# How to find the mixed derivative of the Gaussian copula?

I am having trouble understanding how to use the chain rule to get the mixed partial derivative of the Gaussian copula.

The Gaussian copula is defined as:

$C(x,y)= \Phi_\rho (\Phi^{-1}(x),\Phi^{-1}(y))$

Where $\Phi_\rho$ is the bivariate standard normal distribution function with correlation $\rho$. And $\Phi^{-1}$ is the inverse standard normal distribution function.

And I would like to derive:

$\frac{\partial^2}{\partial x \partial y} C(x,y)$

How can I find this strictly in terms of $\Phi_\rho$, $\Phi^{-1}$, and $\phi_\rho$ (the bivariate normal density).

The biggest problem I'm running into is dealing with the inverse CDF's when using the chain rule. It is not obvious to me what their derivative is. I know that there is a clearly defined answer (as the person's thesis who I'm reading was able to derive it, he just didn't clearly state it).

• Use the formula for the derivative of inverse functions en.m.wikipedia.org/wiki/Inverse_functions_and_differentiation – Dominik Dec 20 '16 at 22:52
• Thanks, that has cleared up a bit. I was wondering how a lot of stuff in his final formula cancelled, and I think this will lead me there. Will see where it goes. – Patty Dec 20 '16 at 22:54
• Your problem is ill-defined unless you say what the $\sigma_i$ are in your bivariate normal density. I believe you meant that both these $\sigma$'s are $1$, right? – Mark Fischler Dec 20 '16 at 23:00
• Yes, sorry. it is the standard bivariate normal distribution with correlation coefficient $\rho$. – Patty Dec 20 '16 at 23:04

$\frac{\partial}{\partial x} \Phi^{-1}(x)= \frac{1}{\phi(\Phi^{-1}(x))}$
$\frac{\partial^2}{\partial x \partial y} C(x,y) = \frac{1}{\phi(\Phi^{-1}(x))} \frac{1}{\phi(\Phi^{-1}(y))} \phi_\rho (\Phi^{-1}(x),\Phi^{-1}(y))$
If $\Phi^{-1}(x)$ is the inverse CDF of a unit normal vairable $x$, then $$\int_{-\infty}^{\Phi^{-1}(x)}\frac{1}{\sqrt{2\pi}} e^{-t^2/2}dt = x\\ \frac{dx}{d\Phi^{-1}(x)} = \frac{1}{\sqrt{2\pi}} e^{-(\Phi^{-1}(x))^2/2}\\ \frac{d\Phi^{-1}(x)} {dx} = \sqrt{2\pi}e^{(\Phi^{-1}(x))^2/2}$$