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Bivariate density on a unit circle

  1. Oct 24, 2013 #1
    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. jcsd
  3. Oct 24, 2013 #2


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    hi gajohnson! :smile:

    (do yo mean f(x, y) = 2/π ? :confused:)
    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 :wink:
  4. Oct 24, 2013 #3


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    It will be much easier if you rotate coordinates 45 degrees.
  5. Oct 24, 2013 #4


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    Start by drawing a picture. [itex]x^2+ y^2\le 1[/itex] 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 [itex]\left(\sqrt{2}/2, \sqrt{2}/2\right)[/itex] and [itex]\left(-\sqrt{2}/2, -\sqrt{2}/2\right)[/itex]. Clearly the smallest value x takes on is [itex]-\sqrt{2}/2[/itex] and the largest is [itex]\sqrt{2}/2[/itex]. Now, for each x, y goes from the line [itex]y= x[/itex] to the upper semi-circle [itex]y= \sqrt{1- x^2}[/itex].
  6. Oct 24, 2013 #5

    Ray Vickson

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    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.
  7. Oct 26, 2013 #6
    I was visualizing it properly but not connecting the obvious dots, thanks everyone!
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