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[PROBABILITY] Conditional probability for random variable

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    X and Y two independent random variables with distribution U(0, 1/2). Find the density of (X + Y)2|X - Y > 0

    3. The attempt at a solution

    I was hoping this would be simpler, but somehow I always end up with nothing.

    The only thing I can work out just fine is that P(X - Y > 0) = 1/2.

    Then I tried to find the conditional probability with P((X + Y)2|X - Y > 0) = P((X + Y)2, X - Y > 0)/P(X - Y > 0), and then stop dead.

    Any guidance here? Because I have to do many of these problems.
  2. jcsd
  3. Apr 24, 2010 #2
    X and Y are uniformly distributed between 0 and 0.5

    density of x-y>0, x>y is just 1/2*uniform density which is 1/0.5*0.5

    for finding density of (x+y)^2 AND x-y>0
    define one random variables
    h = (x+y)^2
    g = (x-y)^2

    Using Jocbian find f_hg, and then using appropriate limits (x-y>0 and 0 and 0.5) find f_h

    The conditional will be just be f_h/2

    I am hoping that should work.
    Last edited: Apr 24, 2010
  4. Apr 24, 2010 #3
    It isn't so easy to find the Jacobian, since the inverse transformation is quite complicated. Unless I can just factor like this:

    g = (X+Y)2 ------> g1/2 = X+Y.

    But I would lose information, wouldn't I?
  5. Apr 24, 2010 #4
    OK, if I can do what I just did, I find that

    fhg = fxy((g1/2+h1/2)/2, (g1/2-h1/2)/2)/|-1/(8(hg)1/2)| = 8 (hg)1/2, if 0<(g1/2+h1/2)/2<1/2 and 0<(g1/2-h1/2)/2<1/2.

    Now, I don't understand what you say about taking apropriate limits considering x-y>0 to find f_h.
  6. Apr 24, 2010 #5
    Also, why did you choose those two random variables to do the Jacobian? Why is g that way?
  7. Apr 24, 2010 #6
    Two multiple random variables you need a pseudo variable to take Jacobian (pg 7 and 8):

    That's one the methods to solve. G(x,y) can be anything. In the end you will get same f_H.

    Also, Note that |J_gh| = 1/|J_xy|

    When I said limits, I meant to find f_H you need to integrate f_GH and you need limits for that definite integral. That's what I meant by those limits.

    There are other approaches but I only use above for majority of the times.
  8. Apr 24, 2010 #7
    No, I know about the Jacobian method. I meant, how did you come up with h = (X-Y)2 to make the transformation one-to-one?
  9. Apr 24, 2010 #8
    Randomly picked, h =x+y might have been easier
  10. Apr 24, 2010 #9
    But I can't randomly pick any given variable, I could end up with a 2-to-1 or 3-to-1 function.

    Anyway, I'm having troubles setting up the limits of the function f_HG (that is, where it isn't zero).

    I have that h is in [0,1] and g is in [0,1/4]. Then, with the inverse transformation and considering that both X and Y are uniform in (0, 1/2):

    0< (g1/2 + h1/2)/2 < 1/2
    0< (g1/2 - h1/2)/2 < 1/2


    0< g1/2 + h1/2 < 1
    0< g1/2 - h1/2 < 1

    -h1/2< g1/2 < 1 - h1/2
    h1/2< g1/2 < 1 + h1/2

    I can see that this ultimately tells me that

    -h1/2< g1/2 < 1 + h1/2

    But how do I get rid of the sqrt?
  11. Apr 24, 2010 #10
    Do you have the answer?
    I managed to get
    1/4 (1/sqrt(u)-1)
    I found it bit simpler to solve without any complicated equations..

    I don't understand
  12. Apr 25, 2010 #11
    How did you do that?

    This is what I did: I need the inverse transformation first (the one that turns h and g into x and y).

    g = (X+Y)2 ----------> g1/2 = X+Y
    h = (X-Y)2 ----------> h1/2 = X-Y

    g1/2 + h1/2 = 2X ----------> X = (g1/2 + h1/2)/2
    g1/2 - h1/2 = 2Y ----------> Y = (g1/2 - h1/2)/2

    Which ends this. The Jacobian is then -1/(8h1/2g1/2) and so:

    fHG(h,g) = fXY ((g1/2 + h1/2)/2, (g1/2 - h1/2)/2)*(1/(8h1/2g1/2)).

    Since fXY is uniform (4), all I have is 1/(2h1/2g1/2).

    And I couldn't work the limits out.

    How did you get to that result so fast?
  13. Apr 25, 2010 #12
    Using the assumption that other random variable can be defined as x.
    h = (x+y)^2
    g = x

    |J_xy| = |-2 (x+y)|
    |J_hg| = 1/ (2(x+y))

    Joinit distribution:
    f_gh = f_xy/[2(x+y)]
    f_gh = 2/[2(x+y)] {Using 2 for f_xy, it should be 8 I believe 1/(0.5*0.5*0.5) given x>y distribution doubles}
    = 1/(x+y)
    = 1/sqrt(h)

    Marginal distribution:
    x = v
    y = sqrt(u)-v

    f_h = integrate f_gh from sqrt(u)/2 to 1/2
  14. Apr 25, 2010 #13
    That I didn't understand. What did you mean with that?
  15. Apr 25, 2010 #14
    You are not interested in f_gh but only in f_h. So, I found f_h knowing that
    x = ..
    y = ..
    and x>y
    in terms of u and v (used u for h and v for g)
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