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 $\max(X_1,X_2)$?
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$\begingroup$ You may also want to see this (joint distribution of order statistics): en.wikipedia.org/wiki/…. $\endgroup$– Minus One-TwelfthJun 1, 2019 at 5:05
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2 Answers
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Generically, \begin{align} f(x_1,x_2)&=f(x_1,x_2)\Bbb I_{x_1<x_2}+f(x_1,x_2)\Bbb I_{x_1>x_2}\\ &=\Bbb P(X_1<X_2)\frac{f(x_1,x_2)\Bbb I_{x_1<x_2}}{\Bbb P(X_1<X_2)}+\Bbb P(X_1>X_2)\frac{f(x_1,x_2)\Bbb I_{x_1>x_2}}{\Bbb P(X_1>X_2)}\\ &=\Bbb P(X_1<X_2)\frac{f(x_{(1)},x_{(2)})}{\Bbb P(X_1<X_2)}+\Bbb P(X_1>X_2)\frac{f(x_{(2)},x_{(1)})}{\Bbb P(X_1>X_2)}\\\end{align}
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2$\begingroup$ Thank you! But I did not get why you multiply and divide $P(X_1 < X_2)$ at the second step. :( $\endgroup$ Jun 1, 2019 at 4:58
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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.
Am I correct ?
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2$\begingroup$ 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. $\endgroup$– whuber ♦Jun 1, 2019 at 16:38