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Joint Probability FunctionStatistical Independence

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    If X and Y are statistically independent, then f(x,y) = g(x)h(y) where
    [itex]g(x) = \int f(x,y) dy[/itex]
    [itex]h(y) = \int f(x,y) dx[/itex]

    3. The attempt at a solution

    [itex]g(x) = \int f(x,y) dy = \int_{y=0}^{1-x} 6x\, dy[/itex]
    [itex]\Rightarrow g(x)=6x(1-x)[/itex]


    [itex]h(y) = \int f(x,y) dx = \int_{x=0}^{1} 6x \,dx[/itex]
    [itex]\Rightarrow h(y)=3[/itex]

    Thus h(y)g(x) [itex]\ne[/itex] f(x,y) and thus X and Y are NOT statistically independent.

    Now before I move onto (b) look at the solution that the text gives.


    I have no idea what is going on in the upper bound for the h(y) integral? They also went a different route with the solution, but I think that my way should work since it is a definition of independence. But clearly our h(y) functions should be the same. What am I missing?

  2. jcsd
  3. Jun 2, 2010 #2


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    If you sketch the region over which f(x,y)=6x, you'll see it's a triangle with vertices at (0,0), (0,1), and (1,0). From the sketch, you can see that for a given value of y, f(x,y)=6x only for values of x between 0 and 1-y; therefore, the upper limit of the integral should be 1-y, not 1.
  4. Jun 3, 2010 #3
    Excellent! I should have sketched it. Thanks Vela. :smile:
  5. Jun 3, 2010 #4


    Staff: Mentor

    I can't think of any situations where this isn't good to do.
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