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Math Help - find the fisher information

  1. #1
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    find the fisher information

     X  \ \ \ has  \ \ \  distribution  \ \ \  Exp(\lambda) \ \ \ \ p(x| \lambda)=\lambda exp(-\lambda x) \ \ \ \ \ \ and \ \ \ \ \phi=Prob(X>x|\lambda)

    compute the fisher information for phi.


    I think  \phi=e^{-\lambda x}

    and therefore the  p(x|\phi)=\phi\frac{\log{\phi}}{-x}, \ \ \ L(x,\phi)=\log{\phi}+\log{\frac{<br />
\log{\phi}}{-x}}

    then the calculation becomes too complex.
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by amberxinya View Post
     X  \ \ \ has  \ \ \  distribution  \ \ \  Exp(\lambda) \ \ \ \ p(x| \lambda)=\lambda exp(-\lambda x) \ \ \ \ \ \ and \ \ \ \ \phi=Prob(X>x|\lambda)

    compute the fisher information for phi.


    I think  \phi=e^{-\lambda x}

    and therefore the  p(x|\phi)=\phi\frac{\log{\phi}}{-x}, \ \ \ L(x,\phi)=\log{\phi}+\log{\frac{<br />
\log{\phi}}{-x}}

    then the calculation becomes too complex.
    You want the second derivative w.r.t. phi?
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  3. #3
    Super Member Anonymous1's Avatar
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    L(x,\phi)=\log{\phi}+\log{\log{\phi}}\underbrace{ - \log{(-x)}} \color{red}{\text{[]EDIT[] You have to incorporate the -log(-x) part since:}} \color{red}{\phi=e^{-\lambda x} \implies x = \frac{\log\phi}{-\lambda}}

    \frac{d}{d\phi}L(x,\phi)=\frac{1}{\phi}+\frac{1}{\  phi}\cdot\frac{1}{\log{\phi}}\color{red}{+...}

    \frac{d^2}{d\phi^2}L(x,\phi)=-\frac{1}{\phi^2}+\left\{\frac{1}{\phi}\cdot\frac{d  }{d\phi}[\frac{1}{\log{\phi}}] + \frac{1}{\log{\phi}}\cdot\frac{d}{d\phi}[\frac{1}{\phi}] \right\}\color{red}{+...}

    ...
    Last edited by Anonymous1; May 10th 2010 at 09:26 PM.
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