# Derivatives for box integral of a bivariate normal distribution

I'm having quite a bit of trouble trying to understand how to to calculate partial derivatives of a specific function.

Suppose I have the standard bivariate normal density function:

$$f(x,y,\rho)=\frac{1}{2\pi\sqrt{1-\rho^2}}\exp \left(-\frac{x^2-2\rho x y + y^2}{2(1-\rho^2)}\right)$$

where $$\rho$$ is the correlation term between $$x$$ and $$y$$.

Now, I'm going to create a function $$F$$ that calculates the box integral of $$f$$:

$$F(x_{\text{L}},x_{ \text{U}},y_{ \text{L} },y_{ \text{U} },\rho)=\int_{x_{ \text{L} }}^{x_{ \text{U} }}\int_{y_{ \text{L} }}^{y_{ \text{U} }}f(x,y,\rho)dydx$$

The derivatives I'm interested in finding are:

$$\frac{\partial F}{\partial x_{ \text{L} }}, \frac{\partial F}{\partial x_{ \text{U} }}, \frac{\partial F}{\partial y_{ \text{L} }}, \frac{\partial F}{\partial y_{ \text{U} }}, \frac{\partial F}{\partial \rho}$$

How on earth do I go about finding these derivatives?

I know that closed form integrals for the univariate, bivariate or multivariate normal density functions simply do not exist. But does that mean that I cannot calculate these partial derivatives either?

Thank you!

Edit

I've been able to make a lot of progress using the Leibniz Integral Rule, aka differentiation under the integral sign. Shout out to @eyeballfrog for the tip that I only need to use the one-dimension case.

I've been able to find values for $$\frac{\partial F}{\partial x_{ \text{L} }}$$, $$\frac{\partial F}{\partial x_{ \text{U} }}$$ , $$\frac{\partial F}{\partial y_{ \text{L} }}$$ and $$\frac{\partial F}{\partial y_{ \text{U} }}$$. However, I seem to have run into a snag for the partial derivative with respect to the correlation term, $$\frac{\partial F}{\partial \rho}$$.

Here is how far I've gotten:

$$\frac{\partial F}{\partial \rho} = \int_{x_{ \text{L} }}^{x_{ \text{U} }}\int_{y_{ \text{L} }}^{y_{ \text{U} }} \frac{\partial}{\partial \rho} f(x,y,\rho)dydx$$

The tricky thing is that $$\frac{\partial F}{\partial \rho}$$ is SUPER messy. Here's what it looks like:

$$\frac{\partial F}{\partial \rho} = \int_{x_{ \text{L} }}^{x_{ \text{U} }}\int_{y_{ \text{L} }}^{y_{ \text{U} }} \frac{\rho^3 - \rho^2 x y +\rho x^2 + \rho y^2 - \rho - xy}{2 \pi (\rho-1)(\rho + 1) (1 - \rho^2)^{3/2}} \cdot exp \left( -\frac{x^2}{2(1-\rho^2)} + \frac{\rho x y}{1-\rho^2} -\frac{y^2}{2(1-\rho^2)} \right)$$

So it seems very unlikely that I'll get a nice "closed-form" solution for the double integral of $$\frac{\partial F}{\partial \rho}$$.

Does anyone see some sort of simplification that can be done here? For example, does $$\frac{\partial F}{\partial \rho}$$ contain the kernel of a modified bivariate normal distribution, or something like that? Because I honestly cannot see it here.

Thanks!!!

• The Leibniz integral rule should give you all of them, but I doubt any will have an elementary form. – eyeballfrog Apr 17 at 20:58
• Since I posted this question, I've found my way to the Leibniz integral rule, but the section on "higher dimensions" seems to be specific to fluid mechanics and moving boundaries. So I'm having a really hard time adapting that process to this problem. – Felipe D. Apr 17 at 21:00
• Yeah, ignore that. Consider that $\int_{x_-}^{x_+} \int_{y_-}^{y_+} f(x,y)dydx = \int_{x_-}^{x_+}\left[ \int_{y_-}^{y_+} f(x,y)dy\right]dx = \int_{y_-}^{y_+}\left[ \int_{x_-}^{x_+} f(x,y)dx\right]dy$ and use the one-dimensional version on the outer integrals as needed. – eyeballfrog Apr 17 at 21:05
• Oh, sweet! Ok, I'll give that a try, then. Thanks! – Felipe D. Apr 17 at 21:07