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Math Help - Maximum Likelihood Estimator

  1. #1
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    Maximum Likelihood Estimator

    Let Y1<Y2<...<Yn be the order statistics of a random sample from a distribution with pdf f(x; \theta) = 1, \theta - 0.5 < x < \theta + 0.5. Show that every statistic u(X1,X2,...,Xn) such that Y_n - 0.5<u(X_1,X_2,...,X_n)<Y_1 + 0.5 is a mle of theta. In particular (4Y_1 + 2Y_n + 1)/6, (Y_1 + Y_n)/2, (2Y_1 + 4Y_n - 1)/6 are three such statistics.

    I don't even know how to start this problem. Can someone please point me in the right direction?
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  2. #2
    MHF Contributor matheagle's Avatar
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    The likelihood function is

    I(\theta-.5<X_{(1)}<X_{(n)}<\theta+.5)

    (which shows that the MLEs are the min and the max).

    This fuction is either 0 or 1.
    It's not calculus, you just need to make this equal to one.

    JUST take your three \hat \thetas and make sure that these holds...

    \hat\theta-.5<Y_1 and Y_n<\hat\theta+.5

    SOLVE for \hat\theta and you get that U!!!

    I plugged in \hat\theta={Y_n+Y_1\over 2} and these are equivalent to

    Y_n-Y_1<1 which is true!
    Last edited by matheagle; December 3rd 2009 at 10:14 PM.
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  3. #3
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    I have a question about
    I(\theta-.5<X_{(1)}<X_{(n)}<\theta+.5)

    Is that the same thing as L( \theta) \theta-.5<X_{(1)}<X_{(n)}<\theta+.5)= (for a likelihood function) because I thought I( \theta) was the fisher information function.

    and what does the 'u' in u(X1,X2,...,Xn) suppose to represent?
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by statmajor View Post
    I have a question about
    I(\theta-.5<X_{(1)}<X_{(n)}<\theta+.5)

    Is that the same thing as L( \theta) \theta-.5<X_{(1)}<X_{(n)}<\theta+.5)= (for a likelihood function) because I thought I( \theta) was the fisher information function.

    and what does the 'u' in u(X1,X2,...,Xn) suppose to represent?
    NO, I use I as an indicator function
    I(A)=1 if the event A occurs
    I(A)=0 if A does not

    U(X1,...,Xn) is just a statistic, it's a function of the data
    (That's the definition of a stat)

    BY the way, this was a fun problem,
    I would love to use it in a class some day,
    but I doubt my students would understand it.
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    Sorry to bother you once again, but it seems that I got stuck once more.

    You said to plug in \hat\theta={Y_n+Y_1\over 2}, and I plugged it into \hat\theta-.5<Y_1 which gave me {Y_n+Y_1\over 2} -.5<Y_1

    I`m not not sure how you got Y_n-Y_1<1[/tex](Probably doing something stupid).
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by statmajor View Post
    Sorry to bother you once again, but it seems that I got stuck once more.

    You said to plug in \hat\theta={Y_n+Y_1\over 2}, and I plugged it into \hat\theta-.5<Y_1 which gave me {Y_n+Y_1\over 2} -.5<Y_1

    I`m not not sure how you got Y_n-Y_1<1[/tex](Probably doing something stupid).
    multiply that by two.....

    Y_n+Y_1 -1<2Y_1

    do I need to continue?
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  7. #7
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    Quote Originally Posted by matheagle View Post
    multiply that by two.....

    Y_n+Y_1 -1<2Y_1

    do I need to continue?
    *slams head on keyboard*

    At least I was right about it being a stupid mistake *sigh*

    Thanks again....
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