$\displaystyle 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 $\displaystyle \phi=e^{-\lambda x} $

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

\log{\phi}}{-x}} $

then the calculation becomes too complex.