# Thread: find the fisher information

1. ## 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 $p(x|\phi)=\phi\frac{\log{\phi}}{-x}, \ \ \ L(x,\phi)=\log{\phi}+\log{\frac{
\log{\phi}}{-x}}$

then the calculation becomes too complex.

2. Originally Posted by amberxinya
$X \ \ \ has \ \ \ distribution \ \ \ Exp(\lambda) \ \ \ \ p(x| \lambda)=\lambda exp(-\lambda x) \ \ \ \ \ \ and \ \ \ \ \phi=Prob(X>x|\lambda)$

compute the fisher information for phi.

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

and therefore the $p(x|\phi)=\phi\frac{\log{\phi}}{-x}, \ \ \ L(x,\phi)=\log{\phi}+\log{\frac{
\log{\phi}}{-x}}$

then the calculation becomes too complex.
You want the second derivative w.r.t. phi?

3. $L(x,\phi)=\log{\phi}+\log{\log{\phi}}\underbrace{ - \log{(-x)}}$ $\color{red}{\text{[]EDIT[] You have to incorporate the -log(-x) part since:}}$ $\color{red}{\phi=e^{-\lambda x} \implies x = \frac{\log\phi}{-\lambda}}$

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

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

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