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Math Help - another order stats question

  1. #1
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    another order stats question

    Let Y_1 and Y_2 be independent and uniformly distributed over the interval (0, 1). Find P(2Y_{(1)} < Y_{(2)}). The solution is 0.5.

    Originally I thought it was:

    g_{(2)} = n[1-F(y)]^{n-1} f(y)= 2y \times 1 = 2y

    g_{(1)}=1-[1-F(y)]^n = 1-[1-y]^2 = 2y-y^2

    2(2y-y^2) < 2y

    2y<y^2

    but this doesn't seem to right. If anybody has any input, it would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by lllll View Post
    Let Y_1 and Y_2 be independent and uniformly distributed over the interval (0, 1). Find P(2Y_{(1)} < Y_{(2)}). The solution is 0.5.

    Originally I thought it was:

    g_{(2)} = n[1-F(y)]^{n-1} f(y)= 2y \times 1 = 2y

    g_{(1)}=1-[1-F(y)]^n = 1-[1-y]^2 = 2y-y^2 Mr F says: This is not the pdf for {\color{red}Y_{(1)}}, it's the cdf ..... The pdf is {\color{red}n [1 - F(y)]^{n-1} f(y)} where n = 2 in this case.

    Also, there should be a different symbol used for the variable in each pdf:

    {\color{red}g_{(2)}(y_2) = n[1-F(y_2)]^{n-1} f(y_2)= 2y_2 \times 1 = 2y_2}

    {\color{red}g_{(1)}(y_1) = n [1-F(y_1)]^{n-1} f(y_1) = 2(1 - y_1)}

    [snip]
    The key is to know that if Y_1 and Y_2 are i.i.d. random variables with density function f(y) then the joint density function of Y_{(1)} and Y_{(2)} is

    g_{12}(y_1, \, y_2) = 2 f(y_1) f(y_2) for y_1 \leq y_2

    and zero elsewhere. I can prove this result if you want, but you should have a copy of the proof somewhere.

    For your problem g_{12}(y_1, \, y_2) = 2 (1) (1) = 2 for y_1 \leq y_2 and 0 \leq y_2 \leq 1.

    Then \Pr(2Y_{(1)} < Y_{(2)}) = \Pr \left(Y_{(1)} < \frac{1}{2} Y_{(2)}\right) = \int_{y_2 = 0}^{y_2 = 1} \int_{y_1 = 0}^{y_1 = y_2/2} 2 \, dy_1 \, dy_2 = \frac{1}{2}.
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