I'm stumped by a number of problems that come up in my probability studies: when there is an integral involving two random variables (X,Y) and the upper bound on one of the random variables (Y) is constrained by more than one variable expressions. Here's a typical scenario:
Select a point
(X, Y)at random from the circle: $$ \{\ (X, Y)\ |X^2 + Y^2| < 1\ \} $$ ComputeE(|X − Y|), i.e. the expectation of the absolute value of the difference of X and Y.
It seems to me as if I need:
$$ \iint\limits_{x^2+y^2<1}{|x-y|f(x,y)dxdy} $$
...which I translate (perhaps myopically) into the following:
$$ \iint\limits_{x^2+y^2<1,\ \ y<x}{(x-y)f(x,y)dxdy} + \iint\limits_{x^2+y^2<1,\ \ x<y}{(y-x)f(x,y)dxdy} $$
I don't know if that last step was the right move, but when I try to proceed with the first half of the previous expression, I have no idea how to set the upper bound because y < x and y^2 < 1 - x^2:
$$ \int_{-1}^1dx \int_{-\sqrt{1-x^2}}^{?}(x-y)f(x,y)dy $$
$\{(x,y):-1{\le}y{\le}x{\le}1, x^2+y^2{\le}1\}$ is $\{(x,y):-\sqrt{\tfrac 12}{\le}x{\le}1, -\sqrt{1-x^2}{\le}y{\le}\min\{x,\sqrt{ 1-x^2}\}\}$
This is the union: $$\{(x,y):-\sqrt{\tfrac 12}{\le}x{\le}\sqrt{\tfrac 12}, -\sqrt{1-x^2}{\le}y\le x\}\cup\{(x,y):\sqrt{\tfrac 12}{<}x{\le}1, -\sqrt{1-x^2}{\le}y\le\sqrt{1-x^2}\}$$
View the interval in question.
Thus what you want is the sum of two integrals. $$\begin{align} \iint_{-1\le y\le x\le 1, x^2+y^2\le 1}(x-y)f_{X,Y}(x,y) \operatorname d(x,y) & = \int_{-\sqrt{1/2}}^{\sqrt{1/2}}\int_{-\sqrt{1-x^2}}^x (x-y)f_{X,Y}(x,y)\operatorname dy \operatorname dx \\ & +\int_{\sqrt{1/2}}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x-y)f_{X,Y}(x,y)\operatorname dy \operatorname dx\end{align}$$
Do likewise for the other interval , $\{(x,y):-1{\le}x{<}y{\le}1, x^2+y^2{\le}1\}$
Or just declare an equality via symmetry.
PS: Have you worked out what the density function, $f_{X,Y}(x,y)$, is yet?