Integral of a bivariate normal cdf

Question

Let $$ \Phi_2(x,y;\rho):=\int_{-\infty}^y\int_{-\infty}^x \frac{1}{2\pi\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho^2)}(s^2+t^2-2st\rho)} \, ds \, dt $$ be the joint cdf of bi-variate normal random variables with $X\sim N(0,1)$ and $Y\sim N(0,1)$ and correlation coefficient $\rho$. Is there a way of simplifying $$ \int_0^Te^{\kappa t}\Phi_2\left(ct+d,ct+d;\frac{t}{T}\right) \, dt $$ in terms of some well known `special functions'? Here $c,d,T$ and $\kappa$ are constants.

Answer 1

I cannot say whether you will consider this as a "simplification", but you can start by writing the standard normal bivariate integral as

$$\Phi_2(x,y;\rho)=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^y\frac{1}{\sqrt {2\pi}\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho^2)}t^2}\int_{-\infty}^x e^{-\frac{1}{2(1-\rho^2)}(s^2-2st\rho)}dsdt$$

$$=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^y\phi\left(\frac {t}{\sqrt{1-\rho^2}}\right)\int_{-\infty}^x e^{-\frac{1}{2(1-\rho^2)}(s^2-2t\rho s)}dsdt$$

For the internal integral we have

$$I_\mathrm{int} = \int_{-\infty}^x e^{-\frac{1}{2(1-\rho^2)}(s^2-2t\rho s)}ds = \int_{-\infty}^{\infty} e^{-\frac{1}{2(1-\rho^2)}(s^2-2t\rho s)}ds-\int_{x}^{\infty} e^{-\frac{1}{2(1-\rho^2)}(s^2-2t\rho s)}ds$$

$$=\int_{0}^{\infty} e^{-\frac{1}{2(1-\rho^2)}(s^2+2t\rho s)}ds+\int_{0}^{\infty} e^{-\frac{1}{2(1-\rho^2)}(s^2-2t\rho s)}ds-\int_{x}^{\infty} e^{-\frac{1}{2(1-\rho^2)}(s^2-2t\rho s)}ds$$

$$=2\int_0^\infty e^{-\frac{1}{2(1-\rho^2)}s^2} \cosh\left[\frac {t\rho}{1-\rho^2}s\right] ds-\int_{x}^{\infty} e^{-\frac{1}{2(1-\rho^2)}(s^2-2t\rho s)}ds$$

Then you can use the formulas (matching the coefficients appropriately)

where please note that here $\Phi$ is NOT the standard normal CDF but the error function (erf).

Obviously, it is very easy to make an algebraic mistake here, especially when matching coefficients, so go very slow. Then bring it all together (some things may simplify), and see how the bivariate standard normal CDF looks like (check also the expression for the error function in terms of the univariate standard normal CDF). Then add one more integral layer and the $cd+t$ arguments you want, and see what further simplifications you can achieve.