If I have a Fisherinformation as:
$i_n(\lambda)=\frac{n}{\lambda}+4n$ then I need the inverse of the Fisherinformation to find the variance.
But suddenly I have doubts if they mean the inverse function or the reciprocal value
So is the answer $i_n(\lambda)^{-1}=\frac{1}{\frac{n}{\lambda}+4n}$?
The Theorem of Maximum Likelihood states that (under some regularity conditions) $$ \sqrt{i_n(\hat{\theta}_n)}(\hat{\theta}_n-\theta_0)\overset{d}{\longrightarrow}\mathcal{N}(0,1), $$ where $\theta_0\in C\subseteq\mathbb{R}$ is the true parameter and $\hat{\theta}_n\in C$ the MLE. This s stated loosely as follows: for big $n$ $$ \hat{\theta}_n\sim\mathcal{N}\left(\theta_0,\dfrac{1}{i_n(\hat{\theta}_n)}\right). $$ Note that the notation is not consistent, but is a practical way to write it. This indicates that the variance of $\hat{\theta}_n$ is approximately $$ \dfrac{1}{i_n(\hat{\theta}_n)}. $$ (Consider $\hat{\theta}_n$ as the value you obtained from the sample). Therefore, yes you are always interested in the reciprocal.