# Partial derivative of bivariate normal cdf and pdf

How do I take the partial derivative of bivariate normal cdf and bivariate normal pdf with its arguments (i.e. $x_{1}$ ,$x_{2}$ , and $\rho$ in the following equations)?

$$y=\Phi(x_{1},x_{2},\rho)$$

$$z=\phi(x_{1},x_{2},\rho)$$

where $x_{1}$ is normally distributed with mean 0 and variance 1 and $x_{2}$ is normally distributed with mean 0 and variance 1. $\rho$ is the correlation between $x_{1}$ and $x_{2}$.

• If you were given just the formulas for these functions and not told that they were cdfs or pdfs (in fact, suppose that you had no knowledge whatsoever of probability and/or statistics), could you find the partial derivatives using standard calculus techniques such as the chain rule? – Dilip Sarwate Oct 5 '13 at 13:40
• Thanks. Yes, I know how to do that from high school calculus. My question is how to do when we have cdf and pdf. – user227710 Oct 5 '13 at 15:20
• I know how to do that from high school calculus. My question is how to do when we have cdf and pdf. Ignore the information that these are pdfs or cdfs and proceed. The bivariate cdf will be given as a double integral with integrand the bivariate pdf and upper limits $x_1$ and $x_2$. So the partial derivative w.r.t. $x_i$ will be a single integral of the bivariate pdf, etc. – Dilip Sarwate Oct 5 '13 at 16:27

\begin{align} y &= \Phi(x_1,x_2,\rho) = \int_{-\infty}^{x_1}\left[\int_{-\infty}^{x_2} \phi(a,b,\rho)\,\mathrm db\right]\,\mathrm da\\ \frac{\partial y}{\partial x_1} &= \frac{\partial}{\partial x_1}\Phi(x_1,x_2,\rho) = \frac{\partial}{\partial x_1}\int_{-\infty}^{x_1} \left[\int_{-\infty}^{x_2} \phi(a,b,\rho)\,\mathrm db\right]\,\mathrm da\\ &= \int_{-\infty}^{x_2} \phi(x_1,b,\rho)\,\mathrm db \end{align} via the rule for differentiating under the integral sign. Similarly, $$\frac{\partial y}{\partial x_2} = \int_{-\infty}^{x_1} \phi(a,x_2,\rho)\,\mathrm da.$$ If you don't recall the rule for differentiating integrals, see for example, the comments following this answer on math.SE.
The derivative with respect to $\rho$ is straightforward to find but messy in its details. We have that $$\phi(x_1,x_2,\rho)=\frac{1}{2\pi\sqrt{1-\rho^2}} \exp\left[-\frac{x^2 -2\rho xy + y^2}{2(1-\rho^2)}\right]$$ whose partial derivative with respect to $\rho$ is left to the OP to find. If $g(x_1,x_2,\rho)$ denotes this partial derivative, then $$\frac{\partial}{\partial \rho}\Phi(x_1,x_2,\rho) = \frac{\partial}{\partial \rho}\int_{-\infty}^{x_1} \left[\int_{-\infty}^{x_2} \phi(a,b,\rho)\,\mathrm db\right]\,\mathrm da = \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} g(a,b,\rho)\,\mathrm db\,\mathrm da$$
• An anonymous user just proposed an edit: "I'm interested in the derivative with respect to $\rho$", which I declined, because it should have been a comment, which I am posting. – Stephan Kolassa Oct 2 '15 at 8:23
• @StephanKolassa Thanks. I added a few lines to the answer describing how to go about finding the derivative with respect to $\rho$. – Dilip Sarwate Oct 2 '15 at 19:12