Partial Derivative of Standard Trivariate Normal CDF

Question

I am trying to get partial derivative of standard trivariate normal cdf with respect to $x_1$. i.e. I would like to get $ \frac{\partial}{\partial x_1}\Phi(x_1,x_2,x_3;r_{12},r_{13},r_{23})$. $\Phi(x_1,x_2,x_3;r_{12},r_{13},r_{23}) = \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} \int_{-\infty}^{x_3} \frac{1}{(2\pi)^{3/2} \sqrt{|R|}} exp \left (-\frac{w}{2|R|} \right) dc\ db\ da$ $|R| = (1-r_{12}^2-r_{13}^2-r_{23}^2+2r_{12}r_{13}r_{23})$ $w = a^2(1-r_{23}^2) + b^2(1-r_{13}^2) + c^2(1-r_{12}^2) - 2[ab(r_{12} - r_{13}r_{23}) + ac(r_{13} - r_{12}r_{23}) + bc(r_{23} - r_{12}r_{13})]$

Can someone help me?

Answer 1

Consider an 'integral function' of the form $$F(x_1) = \int_{-\infty}^{x_1} g(r) dr$$ then as long as the inner function $g$ is continuous at some point $x$, $F'(x) = g(x)$. In your case, one can show that the inner function is continuous at every point (the only reasonable explanation I can come up with is to consider the three integrals as one single integral and then apply the mean value theorem to the three dimensional inner function which is basically just the $e^{-...}$ stuff and this is $L^1$). Hence one can apply the fundamental theorem of calculus and one gets that

$$\partial_{x_1} \Phi = \int_{-\infty}^{x_2} \int_{-\infty}^{x_3} \frac{1}{(2\pi)^{3/2}\sqrt{|R|}} e^{-w/(2R)} dc ~ db$$

i.e. the same function as before just with the integral over $x_1$ being gone and $x_1$ placed in the inner function as a constant.