Joint distribution of maximum and minimum of a bivariate normal distribution

Suppose $$X = (X_1,X_2)^T \sim N(\mu, \Sigma)$$, where $$\mu =(\mu_1,\mu_2)^T$$ and $$\Sigma= \begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{pmatrix}$$. The pdf of $$\min(X_1,X_2)$$ and $$\max(X_1,X_2)$$ is well known. But what is the joint distribution (or pdf) of $$\min(X_1,X_2)$$ and $$\max(X_1,X_2)$$?

Generically, \begin{align} f(x_1,x_2)&=f(x_1,x_2)\Bbb I_{x_1x_2}\\ &=\Bbb P(X_1X_2)\frac{f(x_1,x_2)\Bbb I_{x_1>x_2}}{\Bbb P(X_1>X_2)}\\ &=\Bbb P(X_1X_2)\frac{f(x_{(2)},x_{(1)})}{\Bbb P(X_1>X_2)}\\\end{align}
• Thank you! But I did not get why you multiply and divide $P(X_1 < X_2)$ at the second step. :( – Statisfun Jun 1 '19 at 4:58
Let $$Y_1$$ and $$Y_2$$ be the min and max. $$Pr(Y_1 \le s, Y_2 \le t) = Pr (X_1 \le t, X_2 \le t)- Pr(s \le X_1 \le t, s \le X_2 \le t) = Pr(X_1 \le s, X_2 \le t) +Pr(X_1 \le t, X_2 \le s) - Pr (X_1 \le s, X_2 \le s)$$ Taking derivatives, $$f(s,t) = \phi(s,t)+\phi(t,s),$$ where $$\phi$$ is the pdf of the bivariate normal distribution.
• This cannot possibly be correct, because if $t$ represents the max and $s$ the min, then the density must be zero wherever $t \lt s.$ This is why @Xi'an explicitly includes indicator functions $\mathbb{I}$ in his formula. – whuber Jun 1 '19 at 16:38