# Derivative of bivariate normal CDF with common mean parameters

I am trying to calculate a derivative of the form $$\frac{d}{dz}\Phi_2(\mu_1(z),\mu_2(z),\rho)$$, where $$\Phi_2$$ is the standard bivariate normal CDF.

I am thinking it might be an application of a double Leibniz rule (example here) but the lower limit of integration in the CDF is not finite.

Edit: To be clear, I define the standard bivariate normal CDF as $$\Phi_2(\mu_1(z),\mu_2(z),\rho) = \int_{-\infty}^{\mu_1(z)}\int_{-\infty}^{\mu_2(z)}\frac{1}{2 \pi \sqrt{1-\rho^2}}\exp[-\frac{1}{2(1-\rho^2)}(x^2+y^2-2\rho x y)]\,dy\,dx$$ where $$\mu_1(z)$$ and $$\mu_2(z)$$ are functions differentiable with respect to z.

• Can you specify in more details what $\Phi_2(\mu_1(z),\mu_2(z),\rho)$ stands for. In particular, is it $\int_{-\infty}^z \int_{-\infty}^z f(x,y|\mu_1,\mu_2,\rho)\ dx\ dy$, where $f$ is the bivariate normal density. Apr 6, 2018 at 23:43
• My bad, I guess it is $\int_{-\infty}^x \int_{-\infty}^y f(u,v|\mu_1(z),\mu_2(z),\rho)\ dv\ du$ ? Apr 7, 2018 at 0:12
• Please clarify since it is not standard notation. Is $$\Phi_2(\mu_1(z),\mu_2(z),\rho)=\frac{1}{2\pi\sqrt{1-\rho^2}}\int_{-\infty}^{\mu_1(z)}\int_{-\infty}^{\mu_2(z)}\exp\left[-\frac{1}{2(1-\rho^2)}\left(x^2-2pxy+y^2\right)\right]\,dx\,dy$$ ? Apr 7, 2018 at 12:06
• @StubbornAtom that is correct. Thank you. Post is updated.
– mike
Apr 9, 2018 at 13:23

This post here seems relevant. For the case $$\mu_1(z)=\mu_2(z)=z$$, what you are asking for is essentially the density of $$\max(X,Y)$$ as is shown here.

Let $$(X,Y)\sim\mathcal{B\,N}(\mu_x=0,\mu_y=0,\sigma^2_x=1,\sigma^2_y=1,\rho)$$ having joint pdf $$f_{X,Y}$$.

Using $$\Phi$$ and $$\phi$$ to denote CDF and PDF respectively of a standard normal variate as usual.

Then,

\begin{align} \Phi_2(\mu_1(z),\mu_2(z))&=\int_{-\infty}^{\mu_2(z)}\int_{-\infty}^{\mu_1(z)}f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y \\&=\Pr(X\leqslant\mu_1(z),Y\leqslant\mu_2(z)) \\&=\int_{-\infty}^{\mu_1(z)}\Pr(Y\leqslant\mu_2(z)\mid X=x)\,\phi(x)\,\mathrm{d}x \\&=\int_{-\infty}^{\mu_1(z)}\Phi\left(\frac{\mu_2(z)-\rho x}{\sqrt{1-\rho^2}}\right)\,\phi(x)\,\mathrm{d}x \end{align}

The last equality follows from the fact that $$[Y\mid X=x]\sim\mathcal N(\rho x,1-\rho^2)$$.

Using Leibniz rule, $$\frac{\mathrm{d}}{\mathrm{d}z}\Phi_2(\mu_1(z),\mu_2(z))$$ equals $$\int_{-\infty}^{\mu_1(z)}\frac{\partial}{\partial z}\Phi\left(\frac{\mu_2(z)-\rho x}{\sqrt{1-\rho^2}}\right)\phi(x)\,dx+\mu_1'(z)\Phi\left(\frac{\mu_2(z)-\rho \mu_1(z)}{\sqrt{1-\rho^2}}\right)\phi(\mu_1(z))$$

The 'contribution' due to the lower limit of the integral is zero as the lower limit is not a function of $$z$$. I think the final answer simplifies further in terms of $$\Phi(\cdot)$$ and $$\phi(\cdot)$$.

We can obtain a nice closed-form answer simply by applying definitions and the most basic result of linear regression theory: no calculation is needed.

First consider more generally what happens to $$\Phi_2(x,y,\rho)$$ when $$x$$ is changed to $$x+\mathrm{d}x$$ for a positive infinitesimal $$\mathrm{d}x.$$ By definition, $$\Phi_2(x,y,\rho)$$ is the total probability in the square where $$X\le x$$ and $$Y\le y.$$ The difference therefore is the total probability in the infinitesimal half-infinite vertical strip $$S(x,\mathrm{d}x,y)$$ delimited at the left by $$X=x,$$ the right by $$X=x+\mathrm{d}x,$$ and the top by $$y.$$ Colors denote values of the bivariate density with $$\rho=2/5.$$ Its regression line $$y=\rho x$$ is shown in white. The colored vertical strip on the right is $$S(x,\mathrm{d}x,y).$$ The colored horizontal strip at the top is the analog of this strip after exchanging $$X$$ and $$Y.$$

The theory of linear regression teaches us that in the linear regression line for $$\Phi_2,$$ the conditional distribution of $$Y\mid X=x$$ has mean $$\rho x$$ and variance $$1-\rho^2.$$ Since the probabilities in this vertical strip do not appreciably change across it from left to right, and must be proportional to the standard Normal density $$\phi(x),$$ then in terms of the standard normal CDF $$\Phi$$ they must therefore equal

\eqalign{ \Pr(S(x,\mathrm{d}x,y)) &= \Pr(Y\le y\mid X\in[x,x+\mathrm{d}x))\Pr(X\in[x,x+\mathrm{d}x)) \\ &= \Phi\left(\frac{y-\rho x}{\sqrt{1-\rho^2}}\right)\,\phi(x)\mathrm{d}x . } \tag{*}

It's clear that the same result holds for negative infinitesimal $$\mathrm{d}x.$$

Reversing the roles of $$X$$ and $$Y$$ changes nothing in the reasoning and only swaps $$x$$ and $$y$$ in the result: from the symmetric expression for the bivariate Normal density, the situation is identical.

Specializing to the question, let $$\mathrm{d}z$$ be an infinitesimal change in $$z.$$ By definition of the derivative, this induces simultaneous infinitesimal changes in $$x$$ and $$y$$ given by

\eqalign{ \mathrm{d}x &= \mathrm{d}\mu_1(z) = \mu_1^\prime(z)\mathrm{d}z; \\ \mathrm{d}y &= \mathrm{d}\mu_2(z) = \mu_2^\prime(z)\mathrm{d}z. }

Together this creates two infinitesimal strips between the rectangles $$X\le \mu_1(z), Y\le \mu_2(z)$$ and $$X\le \mu_1(z+\mathrm{d}z), Y \le \mu_2(z + \mathrm{d}z),$$ as shown in the figure. Their total area is given by two applications of $$(*)$$ upon substituting $$(x,y)=(\mu_1(z),\mu_2(z))$$ and $$(\mathrm{d}x,\mathrm{d}y) = (\mu_1^\prime, \mu_2^\prime)\mathrm{d}z:$$

\eqalign{ & \mathrm{d}\Phi_2(\mu_1(z),\mu_2(z),\rho) \\ &=\Phi\left(\frac{\mu_2(z)-\rho \mu_1(z)}{\sqrt{1-\rho^2}}\right)\phi(\mu_1(z))\mu_1^\prime(z)\mathrm{d}z + \Phi\left(\frac{\mu_1(z)-\rho \mu_2(z)}{\sqrt{1-\rho^2}}\right)\phi(\mu_2(z))\mu_2^\prime(z)\mathrm{d}z. }

Dividing both sides by $$\mathrm{d}z$$ gives the desired derivative.