Bivariate density on a unit circle

[gajohnson]

Homework Statement

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)

Homework Equations

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!

 

[tiny-tim]

hi gajohnson!

(do yo mean f(x, y) = 2/π ? )

The problem I'm having is setting the proper bounds of integration … I'm tempted to make the bounds √2/2 and -1 for the x's and 1 and -√2/2 for the y's.

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

 

[haruspex]

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

 

[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}.

 

[Ray Vickson]

Homework Statement

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)

Homework Equations

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!

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.

 

[gajohnson]

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