For a family of distribution [mathjaxinline]\, \{ \textsf{Poiss}(\lambda )\} _{\lambda >0} \,[/mathjaxinline] , Jeffreys prior is proportional to:
[mathjaxinline]\, \pi _ j(\lambda ) \propto \,[/mathjaxinline]
1 answer
[mathjaxinline]\, \sqrt{I(\lambda)} \propto \sqrt{E\left[ \left( \frac{\partial \log f(X|\lambda)}{\partial \lambda} \right)^2 \right]} \propto \sqrt{E\left[ \left( \frac{\partial \log \left( \frac{e^{-\lambda}\lambda^X}{X!} \right)}{\partial \lambda} \right)^2 \right]} \propto \sqrt{E\left[ \left( -X\frac{e^{-\lambda}\lambda^{X-1}}{\lambda^X} \right)^2 \right]} \propto \sqrt{E\left[ \left( -\frac{X}{\lambda} \right)^2 \right]} \propto \sqrt{E\left[ \frac{X^2}{\lambda^2} \right]} \propto \sqrt{\frac{E[X^2]}{\lambda^2}} \propto \frac{\sqrt{\lambda}}{\lambda} = \frac{1}{\sqrt{\lambda}}[/mathjaxinline]