If $X$ and $Y$ are independent random variables, with $Z = \min(X,Y),$ prove that $Z^2\sim\chi^2(1),$ 
Let $X \sim N (0, 1)$ and $Y ∼ N (0, 1)$ be two independent random variables, and define $Z = \min(X, Y )$. Prove that $Z^2\sim\chi^2(1),$ i.e. Chi-Squared with degree of freedom $1.$

I found the density functions of $X$ and $Y,$ as they are normally distributed. How would one use the fact that $Z = \min(X,Y)$ to answer the question? Thanks!
 A: $1-F_Z(t) = P(Z>t) = P(X>t)P(Y>t) =\frac{1}{2\pi}\left[ \int_t^{\infty}\exp(-x^2/2) \, dx \right]^2$. 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 $f_{\chi^2_1}(t) = \frac{1}{\sqrt{2\pi t}}\exp(-t/2)$.
A: Since $X$ and $Y$ are independent, we can see that $f(x,y) = f(x)f(y)$. 
Now, if we transform our axis from $X,Y$ to $X_1,Y_1$,  where $X_1 = \frac{1}{\sqrt{2}}(X+Y)$ and $X_2 = \frac{1}{\sqrt{2}}(X-Y)$ - essentially, we are rotating the $XY$ space by a +45 degrees. 
Since the distribution $f(x,y)$ has radial symmetry, we can see that $f(x_1,x_2) =f(x,y) $. From this we can see that, even $X_2$  $∼ N (0, 1)$. 
Now, the variable $Z = min(X,Y)$ lies to the space right of the line $\frac{1}{\sqrt{2}}(X-Y)$ in the $XY$ space. 
Therefore, distribution of $Z^2$ is the same as the distribution of $X_2^{2}$ - which is essentially a Chi-square distribution with degrees of freedom = 1. 
A: Let $\Phi$ be the standard normal cdf. First, find the cdf of $Z^2$. For any $z>0$,
$$
\begin{align}
P(Z^2\le z)
&=P(Z>-\sqrt{z})-P(Z>\sqrt{z})\\
&=P(X>-\sqrt{z})P(Y>-\sqrt{z})-P(X>\sqrt{z})P(Y>\sqrt{z})\\
&=(1-\Phi(-\sqrt{z}))^2-(1-\Phi(\sqrt{z}))^2.
\\&=(\Phi(\sqrt{z}))^2-(1-\Phi(\sqrt{z}))^2.
\\&=2\Phi(\sqrt z)-1
\end{align}
$$
On the other hand,
$$
P(X^2\le z)=P(X\le \sqrt{z})-P(X<-\sqrt{z})=\Phi(\sqrt{z})-\Phi(-\sqrt{z})=\Phi(\sqrt{z})-(1-\Phi(\sqrt{z}))
$$
As you can see, $P(X^2\le z)=P(Z^2\le z)$ for all $z$, QED.
