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Math Help - MLE of expected value of pdf e^-(x-a)

  1. #1
    Newbie noname's Avatar
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    MLE of expected value of pdf e^-(x-a)

    Hi guys

    I'm trying to solve this

    I have this pdf

    f(x)=e^{-(x-a)} , x > a

    f(x)=0 , x \leq a

    I must find the maximum likelihood estimator of the expected value

    I did this

    f(x_1|a)...f(x_n|a)=e^{-(x_1-a)}...e^{-(x_n-a)}=e^{\sum_{i=1}^{n}-x_i}e^{na}

    The log of the last expression is

    {\sum_{i=1}^{n}-x_i}+na

    In its derivative the a disappears

    Without taking the log I end maximizing the same expression (the exponent of e)

    How can I proceed?

    I see that I can express the pdf in this way:

    1e^{-1x}e^{a}

    This is the exponential distribution multiplied by a constant (but i know that it is defined for different values of x)

    Is this a wrong path?

    Thanks!
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  2. #2
    Moo
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    Hello,

    You get n as the derivative.
    Which is always positive.
    Thus the likelihood function is always increasing.

    In order to maximise the likelihood function, make a the greatest possible.

    But you know that \forall i, x_i>a

    Since the xi's are observations, they're already realised. So find a with respect to xi. Such that it's the greatest possible.

    Would you agree that a=\min \{x_1,\dots,x_n\} ?



    A similar reasoning is made for the MLE of a uniform distribution
    I've seen that on a wikipedia (can't remember in which language), so you should find an example over the internet !
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  3. #3
    Newbie noname's Avatar
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    Thank you Moo

    So x is defined from a to inf

    E(x)=\int_{a}^{+\infty }xf(x)dx=\int_{a}^{+\infty }xe^{a-x}dx=a+1

    The estimator of a is

    \hat{a}=min(x_1,...,x_n)

    So the estimator of the expected value should be

    \hat{a}+1=min(x_1,...,x_n)+1

    Is it correct?

    Thanks
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  4. #4
    Moo
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    Yes


    And the estimator by the method of moments will be \hat{a}=\frac 1n \cdot \sum_{i=1}^n x_i-1

    By the way, we always write E(X), not E(x).
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