# Probability of $Y>X$ when jointly distributed

Suppose $f(x,y)$ is a bivariate normal distribution, with parameters $\mu_x$, $\mu_y$, $\sigma_x$, $\sigma_y$ and $\rho$, as the correlation ($\rho = \frac{V_{xy}}{\sigma_x \sigma_y}$, where $V_{xy}$ is the covariance of $x$ and $y$).

If I randomly choose a pair $(x',y')$ out of this distribution, what is the probability that $y' > x'$ (or what is $P(y>x)$?

I am sure the relevant integrals can somehow have an elegant expression, but wasn't able to make the necessary simplifications.

Thanks!

• It happens that, for some well chosen real numbers $(a,b,c)$, $(x,y)$ is distributed like $(x,ax+bz+c)$, where $z$ is independent of $x$ and standard normal. Thus, $P(y>x)=P((a-1)x+bz>-c)$. Now, compute the mean $\mu$ and variance $\sigma^2$ of $(a-1)x+bz$ and deduce that $P(y>x)=P(\sigma z>-c-\mu)=P(\sigma z<c+\mu)=\Phi((c+\mu)/\sigma)$ and you are done. – Did May 29 '17 at 17:01

Firstly you can rearrange the equation. $P(Y>X)=P(X-Y<0)$. The expected value of $X-Y$ is $\mu_x-\mu_y$. And the variance is $Var(X-Y)=Var(X)+Var(Y)-2Cov(X,Y)$ Consequently
$$T=X-Y\sim \mathcal N(\mu_{x-y}, \sigma^2_{x-y})= \mathcal N\left(\mu_x-\mu_y, \sigma_x^2+ \sigma_y^2-2\sigma_{xy}\right)$$
Now you can calculate $P(T<0)$
Let us define $z = y - x$. It is distributed as $N(\mu_y - \mu_x, \sigma^2_x + \sigma^2_y - 2V_{xy})$, so $\frac{z - (\mu_y - \mu_x)}{\sqrt{\sigma^2_x + \sigma^2_y - 2V_{xy}}} \sim N(0,1)$
Thus $P(y > x) = P(z > 0) = P\left(\frac{z - (\mu_y - \mu_x)}{\sqrt{\sigma^2_x + \sigma^2_y - 2V_{xy}}} > -\frac{\mu_y - \mu_x}{\sqrt{\sigma^2_x + \sigma^2_y - 2V_{xy}}} \right) = \Phi\left(\frac{\mu_x - \mu_y}{\sqrt{\sigma^2_x + \sigma^2_y - 2V_{xy}}} \right)$ where $\Phi$ is distribution function of $N(0,1)$.