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Math Help - uniform order stats question

  1. #1
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    uniform order stats question

    I'm having some difficulty finding the mean and the variance of the following question:

    Let Y_1, Y_2,... \ ,Y_n be independent uniformly dist. r.v. on the interval [0, \theta]

    I found my density function for kth order stat between 1 and n, which is:

    \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} where 0 \leq y  \leq \theta


    now using what I found if I try to find the E(Y_{(k)}) I get:

    \int_{0}^{\theta} \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} \times y \ dy where 0 \leq y  \leq \theta

    = \frac{n!}{(k-1)!(n-k)!} \int_{0}^{\theta} \left( \frac{1}{\theta} y \right)^{k} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \ dy where 0 \leq y  \leq \theta


    now the integral looks a lot like a Beta Function, but what puzzles me is that the beta function goes from 0 to 1, and I have 0 to \theta. If I were to make the substitution. I would get:

    \frac{n!}{(k-1)!(n-k)!} B[k+1, \ n-k+1] where 0 \leq y  \leq \theta

    And for the E(Y_{(k)}^2) I get:

    \int_{0}^{\theta} \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} \times y^2 \ dy where 0 \leq y  \leq \theta

    = \frac{n!}{(k-1)!(n-k)!} \int_{0}^{\theta} \left( \frac{1}{\theta} y \right)^{k} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times y \ dy where 0 \leq y  \leq \theta

    Again the first part of the integral looks like the Beta Function, but since there's an additional y within it, I'm a little confused on how to integrate this out, since I don't think that a closed form of this function exists.
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  2. #2
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    Quote Originally Posted by lllll View Post
    I'm having some difficulty finding the mean and the variance of the following question:

    Let Y_1, Y_2,... \ ,Y_n be independent uniformly dist. r.v. on the interval [0, \theta]

    I found my density function for kth order stat between 1 and n, which is:

    \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} where 0 \leq y \leq \theta


    now using what I found if I try to find the E(Y_{(k)}) I get:

    \int_{0}^{\theta} \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} \times y \ dy where 0 \leq y \leq \theta

    = \frac{n!}{(k-1)!(n-k)!} \int_{0}^{\theta} \left( \frac{1}{\theta} y \right)^{k} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \ dy where 0 \leq y \leq \theta


    now the integral looks a lot like a Beta Function, but what puzzles me is that the beta function goes from 0 to 1, and I have 0 to \theta. If I were to make the substitution. I would get:

    \frac{n!}{(k-1)!(n-k)!} B[k+1, \ n-k+1] where 0 \leq y \leq \theta

    And for the E(Y_{(k)}^2) I get:

    \int_{0}^{\theta} \frac{n!}{(k-1)!(n-k)!} \times \left( \frac{1}{\theta} y \right)^{k-1} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times \frac{1}{\theta} \times y^2 \ dy where 0 \leq y \leq \theta

    = \frac{n!}{(k-1)!(n-k)!} \int_{0}^{\theta} \left( \frac{1}{\theta} y \right)^{k} \times \left( 1-\frac{1}{\theta}y \right)^{n-k} \times y \ dy where 0 \leq y \leq \theta

    Again the first part of the integral looks like the Beta Function, but since there's an additional y within it, I'm a little confused on how to integrate this out, since I don't think that a closed form of this function exists.
    In both cases you should make the substitution u = \frac{y}{\theta} \Rightarrow y = u \theta and dy = \theta \, du.

    Then you have:


    E(Y_{(k)}) = \frac{\theta \, n!}{(k-1)! \, (n-k)!} \int_0^1 u^k \, (1 - u)^{n-k} \, du = \frac{\theta \, n!}{(k-1)! \, (n-k)!} B(k+1, \, n - k + 1)


    and


    E(Y_{(k)}^2) = \frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} \int_0^1 u^{k+1} \, (1 - u)^{n-k} \, du = \frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} B(k+2, \, n - k + 1).


    (I haven't checked your pdf calculation but it integrates to 1 so looks OK).
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  3. #3
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    would it be possible to find the Variance, since you would get:

    <br />
\frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} B(k+2, \, n - k + 1) -\left( \frac{\theta \, n!}{(k-1)! \, (n-k)!} B(k+1, \, n - k + 1) \right)^2

    this might sound like a stupid question, but is it even possible to square a binomial representation or a Beta function?
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  4. #4
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    Quote Originally Posted by lllll View Post
    would it be possible to find the Variance, since you would get:

    <br />
\frac{\theta^2 \, n!}{(k-1)! \, (n-k)!} B(k+2, \, n - k + 1) -\left( \frac{\theta \, n!}{(k-1)! \, (n-k)!} B(k+1, \, n - k + 1) \right)^2

    [snip]
    Yes.

    Note that if x and y are positive integers: B(x, y) = \frac{(x-1)! \, (y-1)!}{(x + y - 1)!}.
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