2
$\begingroup$

How does one calculate the derivative of a multivariate normal CDF with respect to a given correlation coefficient?

I have started with the bivariate case but couldn't work it out.

$\endgroup$
5
  • $\begingroup$ Do you really mean CDF? Or do you rather mean PDF? $\endgroup$
    – ocram
    Jan 13, 2013 at 8:50
  • $\begingroup$ No, I really mean CDF! Differentiating the PDF would be easy. I have tried this for the bivariate case, by taking the derivative inside the integral and then differentiating the PDF. But then the algebra gets really hard when I try to solve the integral. $\endgroup$
    – Stefan
    Jan 13, 2013 at 9:39
  • 1
    $\begingroup$ just a remak in the case where you do not know this: the cdf does not exist in closed form $\endgroup$
    – ocram
    Jan 13, 2013 at 10:02
  • $\begingroup$ I'm not sure what you're trying to do or why (I'm not sure what it means to take a derivative "with respect to a given correlation coefficient")... can you explain? $\endgroup$ Jan 14, 2013 at 3:13
  • $\begingroup$ The what is to calculate $dF(x)/d\rho_{ij}$, the why is to be able to learn the parameters of a product of Gaussian copulae by gradient ascent. $\endgroup$
    – Stefan
    Jan 14, 2013 at 4:01

1 Answer 1

5
$\begingroup$

I have found the answer to this.

There is a result,

$\frac{\partial}{\partial\rho_{ij}}f(x;0,\Sigma)=\frac{\partial^2}{\partial x_i\partial x_j}f(x;0,\Sigma)$ ("A reduction formula for normal multivariate integrals", Plackett 1954).

So using this result by exchanging the integral and derivative, we just need to be able to differentiate a normal CDF with respect to two of the variables. We can do this by first applying the fundamental theorem of calculus, then conditioning on the two variables,

$\frac{\partial}{\partial\rho_{ij}}F(x;0,\Sigma)=f(x_i,x_j;0,\Sigma_{ij})F(x-\{x_i,x_j\}\ |\ x_i,x_j)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.