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


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!


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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.