Differentiating under the integral sign (multiple integrals)

Question

$\newcommand{\bexp}[1]{\exp{\left(#1\right)}}$ I am trying to derive the pdf of a certain random variable $U$ and I have something that looks like this: $$f_{U}(u) = -\frac{d}{du}\int_{x=u}^{\infty}\int_{y=u}^{\infty}f_{XY}(x,y)dxdy,$$ where $$f_{XY}(x,y)= \frac{1}{2\pi \sigma_x\sigma_y \sqrt{1-\rho^2_{xy}}} \times \\ \bexp{\frac{-1}{2(1-\rho_{xy}^2)} \left[\left(\frac{x-\mu_x}{\sigma_x}\right)^2-2\rho_{xy}\left(\frac{x-\mu_x}{\sigma_x}\right)\left(\frac{y-\mu_y}{\sigma_y}\right) +\left(\frac{y-\mu_y}{\sigma_y}\right)^2 \right]}.$$ Here $U= \min\{X,Y\}$ in case someone is interested. Now I am not sure how to differentiate the double integral and proceed further with this problem. Any ideas will be much appreciated.

Answer 1

If we write $$F(s,t)=-\int_s^\infty\int_t^\infty f(x,y)\mathrm dx\mathrm dy,$$ we can see that the partial derivatives are $$ \partial_s F(s,t)=\int_t^\infty f(x,s)\mathrm dx\ \ \text{ and }\ \ \partial_tF(s,t)=\int_s^\infty f(t,y)\mathrm dy $$ by Fubini's theorem and fundamental theorem of calculus. In your case, we want to calculate $$ \frac{d}{du}F(u,u)=\partial_sF(u,u)+\partial_tF(u,u), $$ which follows from the chain rule of differentiation and thus $$ \frac{d}{du}F(u,u)=\int_u^\infty f(x,u)\mathrm dx+\int_u^\infty f(u,y)\mathrm dy. $$