How can one calculate the following integral? $$ \int_0^1{\int_0^1 xy \quad d[min(x,y)]} $$
I have no idea how to hande the $d(min(x,y))$.
Does anyone have an idea for this problem?
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I'm not sure how to interpret this; am I supposed to be thinking Stieltjes type integration with respect to $g(x,y) = \min(x,y)$? What would the integral be if you looked at $\int_0^1 \int_y^1 xy , d[\min(x,y)]$? – Tom Nov 01 '16 at 11:32
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I'm fairly certain that this is a Stieltjes integral. – Sean Roberson Nov 01 '16 at 11:35
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Indeed, it is a Stieltjes integral. – DeMerlo Nov 02 '16 at 10:18
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Why do you say to look at the integral with x going from y to 1? I have problems with the $d[min(x,y)]$... – DeMerlo Nov 02 '16 at 10:26
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Certainly this is a (Lebesgue-)Stieltjes integral. Nevertheless, fortunately $C_1=xy$ and $C_2=\min(x,y)$ are copulas and for copulas the following holds (partial dderivatives exist a.e. of course): $$\iint_{[0,1]^2}C_1(x,y)\,\mathrm{d}C_2(x,y)=\frac{1}{2}-\iint_{[0,1]^2}\frac{\partial}{\partial x}C_1(x,y)\frac{\partial}{\partial y}C_2(x,y)\,\mathrm{d}(x,y).$$ Therefore $\iint_{[0,1]^2}xy\,\mathrm{d}\min(x,y)=\frac{1}{2}-\iint_{T}y\,\mathrm{d}(x,y),$ where $T=\{(x,y)\in[0,1]^2: ~0\leq x\leq 1, ~0\leq y<x\}$. Finally $\iint_{[0,1]^2}xy\,\mathrm{d}\min(x,y)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$.
