Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Oddly Worded uniform probability Question

  1. Jul 4, 2004 #1
    X has a uniform distribution of [0,1]
    Y = h(x) = max(X,1-X)

    question: pdf is what?

    i think it is (x-a)/(B-A) = x , so for
    pdf = x for x>0.5 and 1-x for x<0.5,

    then part b is median, and part c is expected value and variance.

    This is not a homework, but this is on a review sheet.
  2. jcsd
  3. Jul 5, 2004 #2
    When thinking about this type of question, one generally works with the distribution function first.

    Note that max(M, N) <= y if and only if M<=y and N <=y. Thus

    Pr(max(1-X, X)<=x)
    =Pr(1-X<=x and X<=x)

    note that when x<0.5, the RHS is larger than the LHS such that the probability equals zero. For 0.5<=X<=1, the probability is easily seen to be x-(1-x) = 2x-1. When x>1, the statement "1-x<=X<=x" is trivially true (as 0<=X<=1) so the probabiluity equals 1. Combining, (the F(y) be the distribution function of Y)

    F(y) = 0 when y<0.5
    = 2x-1 when 0.5<=y<=1
    = 1 when y>1

    From this one may obtain the density function and so on.
  4. Jul 5, 2004 #3


    User Avatar
    Science Advisor

    It looks to me that Y is uniformly distributed between .5 and 1. There are 2 cases, depending on x<.5 or x>.5. In the first case 1-x (the max) is uniform between .5 and 1, while in the second case x (the max) is uniform between .5 and 1.
  5. Jul 6, 2004 #4
    If I've followed right you're saying that the pdf is uniform and is limited to [0.5,1]

    Wong, I followed your logic after the first line.
    I think i get what mathman is saying as well.

    Thanks for your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook