1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Joint Probability FunctionStatistical Independence

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Picture1-45.png


    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

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

    and

    [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.

    Picture2-26.png

    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?

    Thanks,
    Casey
     
  2. jcsd
  3. Jun 2, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    Mark44

    Staff: Mentor

    I can't think of any situations where this isn't good to do.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook