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Math Help - MLE

  1. #1
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    MLE

    Find the MLE of the unknown parameter \theta when X_{1},...X_{n} is a sample from the following distribution:

    f_{X}(x) = \frac{1}{2}e^{-|x-\theta|}, -\infty < x < \infty

    I have the following, not sure if it's correct.

    f(x_{1}...x_{n}|\theta) = \prod_{i=1}^{n} \frac{1}{2}e^{-|x_{i}-\theta|}

    =\frac{1}{2^{n}}e^{-[\sum_{i=1}^{n}x_{i}] - n\theta}

    \log f(x_{1}...x_{n}|\theta) = \frac{1}{2^{n}} [-\sum_{i=1}^{n}x_{i} - n\theta]

    \frac{d}{d\theta}\log f(x_{1}...x_{n}|\theta) = \frac{-n}{2^{n}}

    Is this correct?
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  2. #2
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    Seems to me that maximizing the likelihood function of this guy is equivalent to minimizing \sum _ {i = 1} ^ n |x_i - \theta|. The value that accomplishes this is a well known summary statistic, albeit not the sample mean.
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  3. #3
    MHF Contributor matheagle's Avatar
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    No the absolute value is around the Xi-theta
    Use logic not calc here

    You want to maximize 2^{-n}e^{-\sum_{i=1}^n|X_i-\theta|} wrt theta

    that the same as minimizing \sum_{i=1}^n|X_i-\theta| wrt theta
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  4. #4
    Moo
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    Hello,

    Note that you can simplify the writing in latex : when there's a single element in the exponent or in the indice, you don't have to put { }
    X_1, X_2, ... will produce the same result
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  5. #5
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    Quote Originally Posted by theodds View Post
    Seems to me that maximizing the likelihood function of this guy is equivalent to minimizing \sum _ {i = 1} ^ n |x_i - \theta|. The value that accomplishes this is a well known summary statistic, albeit not the sample mean.
    Which statistic is that?

    Also, if I were to try to estimate theta through method of moments...how would I do it?
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  6. #6
    MHF Contributor matheagle's Avatar
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    the median
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  7. #7
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    ^ what he said.
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  8. #8
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    What about using method of moments?
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  9. #9
    MHF Contributor matheagle's Avatar
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    E(X)= .5\int_{-\infty}^{\theta}xe^{x-\theta}dx +.5\int_{\theta}^{\infty}xe^{\theta -x}dx

    and set that equal to the sample mean.

    I rushed but I got theta as I expected

    Let u=\theta -x in the first integral and u=x-\theta
    in the second

    =.5(\theta\Gamma(1)-\Gamma(2))+.5(\Gamma(2)+\theta\Gamma(1))=\theta
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