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)?
\begin{equation}
y=\Phi(x_{1},x_{2},\rho)
\end{equation}
\begin{equation}
z=\phi(x_{1},x_{2},\rho)
\end{equation}
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 $x_{2}$.
Best Answer
\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 $$\frac{\partial}{\partial \rho}\Phi(x_1,x_2,\rho) = \frac{\partial}{\partial \rho}\int_{-\infty}^{x_1} \left[\int_{-\infty}^{x_2} \phi(a,b,\rho)\,\mathrm db\right]\,\mathrm da = \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} g(a,b,\rho)\,\mathrm db\,\mathrm da$$