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Thread: Order statistics

  1. #1
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    Order statistics

    Let Y1<Y2<Y3 be the order statistics of a random sample from a Uniform $\displaystyle (\theta,1)$ distribution.

    1) Show that Y1 is an unbiased estimator of $\displaystyle \theta$
    2)Find an unbiased estimator of $\displaystyle \theta$ and that it is unbiased.

    Thanks for the help, biased and unbiased have always confused me
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  2. #2
    MHF Contributor matheagle's Avatar
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    well, without doing any math I'm sure Y1 is BIASED for theta

    unbiased is $\displaystyle E(Y_1)=\theta$

    But since the density of Y1 will be between theta and 1 there is no way it's expected value can be either endpoint.
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    How can I show that Y1 is a biased estimator of $\displaystyle \theta$?
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  4. #4
    MHF Contributor matheagle's Avatar
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    obtain it's density and compute it's expected value
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  5. #5
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    attempt of an answer to show that Y1 is a biased estimator of $\displaystyle \theta$

    Y1 being the minimum of the order statistic, i get that the distribution of the minimum is $\displaystyle f(y1)=n*[1-F(y)]^(n-1)*f(y)$

    after plugging in $\displaystyle f(y1)=3*[1-\frac{y-\theta}{1-\theta}]^2*\frac{1}{1-\theta}$

    and then to get the expected value of Y1 i get the $\displaystyle \int$ from $\displaystyle \theta$ to 1 of Y1 times f(y1)

    am i in the right direction or completely off?
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  6. #6
    MHF Contributor matheagle's Avatar
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    $\displaystyle f_{Y_1}(y)=3*\left(1-\frac{y-\theta}{1-\theta}\right)^2*\frac{1}{1-\theta} $

    is the density of the first order stat. Next integrate

    $\displaystyle yf_{Y_1}(y)$ wrt y from theta to 1.

    $\displaystyle {3\over 1-\theta}\int_{\theta}^1 y\left(1-\frac{y-\theta}{1-\theta}\right)^2 dy$
    Last edited by matheagle; Mar 1st 2010 at 08:34 PM.
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  7. #7
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    non-standard interval

    Hi,
    I have gone through the steps on a similar type of problem - the main difference is that the distribution is $\displaystyle U(\theta_1, \theta_2) $ with n > 10 and $\displaystyle \theta_1 < \theta_2 $ .

    I found the MLE's to be
    $\displaystyle
    \theta_1 = X_{(1)}
    $
    and
    $\displaystyle
    \theta_2 = X_{(n)}
    $

    I'm trying to show whether they are biased on unbiased. I get the density and it looks like Beta (1, n) and Beta (n,1) for $\displaystyle \theta_1 $ and $\displaystyle \theta_2 $, respectively, except that I have an extra term of
    $\displaystyle
    \frac {1} {\theta_2 - \theta_1}
    $
    in the function.

    I am not sure what to do with that - do I just multiply the mean of the beta function with that fraction? It seems like I would need to "do something" with it in order for the equation to integrate to 1?
    I would love a hint... Thanks!
    Last edited by Statistik; Apr 3rd 2010 at 05:59 PM.
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  8. #8
    Moo
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    Hello Statistik,

    Please show your working, because it's impossible to see if you've made a mistake somewhere or what you call "looks like a Beta"
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  9. #9
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    non-standard interval

    Moo,

    Sorry. Here it is:

    $\displaystyle
    iid: U(\theta_1, \theta_2) => MLE (\theta_1, \theta_2) = (X_{(1)}, X_{(n)})
    $

    $\displaystyle
    f_{X_{(1)}}(x) = n * (1 - \frac {x-\theta_1} {\theta_2 - \theta_1})^{(n-1)} * \frac {1} {\theta_2 - \theta_1} $ ~ $\displaystyle
    \frac {1} {\theta_2 - \theta_1} * Beta (1,n) $

    I am then trying to get $\displaystyle E[X_{(1)}] $ = ? would this be $\displaystyle \frac{1} {\theta_2 - \theta_1} * \frac {1} {n+1} $ ?

    For $\displaystyle X_{(n)} $, I get the same thing except the distribution is Beta (n,1) and gives $\displaystyle E[X_{(n)}] = \frac{1} {\theta_2 - \theta_1} * \frac {n} {n+1} $ ?

    Thanks!
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  10. #10
    Moo
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    Quote Originally Posted by Statistik View Post
    Moo,

    Sorry. Here it is:

    $\displaystyle
    iid: U(\theta_1, \theta_2) => MLE (\theta_1, \theta_2) = (X_{(1)}, X_{(n)})
    $

    $\displaystyle
    f_{X_{(1)}}(x) = n * (1 - \frac {x-\theta_1} {\theta_2 - \theta_1})^{(n-1)} * \frac {1} {\theta_2 - \theta_1} $ ~ $\displaystyle
    \frac {1} {\theta_2 - \theta_1} * Beta (1,n) $

    I am then trying to get $\displaystyle E[X_{(1)}] $ = ? would this be $\displaystyle \frac{1} {\theta_2 - \theta_1} * \frac {1} {n+1} $ ?

    For $\displaystyle X_{(n)} $, I get the same thing except the distribution is Beta (n,1) and gives $\displaystyle E[X_{(n)}] = \frac{1} {\theta_2 - \theta_1} * \frac {n} {n+1} $ ?

    Thanks!
    This is not a Beta distribution... Especially with the support ! With a substitution, you can get something like n(1-t)^(n-1) (the 1/(theta2-theta1) simplifies with the du).
    But it's not a Beta distribution.

    Your pdf is correct though you have to state the support, in which interval it belongs. Namely $\displaystyle [\theta_1,\theta_2]$

    In order to find its expectation, just calculate $\displaystyle \int_{\theta_1}^{\theta_2} x ~f_{X_{(1)}}(x) ~dx$, like you would do for any other random variable.
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  11. #11
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    Thanks!

    Moo,

    THANK YOU! I won't forget to consider the support again, either. ;-)
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