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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.

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    $\begingroup$ 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. $\endgroup$ – Perochkin Apr 6 '18 at 23:43
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    $\begingroup$ My bad, I guess it is $\int_{-\infty}^x \int_{-\infty}^y f(u,v|\mu_1(z),\mu_2(z),\rho)\ dv\ du$ ? $\endgroup$ – Perochkin Apr 7 '18 at 0:12
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    $\begingroup$ Please clarify since it is not standard notation. Is $$\Phi_2(\mu_1(z),\mu_2(z),\rho)=\frac{1}{2\pi\sqrt{1-\rho^2}}\int_{-\infty}^{\mu_1(z)}\int_{-\infty}^{\mu_2(z)}\exp\left[-\frac{1}{2(1-\rho^2)}\left(x^2-2pxy+y^2\right)\right]\,dx\,dy$$ ? $\endgroup$ – StubbornAtom Apr 7 '18 at 12:06
  • $\begingroup$ @StubbornAtom that is correct. Thank you. Post is updated. $\endgroup$ – mike Apr 9 '18 at 13:23
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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 $\phi(\cdot)$.

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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 $y.$

Figure

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.

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