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    Computing MLE estimate




    So I have



    so i'm stuck at taking the log

    can somoene help me ?

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    Re: Computing MLE estimate

    Well (\prod{x^{-(\theta+2)}}(\theta +1)^n get the log of this...nln(\theta) + -(\theta +2)\sum{lnx}. Get the derivative of this and then let it equal to 0.. And you should get..\theta=\frac{n}{\sum\ln(x)}

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    Re: Computing MLE estimate

    Quote Originally Posted by MathJack View Post
    Well (\prod{x^{-(\theta+2)}}(\theta +1)^n get the log of this...nln(\theta) + -(\theta +2)\sum{lnx}. Get the derivative of this and then let it equal to 0.. And you should get..\theta=\frac{n}{\sum\ln(x)}
    I had some probs with the calculation of the derivative, but atlast i found it

    \frac{n}{\theta} - \sum ln(x) = 0


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    Re: Computing MLE estimate

    Quote Originally Posted by MathJack View Post
    Well (\prod{x^{-(\theta+2)}}(\theta +1)^n get the log of this...nln(\theta) + -(\theta +2)\sum{lnx}. Get the derivative of this and then let it equal to 0.. And you should get..\theta=\frac{n}{\sum\ln(x)}
    how come the log of (\theta+1) = n ln(\theta)???

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    Re: Computing MLE estimate

    That is a slight typo in the calculation and the final answer will be shifted by 1.

    BTW just curious to ask: Is the distribution continuous or discrete? If it is a continuous distribution, it looks like the pdf of a type I pareto distribution and the support will be on (1, +\infty) which will not yield a sample of 0.5

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    Re: Computing MLE estimate


    Aaaaah so it should be

    (\theta+1) = n ln(\theta+1)

    Now the world makes sense again

    The distribution isn't indentified, but in the question it indeed says that x should be larger then 1

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