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Math Help - maximum likelihood estimator!

  1. #1
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    maximum likelihood estimator!

    tried to solve a problem on MLE of the this type below;

    to find the MLE of  \theta

     f(x;\theta) = \theta x^{\theta-1}  0<x<1, 0<\theta<\infty

     L(x;\theta) = \prod_{i=1}^{n}\theta x^{\theta-1}

    if i were to take the In of the above, how will the next step look like? cause am confuse here

    thanks
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  2. #2
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    Re: maximum likelihood estimator!

    Quote Originally Posted by lawochekel View Post
    tried to solve a problem on MLE of the this type below;

    to find the MLE of  \theta

     f(x;\theta) = \theta x^{\theta-1}  0<x<1, 0<\theta<\infty

     L(x_1,...,x_n;\theta) = \prod_{i=1}^{n}\theta x_i^{\theta-1}

    if i were to take the In of the above, how will the next step look like? cause am confuse here

    thanks

     LL(x_1,...,x_n;\theta) = \sum_{i=1}^{n}\log(\theta)+ (\theta-1)\log(x_i) = n\log(\theta) + (\theta-1)\sum_{i=1}^n\log(x_i)

    .
    Last edited by zzephod; May 10th 2013 at 08:51 AM.
    Thanks from lawochekel
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  3. #3
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    Re: maximum likelihood estimator!

     f(x;\theta) = \frac{1}{2} \exp^{-{\mid x-\theta \mid}}

    pls the mod in the above equation seem to confuse me, if i take In of both sides, what will it be?

    thanks
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  4. #4
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    Re: maximum likelihood estimator!

    Quote Originally Posted by lawochekel View Post
     f(x;\theta) = \frac{1}{2} \exp^{-{\mid x-\theta \mid}}

    pls the mod in the above equation seem to confuse me, if i take In of both sides, what will it be?

    thanks
    The log of a single term in the log-likelihood will be \log(1/2) -|x-\theta|

    .
    Thanks from lawochekel
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  5. #5
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    Re: maximum likelihood estimator!

    pls to differentiate this function  f(x;\theta) = -\mid x-\theta \mid w.r.t  \theta

    how do i go about it to remove the modulus sign.

    thanks
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