If $X$ and $Y$ are independent random variables, with $Z = \min(X,Y),$ prove that $Z^2\sim\chi^2(1),$

Solution 1:

$1-F_Z(t) = P(Z>t) = P(X>t)P(Y>t) =\frac{1}{2\pi}\left[ \int_t^{\infty}\exp(-x^2/2) \, dx \right]^2$. Take derivative w.r.t. $t$ and we can get $$ f_Z(t) = -\frac{d}{dt}\frac{1}{2\pi} \left[ \int_t^\infty \exp(-x^2/2)\,dx \right]^2 = \frac{1}{\pi}\exp(-t^2/2)\left[\int_t^{\infty}\exp(-x^2/2)\,dx\right]. $$ Now let $W = Z^2$ \begin{align} 1-F_W(t) = {} & P(W>t) = P(Z>\sqrt{t})+P(Z<-\sqrt{t})\\[10pt] = {} & \int_{\sqrt{t}}^{\infty}\frac{1}{\pi}\exp(-s^2/2) \left[\int_s^\infty \exp(-x^2/2)\,dx\right]\,ds \\[10pt] & {} + \int_\infty^{-\sqrt{t}}\frac{1}{\pi}\exp(-s^2/2) \left[ \int_s^{\infty}\exp(-x^2/2)\,dx\right]\,ds\\[10pt] = {} & \int_{\sqrt{t}}^{\infty}\frac{1}{\pi}\exp(-s^2/2)\left[ \int_s^\infty \exp(-x^2/2)\,dx \right] \, ds \\[10pt] & {} + \int^\infty_{\sqrt{t}}\frac{1}{\pi}\exp(-s^2/2) \left[ \int^{s}_{-\infty}\exp(-x^2/2)\,dx\right]\,ds\\[10pt] = {} & \int_{\sqrt{t}}^\infty \frac{1}{\pi}\exp(-s^2/2)\frac{\sqrt{2\pi}}{2}\,ds \end{align} Taking derivative we can get $f_W(t) = \frac{1}{\sqrt{2\pi t}}\exp(-t/2)$, which is the same as $f_{\chi^2_1}(t) = \frac{1}{\sqrt{2\pi t}}\exp(-t/2)$.