Calculating $p(X_2-X_1+Y>\pi, X_2>\pi)$ with $X_1$ and $X_2$ correlated, jointly Normal and $Y$ independent from them and Normal

Given $X_1$ and $X_2$ as correlated ($\rho$) and jointly Normal random variables (each distributed ~$N(0,\sigma_X^2)$), and also given $Y$ as Normally distributed((~$N(0,\sigma_Y^2)$)) and independent of $X_1$ and $X_2$, how can one compute

$p(X_2-X_1+Y>\pi, X_2>\pi)$

Any help and guidance would be greatly appreciated!

I know this is equivalent to

$p(X_2-X_1+Y>\pi, X_2>\pi)$=$p(X_2-X_1+Y>\pi | X_2>\pi)p(X_2>\pi)$

but I'm not sure if this is the way to proceed and how to do so. Perhaps there is a simple way to do this that I'm missing? I would really appreciate the help!

This is $$E\left(\Phi\left(\frac{X_2-X_1-\pi}{\sigma_Y}\right)\cdot\mathbf 1_{X_2\gt\pi}\right).$$ Now, use the density of $(X_1,X_2)$ to transform this expectation into a double integral on $\mathbb R\times(\pi,+\infty)$.
• Thanks Did, this is very nice and makes it quite manageable! Just a clarification on it. How do you go from $p(Y>X_1-X_2+\pi, X_2>\pi)$ to having $X_2-X_1-\pi$ inside $\Phi$ which I'm assuming is the cumulative distribution function? Secondly, if I also had the condition $X_1>X_2-\pi$, then would I just integrate $X_1$ first from $X_2-\pi$ to $\inf$ and $X_2$ from $\pi$ to $\inf$? Thanks again for your help! – Hakeem May 26 '14 at 7:22
• $[Y\gt X_1-X_2+\pi]=[Z\lt(X_2-X_1-\pi)/\sigma_Y]$ where $Z=-Y/\sigma_Y$ is standard normal. – Did May 26 '14 at 7:26
• Thanks for clearing that up! You're so quick, so you may have missed the second question I was editing in, so I am reposting. If I also had the condition $X_1>X_2-\pi$, then would I just integrate $X_1$ first from $X_2-\pi$ to $\inf$ and $X_2$ from $\pi$ to $\inf$? – Hakeem May 26 '14 at 7:41
• Yes.    – Did May 26 '14 at 7:45