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Expected value of a third order statistic?

  1. Mar 11, 2009 #1
    Hi all. I'm struggling with this HW question. I've searched through the textbook and on the web and have been unable to find a solution

    1. The problem statement, all variables and given/known data
    I've got 4 i.i.d. random variables, X1, X2, X3, X4. Uniformly distributed on [0,1]
    so the pdf = 1
    and cdf F(x_i) = x_i

    Let Y3 = the third highest value of the 4 variables.

    What is E(Y3)? I know the answer is 2/5. But I can't figure out how to show this.

    2. Relevant equations

    I know the formula for E(Y2) (second highest) and E(Y1) (highest), but I cannot find anything about the expectation of the third highest.

    [tex] E(Y_1) = \int_{0}^{1} y \cdot 4y^3 \, dy = 4/5 [/tex]
    [tex] E(Y_2) = \int_{0}^{1} y \cdot (12y^2 - 12y^3) \, dy = 3/5 [/tex]

    based on the pattern, one would assume that E(Y3) = 2/5 (plus I know that's the answer), but I can't derive the formula.

    3. The attempt at a solution
    I've tried a bunch of different approaches, but they've all been dead ends. Maybe I'm misunderstanding the problem. Or maybe just misunderstanding the concept of order statistics entirely.
  2. jcsd
  3. Mar 11, 2009 #2
    alright I figured out my problem. I had it backwards. Our teacher uses (Y1) to denote the highest value of the 4 random variables, but the stuff I found on the web used Y1 to denote the LOWEST value.

    so [tex] f_{Y_3} = \frac{4!}{1! \cdot 2!} y \cdot (1-y)^2 \cdot 1 [/tex]

    integrate to find the expected value, and we get

    [tex] E(Y_3) = \int_{0}^{1} y \cdot (12y(y-1)^2) dy = \frac{2}{5} [/tex]

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