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Homework Help: Limits of Integration for Joint Distribution problems

  1. Jun 23, 2012 #1
    Suppose that (X,Y) is uniformly distributed over the regiondefined by 0≤ y ≤ 1-x2
    and -1≤ x ≤ 1.

    a) find the marginal densities of X and Y

    Attempted solution:

    So first I have to find the joint density function which ends up being fxy(x,y) = 3/4

    and then from that I would solve for the marginal densities. Since there was a solution I was able to do these things, but my issue is finding the limits of integration.

    in finding the joint density, why did they use -1 to 1 as the limit of integration and (1-x2) to find the joint density function. Then to find the marginal densities, they used 0 to 1-x2 to find the marginal density of X and ± (1-y)1/2 to find the marginal density of Y. How and why did these limits occur?
  2. jcsd
  3. Jun 23, 2012 #2


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    Science Advisor

    I assume you took a "multi-variable Calculus" course before this course. That is what is being used here. You are given that the region is "defined by 0≤ y ≤ 1-x2
    and -1≤ x ≤ 1." To cover that region, x has to go from -1 to 1 and, for each x, y goes from 0 to [itex]1- x^2[/tex]. Since you are also given that this is a uniform probability distribution, a constant over the region, whicy we can call "M", and so the integral over the entire region must be
    [tex]\int_{x=-1}^1\int_{y= 0}^{1- x^2} M dydx= 1[/tex]
    Which gives M= 4/3. That's what you did isn't it?

    Now, look at the "inner integral":
    [tex]\frac{4}{3}\int_{y= 0}^{1- x^2} dy[/tex]
    That is the integral with respect to y for fixed x. It is the probability of y for fixed x which is what is meant by "marginal probability".

    Now, look at the region described by "[itex]-1\le x\le 1[/itex]" and "[itex]0\le y\le 1- x^2[/itex]. That is the region under the parabola [itex]y= 1- x^2[/itex] between x= -1 and x= 1. Clearly y can take on values as low as 0 or as high as 1. For each y, then x can go from the parabola on the left to the parabola on the right. Since the parabola is given by [itex]y= 1- x^2[/itex], [itex]x^2= 1- y[/itex] and [itex]x= \pm\sqrt{1- y}[/itex]. The left side is, of course, [itex]x= -\sqrt{1- y}[/itex] and the right side is [itex]x= +\sqrt{1- y}[/itex].

    That is, if you were to integrate
    [tex]\int_{y= 0}^1\int_{x=-\sqrt{1- y}}^{\sqrt{1- y}}\frac{4}{3}dxdy[/tex]
    You would get "1" again. Removing the outside integral gives the "marginal probabilty" of x for a given y:
    [tex]\int_{x= -\sqrt{1- y}}^{\sqrt{1- y}}\frac{4}{3}dx[/tex]
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