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Thread: Constructing one-sided CI from MLE

  1. #1
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    Constructing one-sided CI from MLE

    $\displaystyle X_1,X_2,...,X_n$ is a sample from the distribution whose density is:

    $\displaystyle f_{X}(x) = e^{-x-\theta} \mbox{ if }x \geq \theta$

    Based on the MLE estimator of $\displaystyle \theta$ construct a one-sided confidence interval for the unknown parameter at confidence level $\displaystyle 1-\alpha$

    Here's what I have so far.

    $\displaystyle f(x_1,...,x_n|\theta) = e^{-x_{i} + \theta}\cdot\cdot\cdot e^{-x_{n} + \theta}$

    $\displaystyle =e^{\sum_{1}^{n} -x_i + n\theta}
    $

    $\displaystyle \log{f(x_1...x_n|\theta)}=\sum_{i=1}^{n}-x_i + n\theta$
    $\displaystyle \frac{d}{d\theta}f(x_1...x_n|\theta) = n$

    $\displaystyle n=0 ?$

    I'm not too sure where to go from here. Any help? Thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    YOU are ignoring the indicator function
    BESIDES TWO TYPOs (the i and the negative sign in front of theta)
    THE MLE is the smallestorder stat.
    YOU need to use common sense, not calculus (incorrectly)
    to maximize the likelihood function

    $\displaystyle f(x_1,...,x_n|\theta) = e^{-x_{1} + \theta}\cdot\cdot\cdot e^{-x_{n} + \theta}I(X_{(1)}>\theta)$

    $\displaystyle = e^{-\sum x_{i} + n\theta}I(X_{(1)}>\theta)$

    This is smallest, ZERO, when $\displaystyle X_{(1)}<\theta$

    while you want $\displaystyle e^{-\sum x_{i} + n\theta}I(X_{(1)}>\theta)$ as large as possible WITH RESPECT TO THETA

    So, you will need $\displaystyle X_{(1)}>\theta$

    so to make $\displaystyle e^{-\sum x_{i} + n\theta}I(X_{(1)}>\theta)$ as large as possible
    you want $\displaystyle n\theta$ or $\displaystyle \theta$ as big as possible
    and it CANNOT exceed the data, hence the MLE (is our sufficient statistic by the way) $\displaystyle X_{(1)}$
    Last edited by matheagle; Nov 30th 2009 at 11:03 PM.
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  3. #3
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    Thank you! I was wondering what X_{(1)} means? Is that just X_1?

    Also, what is the sufficient statistic? I looked it up on wikipedia but I don't really understand it.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by BERRY View Post
    Thank you! I was wondering what X_{(1)} means? Is that just X_1?

    Also, what is the sufficient statistic? I looked it up on wikipedia but I don't really understand it.

    $\displaystyle X_{(1)}$ is the minimum of $\displaystyle X_1, X_2,...,X_n$
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