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(Easy?) Probability Q: Uniform in Uniform

  1. Apr 28, 2010 #1

    I have been struggling with a simple probability question.

    1. The problem statement, all variables and given/known data

    We are given that X is a uniformly distributed random variable on [0, 1]. After X is chosen, we take another uniform [0, 1] random variable Y (independent of X) and choose the subinterval that Y falls in. L is the length of this chosen subinterval (either [0, X] or (X, 1]), and we want its expectation.

    2. Relevant equations

    The expectation of a uniform random variable is (a+b)/2, where a and b are the interval bounds. These equations may also be useful:

    [tex]f_{Y}(y|X=x) = \frac{f(x,y)}{f_{X}(x)}[/tex]

    [tex]f_{X}(x|Y=y) = \frac{f_{Y}(y|X=x)f_{X}(x)}{f_{Y}(y)}[/tex]

    [tex]E(g(Y)|X=x) = \int \! g(y)f_{Y}(y|X=x) \,dx[/tex]

    [tex]E(Y) = \int \! E(Y|X=x)f_{X}(x) \,dx[/tex]

    3. The attempt at a solution

    I notice

    [tex]L=\left\{\begin{array}{cc}X,&\mbox{ if }
    Y\leq X\\1-X, & \mbox{ if } Y>Z\end{array}\right[/tex]

    And E(X) is obviously 1/2... but I do not know how to find E(L) :(

    I tried doing double integrals for expectation on both cases, and summing these, but I get 1/2 which doesn't seem right...

    [tex]E_0(L) = \int_{0}^{1} \int_{0}^{x} \! 1 y \, dy \,dx = 1/3[/tex]

    [tex]E_1(L) = \int_{0}^{1} \int_{x}^{1} \! 1 y \, dy \,dx = 1/6 [/tex]

    [tex]E(L) = E_0 + E_1 = 1/3 + 1/6 = 1/2[/tex]

    Any help?
    Last edited: Apr 29, 2010
  2. jcsd
  3. Apr 29, 2010 #2
    Where is the "y" in your expectation integrals coming from?
    The value function is either f(x) = x (when y < x) or f(x) = 1 - x (when y > x). I don't see that in your integrals.

    Recall that [tex] \int f(x)g(x) dx [/tex] is the integral for expectation where f(x) is the value function and g(x) is the density. Make sure when you construct your integrals that you have the right values and densities.
  4. Apr 29, 2010 #3
    I was using y as the value function, with the bounds ranging over all possible values.

    When I do integrals with x*1 from 0 to 1, I still get 1/2.
  5. Apr 29, 2010 #4
    Are these your integrals?

    E_0(L) = \int_{0}^{1} \int_{0}^{x} x \, dy \,dx

    E_1(L) = \int_{0}^{1} \int_{x}^{1} 1 - x \, dy \,dx

    They don't add up to 1/2.
  6. Apr 29, 2010 #5
    Yeah, you're right, I did my integrals wrong. By symmetry, it is obvious that it should be 2/3, which is what those add up to. Thanks!
  7. May 12, 2010 #6
    hi there im new with this stuff but need some help... I've given a question in Finding C,D, F(x). Where the mean is 10 standard deviation is 1. (Uniform Probability distribution) Can you tell me which formula i should used...
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