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?


1 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.

  • $\begingroup$ On your first sentence: Up to substitution, the remaining integral will be $\int_{-\infty}^x e^{-x^2} dx$ and it is known that there is not closed form solution for these integrals! See stats.stackexchange.com/questions/18286/… for this road :-) $\endgroup$ Commented Dec 22, 2017 at 9:18
  • $\begingroup$ @Sun: On the second part: could you elaborate a little more in depth? What is the goal you want to achieve? Do you want to optimize this function w.r.t. $x_1$ or something similar? Just trying to understand what you want to achieve: The equations above give you a form that you want: the remaining double integral should be the cdf of a bivariate normal distribution with skewed parameters similar to the case you mentioned above... $\endgroup$ Commented Dec 22, 2017 at 9:21
  • $\begingroup$ I am maximizing a MLE, involving trivariate normal cdf. I tried to use Newton-Raphson algorithm, but it was too slow. I am planning to use trust-region algorithm instead. But, it requires me to specify the first derivative of trivariate normal cdf. $\endgroup$
    – Sun
    Commented Dec 22, 2017 at 10:56
  • $\begingroup$ If I evaluate the above double intergral using Gauss-Legendre Quadrature, it becomes too slow. Therefore, I am trying to express the first derivative of trivariate cdf using elementary functions. $\endgroup$
    – Sun
    Commented Dec 22, 2017 at 11:00
  • $\begingroup$ According to stats.stackexchange.com/questions/18286/… What I mean by 'closed form' and 'elementary fucntions' are well-known functions. I am really sorry for the confusion. $\endgroup$
    – Sun
    Commented Dec 22, 2017 at 11:20

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