bar{y}-\mu}{\sigma/\sqrt{n}})^2\sim \chi_1^2 \,$?

Question

For $y_i\sim$ i.i.d. $\mathcal{N}(\mu, \sigma^2)$ for $i=1,2,\dots, n$, let $\bar{y}=\frac{\sum_i x_i}{n}$. We have $\bar{y}\sim\mathcal{N}(\mu, \frac{\sigma^2}{n})$. Why do we have $$(\frac{\bar{y}-\mu}{\sigma/\sqrt{n}})^2\sim \chi_1^2 \,$$?

Answer 1

First notice that if $\bar{y}\sim\mathcal{N}(\mu, \frac{\sigma^2}{n})$, then $z:=\frac{\bar{y}-\mu}{\sigma/\sqrt{n}}\sim\mathcal{N}(0, 1)$. It is a well known result that, for a normal random variable, substracting the mean and dividing by the standard deviation yields a standard normal random variable.
To prove it, you can use that for $X\sim \mathcal{N}(\mu, \sigma^2)$, $X + a\sim \mathcal{N}(\mu+a, \sigma^2) $ and $b\cdot X\sim \mathcal{N}\left(b\cdot\mu,\ b^2\cdot\sigma^2\right) $

Once you know that $z$ is standard normal, then you can conclude that $z^2$ is $\chi_1^2$ distributed, as the square of a standard normal random variable is always $\chi_1^2$ distributed, as very nicely shown here.