# Derivative of a Bivariate normal CDF with respect to its variables

Following up on the question (and answers) here, I'm trying to derive $$\frac{\partial \Phi(x_1, x_2|\mathbf{\underline{\theta}})}{\partial x_1}$$ and $$\frac{\partial \Phi(x_1, x_2|\mathbf{\underline{\theta}})}{\partial x_2}$$ where $$\mathbf{\underline{\theta}} = (\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho)$$ and $$\Phi$$ denotes the bivariate normal CDF under examination.

To clarify, I'm not looking for the final answer/expression, the answer is already available online. I just want to know if I'm deriving it correctly, and if not, where am I going wrong?

My work is as follows:

$$F(x_1, x_2|\mathbf{\underline{\theta}}) = \int_{-\infty}^{x_1}\int_{-\infty}^{x_2} \phi(a, b|\mathbf{\underline{\theta}}) \;db\;da$$

$$= \int_{-\infty}^{x_1}\int_{-\infty}^{x_2} \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}.exp \left\{ -\frac{1}{2(1-\rho^2)}\left[\left(\frac{a-\mu_1}{\sigma_1}\right)^2-2\rho\left(\frac{a-\mu_1}{\sigma_1}\right)\left(\frac{b-\mu_2}{\sigma_2}\right)+\left(\frac{b-\mu_2}{\sigma_2}\right)^2\right] \right\}\;db\;da$$

$$= \int_{-\infty}^{x_1}\frac{1}{\sqrt{2\pi}\sigma_1}.exp \left\{ -\frac{1}{2(1-\rho^2)}\left(\frac{a-\mu_1}{\sigma_1}\right)^2\right\}\int_{-\infty}^{x_2}\frac{1}{\sqrt{2\pi}\sigma_2\sqrt{1-\rho^2}}.exp \left\{ -\frac{1}{2(1-\rho^2)}\left[-2\rho\left(\frac{a-\mu_1}{\sigma_1}\right)\left(\frac{b-\mu_2}{\sigma_2}\right)+\left(\frac{b-\mu_2}{\sigma_2}\right)^2\right] \right\}\;db\;da$$

Let:

$$\left(\frac{a-\mu_1}{\sigma_1}\right)=X \;;\;\left(\frac{b-\mu_2}{\sigma_2}\right)=Y$$

Therefore the inner integral is now:

$$= \int_{-\infty}^{x_2}\frac{1}{\sqrt{2\pi}\sigma_2\sqrt{1-\rho^2}}.exp \left\{ -\frac{1}{2(1-\rho^2)}\left(-2\rho XY+Y^2\right) \right\}\;db\;da$$

Focusing on the inner integral, specifically the term inside the exponential; completing the square yields:

$$exp \left\{ -\frac{1}{2(1-\rho^2)}\left(Y-\rho X\right)^2 \right\}.exp \left\{ \frac{1}{2(1-\rho^2)}\left(\rho X\right)^2 \right\}$$

Taking out the second term in the product (which is constant in Y) yields:

$$\int_{-\infty}^{x_1}\frac{1}{\sqrt{2\pi}\sigma_1}.exp \left\{ -\frac{1}{2(1-\rho^2)}X^2\right\}.exp \left\{ \frac{1}{2(1-\rho^2)}\left(\rho X\right)^2 \right\}\int_{-\infty}^{x_2}\frac{1}{\sqrt{2\pi}\sigma_2\sqrt{1-\rho^2}}.exp \left\{ -\frac{1}{2(1-\rho^2)}\left(Y-\rho X\right)^2 \right\}\;db\;da$$

$$= \int_{-\infty}^{x_1}\phi(a)\int_{-\infty}^{x_2}\frac{1}{\sqrt{2\pi}\sigma_2\sqrt{1-\rho^2}}.exp \left\{ -\frac{1}{2(1-\rho^2)}\left(Y-\rho X\right)^2 \right\}\;db\;da$$

Undoing our compact notation in the inner integral yields:

$$= \int_{-\infty}^{x_1}\phi(a)\int_{-\infty}^{x_2}\frac{1}{\sqrt{2\pi}\sigma_2\sqrt{1-\rho^2}}.exp \left\{ -\frac{1}{2}\left(\frac{b-(\mu_2+\frac{\sigma_2}{\sigma_1}a\rho-\frac{\sigma_2}{\sigma_1}\mu_1\rho)}{\sigma_2\sqrt{1-\rho^2}}\right)^2 \right\}\;db\;da$$

Rewriting the inner integral as a CDF yields:

$$= \int_{-\infty}^{x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2|a)\;da$$

Note that I'm using * to denote the fact that the CDF has a different mean and variance than the marginal cdf of $$X_2$$.

Now here's where things get a little confusing for me. The answer I referenced at the beginning of this question implies that I need to use Leibniz's rule for differentiating under the integral sign to find $$\frac{\partial \Phi(x_1, x_2|\mathbf{\underline{\theta}})}{\partial x_1}$$. While I am familiar with Leibniz's rule, I can't help shake this feeling that I'm applying it here incorrectly. Here's the rest of my work:

$$\frac{\partial \Phi(x_1, x_2|\mathbf{\underline{\theta}})}{\partial x_1}=\frac{\partial}{\partial x_1} \int_{-\infty}^{x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2| a)\;da$$

$$= \int_{-\infty}^{x_1}\frac{\partial}{\partial x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2| a)\;da + \phi_{X_1}(x_1)\Phi_{X_2^*}(x_2| x_1)\frac{\partial x_1}{\partial x_1} - \phi_{X_1}(x_1)\Phi_{X_2^*}(x_2| (-\infty))\frac{\partial (-\infty)}{\partial x_1}$$

$$= \int_{-\infty}^{x_1}\frac{\partial}{\partial x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2| a)\;da + \phi_{X_1}(x_1)\Phi_{X_2^*}(x_2| x_1).1 - \phi_{X_1}(x_1)\Phi_{X_2^*}(x_2| (-\infty)).0$$

$$= \int_{-\infty}^{x_1}\frac{\partial}{\partial x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2| a)\;da + \phi_{X_1}(x_1)\Phi_{X_2^*}(x_2| x_1)$$

Now I know the answer is:

$$\phi_{X_1}(x_1)\Phi_{X_2^*}(x_2| x_1)$$

However, the only way I can think of to zero the first term is:

$$\int_{-\infty}^{x_1}\frac{\partial}{\partial x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2| a)\;da$$

$$= \int_{-\infty}^{x_1}0\;da = 0$$

However, the above feels incredibly wrong to me. My source of confusion arises from the integration dummy, a. I'm not sure if I should treat a as $$x_1$$ when taking the derivative with respect to $$x_1$$ or just treat it as a constant. Any help is appreciated.

• @Xi'an yes, thanks for pointing that out, I’ll fix it right away
– tvbc
Jul 13 '21 at 8:22

When computing $$\frac{\partial}{\partial x_1} \int_{-\infty}^{x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2| a)\;\text da$$ the only instance of $$x_1$$ in the differentiated term is the upper bound of the integral, hence $$\frac{\partial}{\partial x_1} \int_{-\infty}^{x_1}\phi_{X_1}(a)\Phi_{X_2^*}(x_2| a)\;\text da=\phi_{X_1}(x_1)\Phi_{X_2^*}(x_2| x_1)$$
• Thanks, I was confused as to whether I should treat the integration dummy as $x_1$ or just as a constant
• The integration dummy is a symbol but does not "exist" as such when considering the integral as a function of $x_1$. Jul 13 '21 at 15:40