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Integration with absolute value

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem is in the context of a probability problem; however, my question in regards to a computation regarding a particular integral. All that is needed to know is that the probability density function is 1 in the range 0 < y < 1 , y-1 < x < 1 - y, and 0 otherwise.

    I need to find E(XY) and E(|XY|).

    2. Relevant equations
    [tex] E(XY) = \int_0^{1} \int_{y-1}^{1-y}\ XY \ dx dy[/tex]
    [tex] E(|XY|) = \int_0^{1} \ \int_{y-1}^{1-y} \ |XY| \ dx dy[/tex]

    3. The attempt at a solution

    E(XY) = 0. Simple calculation.

    For E(|XY|) however, I am getting 1/24, but WolframAlpha computes it out to be 1/12.

    I am doing:
    [tex] E(|XY|) = \int_0^{1}\ \int_{y-1}^{1-y}\ |XY| \, dx dy[/tex] = [tex]

    \int_0^{1} \int_0^{1-y} \ XY \, dx dy[/tex]

    I am not sure why this is incorrect. Making the lower bound of x 0 worked for me when I was computing E(|X|), but is giving me half the answer for E(|XY|).
    Last edited: Oct 13, 2013
  2. jcsd
  3. Oct 13, 2013 #2


    Staff: Mentor

    Fixed your LaTeX, which was garbling your integrals.

    For definite integrals, the LaTeX markup should look like this:

    \int_{lower_bound}^{upper_bound} <integrand> dx dy

    If lower_bound or upper_bound is a single character, you don't need the braces, but for expressions such as y - 1, you do need braces, not parentheses as you used.

    Also, you had stuff like \1 - y, which was causing problems.
  4. Oct 13, 2013 #3
    Thanks! I was surprised it magically fixed itself. I'll keep that in mind for future posts.
  5. Oct 13, 2013 #4


    Staff: Mentor

    I don't think this is right:
    $$ \int_0^{1} \int_0^{1-y} \ XY \, dx dy$$
    You are changing the shape of the region of integration, removing half of it. The region over which integration should be taking place is the triangle whose vertices are (-1, 0), (0, 1), and (1, 0). The region you are using in the integral above is also a triangle, with vertices (0, 0), (0, 1), and (1, 0), that has half the area of the larger triangle. I believe that's why your answer (1/24) is half of what Wolfram shows (1/12).
  6. Oct 13, 2013 #5
    That does make sense geometrically, I rechecked and it turns out.

    $$ 2 * \int_0^{1} \int_0^{1-y} \ X \, dx dy = \int_0^{1} \int_{y-1}^{1-y} \ |X| \, dx dy $$

    So you would need a factor of 2 for |X| as well. Thanks for the help!
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