Suppose that $X$ and $Y$ are independent variables of distribution $N(0,1)$, and $Z=\min(X,Y)$. Show that $Z^2 ∼ χ^2 (1)$

Question

Suppose $X$ and $Y$ are independent of distribution $N(0,1)$, and $Z=\min(X,Y)$. Show that $Z^2 ∼ χ^2 (1)$ (even though it doesn't mean that Z∼N(0,1))

I know that:

F_z(t) = P(min(X,Y)≤t) = 1−P(X>t)P(Y>t) =1−(1−P(X≤t))(1−P(Y≤t)) = 1−(1−F_X(t))(1−F_Y(t))

But I get confused at how to make this equation equal to $χ^2 (1)$

Answer 1

$$ F_{M}(m)=P\left(M\leq m\right)=P\left(\min\left\{X,Y\right\}^{2}\leq m\right)=P\left(Y\leq X,Y^{2}\leq m\right)+P\left(Y\leq X,X^{2}\leq m\right)=2P\left(Y\leq X,Y\leq m^{2}\right)=2\frac{1}{2}P\left(Y^{2}\leq m\right)=P\left(Y^{2}\leq m\right) $$ since $Y\sim\mathcal{N}(0,1)$ then it follows by definition that $M\sim\chi^{1}(1)$.