# Bivariate density on a unit circle

1. Oct 24, 2013

### gajohnson

1. The problem statement, all variables and given/known data

Consider the bivariate density of X and Y,
f(x, y) = pi/2 for x^2 + y^2 ≤1 and y > x
and = 0 otherwise.
(a) Verify that this is a bivariate density (that is, the total volume ∫∫ f(x,y)dxdy = 1)

2. Relevant equations

3. The attempt at a solution

The problem I'm having is setting the proper bounds of integration. It seem to me that this is describing a region on the unit circle that ranges from (√2/2 , √2/2) to (-√2/2 , -√2/2).

So in that case I'm tempted to make the bounds √2/2 and -1 for the x's and 1 and -√2/2 for the y's. But that doesn't seem to be giving me an answer of 1 when I integrate to find test whether or not it is a bivariate density. If the bounds are right, then I'll just keep plugging away at the integration until I get it. But, if my thinking is wrong on the bounds, any help would be greatly appreciated in figuring out how to set the right ones. Thanks!

Last edited: Oct 24, 2013
2. Oct 24, 2013

### tiny-tim

hi gajohnson!

(do yo mean f(x, y) = 2/π ? )
that (ie fixed limits for both variables) would give you a rectangle

you need fixed limits for the first variable (its least and greatest possible values), and limits that are a function of the first variable for the second variable

3. Oct 24, 2013

### haruspex

It will be much easier if you rotate coordinates 45 degrees.

4. Oct 24, 2013

### HallsofIvy

Start by drawing a picture. $x^2+ y^2\le 1$ is, of course, the unit disk. y= x is the line through the origin bisecting that disk and y> x is the region above that line. The circle and line boundaries intersect, as you say, at $\left(\sqrt{2}/2, \sqrt{2}/2\right)$ and $\left(-\sqrt{2}/2, -\sqrt{2}/2\right)$. Clearly the smallest value x takes on is $-\sqrt{2}/2$ and the largest is $\sqrt{2}/2$. Now, for each x, y goes from the line $y= x$ to the upper semi-circle $y= \sqrt{1- x^2}$.

5. Oct 24, 2013

### Ray Vickson

Since the probability density is constant over the relevant (x,y) region, and that region is a semi-circle, all you need to do is figure out the area of a semi-circle.

6. Oct 26, 2013

### gajohnson

I was visualizing it properly but not connecting the obvious dots, thanks everyone!