How to find the mixed derivative of the Gaussian copula?

Question

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).

Answer 1

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

Answer 2

Using the page provided by Dominik I was able to find an answer that fits with the thesis.

$\frac{\partial}{\partial x} \Phi^{-1}(x)= \frac{1}{\phi(\Phi^{-1}(x))}$

And so using the chain rule the mixed partial derivative should be, assuming I'm correct,

$\frac{\partial^2}{\partial x \partial y} C(x,y) = \frac{1}{\phi(\Phi^{-1}(x))} \frac{1}{\phi(\Phi^{-1}(y))} \phi_\rho (\Phi^{-1}(x),\Phi^{-1}(y))$