How to verify a bivariate density? Verify that the following is a bivariate density:
$F(X, Y) = \displaystyle\frac{2}{\pi}$ for $X^2+Y^2\leq1$ and $X < Y$
$F(X,Y) = 0$ otherwise
Now, I know that verifying that this is a bivariate density involves taking a double integral and showing that it is equal to 1. However, I'm not quite sure what the limits are for this integral! Here is the integral I have:
$\displaystyle\int\displaystyle\int \displaystyle\frac{2}{\pi} dx dy$
How can I find out the limits for this integral?
Thank you!
 A: My calculus is probably rusty.
I think it's a good idea to sketch the region over which you will integrate. Firstly, we know that:
$$X^{2}+Y^{2}\leq 1$$
which is the region bound by a unit circle. Furthermore, we have the constraint that:
$$X<Y$$
so the desired region is that bounded by a unit circle where $X$ is less than $Y$. To illustrate:

If you want you can convert to polar coordinates as follows:
$$\begin{align}
x=r\text{cos}(\theta)\\
y=r\text{sin}(\theta)
\end{align}$$
where the Jacobian is given by:
$$\begin{align}
J&=\left|\begin{array}{cc}
\tfrac{\partial x}{\partial r} & \tfrac{\partial x}{\partial \theta}\\
\tfrac{\partial y}{\partial r} & \tfrac{\partial y}{\partial \theta}
\end{array} \right|
\\
&=\left|\begin{array}{cc}
\text{cos}(\theta) & -r\text{sin}(\theta)\\
\text{sin}(\theta) & r\text{cos}(\theta)
\end{array} \right|
\\
&=r
\end{align}$$
You'll be able to determine the appropriate limits of integration from the plot. In polar coordinates, the bounds are:
$$\begin{align}
0&\leq r\leq 1\\
\tfrac{\pi}{4}&\leq \theta \leq \tfrac{5\pi}{4}
\end{align}$$
The integral you want to evaluate is:
$$\frac{2}{\pi}\int_{\pi/4}^{5\pi/4}\int_{0}^{1}r\,dr\,d\theta$$
