# Show the distribution of the minimum of bivariate normal variables

Suppose $X_1,X_2$ have a bivariate normal distribution with pdf

$f_{X_1,X_2}(x_1,x_2)=\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\left[-\frac{1}{2(1-\rho^2)}(x_1^2+x_2^2-2x_1x_2\rho)\right]$

Let $Y=\min(X_1,X_2)$. The question is to show that pdf of $Y$ is

$f_Y(y)=2\phi(y)\Phi(\frac{-y+\rho y}{\sqrt{1-\rho^2}})$ where $\phi$ and $\Phi$ are the pdf and cdf of the $N(0,1)$ distribution.

I am clueless about the expression on the right. How should I tackle this problem? Could any kind soul enlighten me? Thank you!

## 1 Answer

From the joint density you have \begin{align*} \mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 - \mathbb{P}\left\{ X_1 > y, X_2 > y \right\} \\ &= 1 - \int_{y}^{\infty} \int_y^{\infty} f_{X_1,X_2}(s,t)dsdt \\ &= 1 -\int_y^{\infty} f_{X_2}(t)\int_y^{\infty} f_{X_1|X_2}(s,t) ds dt \end{align*} where $$f_{X_2} (t)=\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}, \qquad f_{X_1|X_2}(s,t)=\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(s-\rho t)^2}{2(1-\rho^2)}},$$ and so \begin{align*} \mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 - \int_{y}^{\infty} \varphi(t) \left(1 - \Phi\left( \frac{y -\rho t}{\sqrt{1-\rho^2}}\right) \right) dt. \\ &= 1 - \int_{y}^{\infty} \varphi(t) \Phi\left( \frac{\rho t - y }{\sqrt{1-\rho^2}} \right) dt \end{align*} To get the density we differentiate with respect to $y$ giving \begin{align*} f_{Y}(y) &= -\frac{\partial}{\partial y}\int_{y}^{\infty}\varphi(t)\Phi\left(\frac{\rho t - y}{\sqrt{1-\rho^2}}\right) dt \\ &= \varphi(y)\Phi\left(\frac{\rho y - y}{\sqrt{1-\rho^2}}\right) + \int_{y}^{\infty}\varphi(t) \frac{1}{\sqrt{1-\rho^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\rho t - y)^2}{2(1-\rho^2)}}dt \tag{1} \end{align*} Completing the square of the last term in $(1)$ we have \begin{align*} \varphi(t)\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(\rho t-y)^2}{2(1-\rho^2)}} &= \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left((1-\rho^2)t^2 +(\rho t -y)^2 \right)} \\ &=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left(t^2 - 2 t \rho y + \rho^2 y^2 + (1-\rho^2)y^2 \right)} \\ &= \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}. \end{align*} So putting this back in to $(1)$ we get \begin{align*} f_{Y}(y) &= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) + \varphi(y)\int_y^{\infty} \frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}dt \\ &= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) + \varphi(y)\left( 1 - \Phi\left(\frac{y - \rho y}{\sqrt{1-\rho^2}}\right)\right) \\ &= 2\varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right). \end{align*}