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Thread: MLE

  1. #1
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    MLE

    Find the MLE of the distribution having the following pdf

    $\displaystyle f(x;\theta)=e^-(x-\theta)$, $\displaystyle \theta \le x<\infty$, $\displaystyle -\infty<\theta<\infty$

    Attempt to a solution:

    Likelihood function
    $\displaystyle L(\theta;X_i)=e^[-\sum X_i+\theta]$

    Log likelihood function
    $\displaystyle l(\theta)=-\sum X_i+\theta$

    But then taking the derivative with respect to $\displaystyle \theta$ it makes $\displaystyle \theta$ disappear.

    I feel like i missing something especially with the bounds of x and $\displaystyle \theta$
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  2. #2
    Moo
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    Hello,

    It's like for a uniform distribution. The log likelihood function is always increasing, for any theta (since its derivative is n, always positive). So the maximum is attained at the maximum possible value of theta.
    But since, for any i, $\displaystyle x_i\geq\theta$ (that's the info in the pdf), we have that $\displaystyle \theta\leq\min(x_i)$
    Hence, the MLE is $\displaystyle \min(x_i)$
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  3. #3
    MHF Contributor matheagle's Avatar
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    The likelihood function is....

    $\displaystyle e^{n\theta}e^{-\sum x_i}I(x_{(1)}\ge \theta)$

    you want to maximize this, hence you want $\displaystyle e^{n\theta}$
    as large as possible and also that indicator function to be one.

    The largest theta can be in that case is $\displaystyle X_{(1)}$

    There's no need to take the log here, it only makes it messier.
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