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.


  • $\begingroup$ 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. $\endgroup$ – 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 $N(0,1)$.

  • $\begingroup$ Thanks a lot! A follow up question - what if the copula between those normal random variables is not Gaussian but Clayton or Gumbel copulas - how does it work then? $\endgroup$ – Serb Oct 5 '17 at 8:49

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.