Let me take $Z\sim \mathcal{N}(0,1)$. Then I know that $Z^2\sim \chi^2_1$. Now I want to show that the quantiles $\left(z_{1-\frac{\alpha}{2}}\right)^2=\chi_{1,1-\alpha}^2$ are equal.
We know by definition that $\Bbb{P}(Z^2>\chi^2_{1,1-\alpha})=1-(1-\alpha)=\alpha$. Thus we deduce by taking the square root that $\alpha=\Bbb{P}\left(|Z|>\sqrt{\chi^2_{1,1-\alpha}}\right)=\Bbb{P}\left(Z>\sqrt{\chi^2_{1,1-\alpha}}\right)+\Bbb{P}\left(Z<-\sqrt{\chi^2_{1,1-\alpha}}\right)=2\Bbb{P}\left(Z>\sqrt{\chi^2_{1,1-\alpha}}\right)$. Hence $\Bbb{P}\left(Z>\sqrt{\chi^2_{1,1-\alpha}}\right)=\frac{\alpha}{2}$. But this means by definition that $\sqrt{\chi^2_{1,1-\alpha}}=z_{1-\frac{\alpha}{2}}$ and thus we deduce $\left(z_{1-\frac{\alpha}{2}}\right)^2=\chi_{1,1-\alpha}^2$.
Does this work or am I wrong?
