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Math Help - 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 (\theta,1) distribution.

    1) Show that Y1 is an unbiased estimator of \theta
    2)Find an unbiased estimator of \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 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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  3. #3
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    How can I show that Y1 is a biased estimator of \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 \theta

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

    after plugging in 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 \int from \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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    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

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

    {3\over 1-\theta}\int_{\theta}^1 y\left(1-\frac{y-\theta}{1-\theta}\right)^2  dy
    Last edited by matheagle; March 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  U(\theta_1, \theta_2) with n > 10 and  \theta_1 < \theta_2 .

    I found the MLE's to be
    <br />
\theta_1 = X_{(1)}<br />
    and
    <br />
\theta_2 = X_{(n)}<br />

    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  \theta_1 and  \theta_2 , respectively, except that I have an extra term of
    <br />
\frac {1} {\theta_2 - \theta_1}<br />
    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; April 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:

    <br />
iid: U(\theta_1, \theta_2) => MLE (\theta_1, \theta_2) = (X_{(1)}, X_{(n)})<br />

    <br />
f_{X_{(1)}}(x) = n * (1 - \frac {x-\theta_1} {\theta_2 - \theta_1})^{(n-1)} * \frac {1} {\theta_2 - \theta_1} ~ <br />
 \frac {1} {\theta_2 - \theta_1} * Beta (1,n)

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

    For  X_{(n)} , I get the same thing except the distribution is Beta (n,1) and gives  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:

    <br />
iid: U(\theta_1, \theta_2) => MLE (\theta_1, \theta_2) = (X_{(1)}, X_{(n)})<br />

    <br />
f_{X_{(1)}}(x) = n * (1 - \frac {x-\theta_1} {\theta_2 - \theta_1})^{(n-1)} * \frac {1} {\theta_2 - \theta_1} ~ <br />
 \frac {1} {\theta_2 - \theta_1} * Beta (1,n)

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

    For  X_{(n)} , I get the same thing except the distribution is Beta (n,1) and gives  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 [\theta_1,\theta_2]

    In order to find its expectation, just calculate \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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