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Problem with probability theory and random variables

  1. May 8, 2013 #1
    Hello. I have a problem with probability theory task.
    The task is:
    X and Y is independent random variables with same density function fx=fy=f. What will be probability of P(X>Y).

    This P(X>Y) reminds me a cdf: P(X>Y)=1-P(X<Y)=1-cdf of X.

    Cdf of x is equal to integral ∫f dx from -inf to y.
    But dont know how to continue because Y depends on its dencity. Maybe some one will be able to help me?
  2. jcsd
  3. May 8, 2013 #2


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    It should be clear that if x and y have the same probability distribution, then the probability x< y is the same as the probability y< x. And since this is a is a continuous distribution, the probability that x= y is 0.
  4. May 8, 2013 #3
    So the answer is 1/2. hm :) ok
  5. May 8, 2013 #4

    Ray Vickson

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    You can also get it using integration:
    [tex] P(X>Y) = \int_{-\infty}^{\infty} G(y) f(y)\, dy,[/tex]
    where ##G(y) = P(X>y)##. We have ##f(y) \, dy = -dG(y),## so
    [tex] P(X>Y) = -\frac{1}{2} \left. G(y)^2 \right|_{-\infty}^{\infty} = 1/2. [/tex]
  6. May 8, 2013 #5
    Also i have very similar question. Suppose that X,Y,Z are independent random variables and uniformly distributed in the interval [0;1]. What will be P(XY<Z^2)
    From that i know that density of all thre r.v. is 1 if xε[0,1] and 0 otherwise.
    So fx*fy=1 when x[0;1] and 0 otherwise. cdf of this would be F=x when x[0;1] and 0 otherwise...
    But whats about Z^2?
  7. May 8, 2013 #6

    Ray Vickson

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    Start by writing down the integrations you need to perform.
  8. May 8, 2013 #7
    I want to ask, from where you get this:
    [tex] P(X>Y) = \int_{-\infty}^{\infty} G(y) f(y)\, dy,[/tex]
    G(y) is cdf and f(y) density?
  9. May 8, 2013 #8


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    P[X>Y] = ƩP[X>Y|Y[itex]\in[/itex]I]P[Y[itex]\in[/itex]I], where the sum is over a partition of the range of Y. In the limit of vanishingly small intervals I, that produces the integral.
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