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Function of two random variables

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data

    We have two independent, exponentially distributed random variables X and Y (with parameter a).

    Z = X/(X+Y)

    What is Z:s distributon function?

    2. Relevant equations



    3. The attempt at a solution

    I think I need some intuition to what I'm really doing with these, I'm having a really tough time and would be extremely grateful for a good explanation.

    P(Z < z) = P(X/(X+Y) < z)... Solving this problem corresponds to doing a double integral of f(x,y) over some interval (?) defined by z. Am I getting that correctly? I don't understand how to find that interval. Help??!

    Been at this for days but I'm still struggling. Having a hard time finding good materials that cover this aswell...
     
  2. jcsd
  3. Apr 28, 2013 #2

    Office_Shredder

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    It's going to be an integral over some region in the plane defined by z, not an interval

    The region is exactly what you have written down already. It's defined by the equation x/(x+y) < z
     
  4. Apr 28, 2013 #3
    Ok so we can write the area as: X < Yz/(1-z)

    So ∫∫ f(x,y) dx dy, were we integrate from 0 to Yz/(1-z) with respect to x, and from 0 to ∞ with respect to y?
     
  5. Apr 28, 2013 #4

    Office_Shredder

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    You have to be careful about whether 1-z is positive or negative when you divide both sides of the inequality though (so your region of integration is slightly different). Other than that it looks good
     
  6. Apr 29, 2013 #5

    haruspex

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    The need for care starts before that. You can't even get to X < z(X+Y) without worrying about the signs of factors you're multiplying by.
     
  7. Apr 29, 2013 #6
    Can you elaborate on what "being careful" entails here? Obviously z will have bounds but I guess you're talking about something else?
     
  8. Apr 29, 2013 #7

    Ray Vickson

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    For z > 0, [tex] \{X/(X+Y) \leq z \} = \{ X \leq z(X+Y) \} = \{ (1-z) X \leq z Y \}.[/tex] What do you get if 0 < z < 1? What do you get if z > 1?
     
  9. Apr 29, 2013 #8
    Well... if X and Y are positive, X/(X+Y) < z tells us that z < 1, right?
     
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