How do I calculate this gaussian integral

Question

I am asked to find the expectation of the greater of X and Y where X and Y are normal bivariates with zero individual means and unit variance with correlation coefficient $\rho$. I know the answer is just $$2\frac{1}{2\pi\sqrt{1-\rho^{2}}}\int_{-\infty}^{\infty}\int_{x}^{\infty}y\cdot \exp\left(-\frac{(x^{2}+y^{2}-2xy\rho)}{2(1-\rho^{2})}\right)\,\mathrm{d}y\mathrm{d}x$$ . But how do I calculate this Gaussian integral?. I have tried adding and substracting $x\rho$ so that the integral in terms of y can be calculated but then I run into another integral where the integral in terms of y is of the form $\int \exp(-y^{2})\mathrm{d}y$ . Which has no antiderivative. I tried polar coordinates but it resulted in a really tough trigonometric integral.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{\sigma \equiv \root{1 - \rho^{2}}}$ and $\ds{{\pars{x,y} \over \root{2}\sigma} \mapsto \pars{x,y}}$: \begin{align} &\bbox[5px,#ffd]{{1 \over \pi\root{1 - \rho^{2}}} \int_{-\infty}^{\infty}\int_{x}^{\infty}y\, \exp\pars{-{x^{2} + y^{2} -2xy\rho \over 2\pars{1 -\rho^{2}}}}\,\dd y\,\dd x} \\[5mm] = &\ {2\root{2}\sigma^{2} \over \pi} \int_{-\infty}^{\infty} \int_{\root{2}\sigma x}^{\infty}y\, \expo{-x^{2} - y^{2} + 2\rho xy}\,\dd y\,\dd x \\[5mm] = &\ {4 \over \pi^{3/2}}\, {\pars{1 - \rho^{2}}\pars{1 - \root{2}\rho\root{1 - \rho^{2}}} \over \root{3 - 2\rho\pars{\rho + \root{2}\root{1 - \rho^{2}}}}}\label{1}\tag{1} \end{align} with $\ds{\root{2}\root{1 - \rho^{2}} < \rho < 1}$. (\ref{1}) was found with some CAS.