What is Cov(X1 - X2, D)?

Question

Let X1 and X2 be normally distributed random variables with means m1 and m2 and standard deviations s1 and s2 and correlation coefficient r. Further, let D be an indicator (bivariate) variable taking the value 1 if X1 > X2 and 0 otherwise.

Question: What is Cov(X1 - X2, D)?

I expect the answer to be

q^2 f(a),

where q^2 = s1^2 + s2^2 - 2 r s1 s2, a = (m1 - m2)/q, and f() is the normal probability density function.

The challenge for you experts is to prove that I am wrong - or better that I am right in a compact way, or to provide a reference to the solution

Being a researcher in financial accounting, my statistical knowledge has become rusty over the years, so I will be deeply grateful for any help!

Answer 1

Letting $\mu$ and $\sigma$ denote the mean and standard deviation of $X=X_1-X_2$ and $\mu_I$ the expected value of $I_{X>0}$, \begin{align} \operatorname{Cov}(X_1-X_2,I_{X_1>X_2}) &=\operatorname{Cov}(X,I_{X>0}) \\&=\int_{-\infty}^\infty (x-\mu)(I_{(0,\infty)}(x)-\mu_I)f_X(x)dx \\&=\int_{-\infty}^\infty (x-\mu)I_{(0,\infty)}(x)f_X(x)dx - \int_{-\infty}^\infty (x-\mu)\mu_If_X(x)dx \\&=\int_0^\infty (x-\mu)f_X(x)dx - \underbrace{E((X-\mu)\mu_I)}_{=0} \\&=\sigma\int_0^\infty \frac{x-\mu}\sigma \frac1{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx \\&=\frac{\sigma}{\sqrt{2\pi}}\int_{\mu^2/2\sigma^2}^\infty e^{-u}du \\&=\frac{\sigma}{\sqrt{2\pi}}\int_{\mu^2/2\sigma^2}^\infty e^{-u}du \\&=\frac{\sigma}{\sqrt{2\pi}}e^{-\frac12(\mu/\sigma)^2} \\&=\sigma \phi(\mu/\sigma) \end{align} where $\phi$ is the pdf of the standard normal.