Results 1 to 2 of 2

Thread: another order stats question

  1. #1
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1

    another order stats question

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

    Originally I thought it was:

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

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

    $\displaystyle 2(2y-y^2) < 2y $

    $\displaystyle 2y<y^2$

    but this doesn't seem to right. If anybody has any input, it would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by lllll View Post
    Let $\displaystyle Y_1$ and $\displaystyle Y_2$ be independent and uniformly distributed over the interval $\displaystyle (0, 1)$. Find $\displaystyle P(2Y_{(1)} < Y_{(2)})$. The solution is 0.5.

    Originally I thought it was:

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

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

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

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

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

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

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

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

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

    Then $\displaystyle \Pr(2Y_{(1)} < Y_{(2)}) = \Pr \left(Y_{(1)} < \frac{1}{2} Y_{(2)}\right) = \int_{y_2 = 0}^{y_2 = 1} \int_{y_1 = 0}^{y_1 = y_2/2} 2 \, dy_1 \, dy_2 = \frac{1}{2}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] expected value of order stats for uniform distribution
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Feb 11th 2011, 02:08 PM
  2. Order Stats and sufficiency question
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Dec 8th 2008, 04:25 PM
  3. min order stats and unbiased estimator.
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Nov 16th 2008, 01:13 AM
  4. uniform order stats question
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Oct 18th 2008, 10:19 PM
  5. Stats help (combinations/ order prb)
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Nov 5th 2007, 11:33 AM

Search Tags


/mathhelpforum @mathhelpforum