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

Staff Emeritus
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!