Function of two random variables

  • Thread starter Gauss M.D.
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  • #1
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Homework Statement



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

Z = X/(X+Y)

What is Z:s distributon function?

Homework Equations





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...
 

Answers and Replies

  • #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
 
  • #3
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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?
 
  • #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
 
  • #5
haruspex
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You have to be careful about whether 1-z is positive or negative
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.
 
  • #6
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Can you elaborate on what "being careful" entails here? Obviously z will have bounds but I guess you're talking about something else?
 
  • #7
Ray Vickson
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Can you elaborate on what "being careful" entails here? Obviously z will have bounds but I guess you're talking about something else?

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?
 
  • #8
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Well... if X and Y are positive, X/(X+Y) < z tells us that z < 1, right?
 

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