If $Y=\min\{ X_{1},X_{2} \}$ and $X_{i} \sim N(0,1)$ then $Y^{2} \sim \chi ^{2}_{(1)}$ If $Y=\min\{ X_{1},X_{2} \}$ and $X_{i} \sim N(0,1)$ then $Y^{2} \sim \chi ^{2}_{(1)}$
If $X \sim  N(0,1) \implies X^{2} \sim \chi ^{2}_{1}$
What about the square?
If we suppose that $X_{1} < X_{2}$ we have the previous proposition and is the same for $X_{2} < X_{1}$.

If $X_1,X_2 \stackrel{i.i.d}{\sim} N(0,1)$ and $Y = \min(X_1,X_2)$, then
$Pr(Y > y) = \prod^{2}_{i=1} Pr(X_i > y)=(1-\phi (y))^2 \implies F_Y(y) = 1 - (1-\phi (y))^2$, so
$f_Y(y) = 2(1-\phi(y))\phi'(y)$
Let $T = Y^2$, then
$Pr(T\leq t) = Pr(Y^2 \leq t) = Pr(-\sqrt{t} \leq Y \leq \sqrt{t}) = F_Y(\sqrt{t})-F_Y(-\sqrt{t})$
$ \implies f_T(t) = \frac{1}{\sqrt{t}} (f_Y(\sqrt{t}) - f_Y(-\sqrt{t})$, so
$f_T(t) = \frac{\phi'(\sqrt{t})}{\sqrt{t}}$ for $t>0$, which is the pdf of chi-square with 1 df.
