TITLE: Finding the Fisher's Information in a normal distribution with known $\mu$ and unknown $\sigma^{2}$
QUESTION [8 upvotes]: I have a point statistic $x_{i},...,x_{n} X \in N(\mu,\sigma^{2}), \mu$ is known. 
I have to apply the Rao-cramer theorem but calculating the Fisher's information I stumbled upon this problem:
$I(\sigma)=-E(\frac{n}{\gamma}+3\sum(\frac{x_{i}-\mu)^{2}}{\sigma^{2}})=\frac{n}{\gamma}+ 3\frac{E(\sum(x_{i}-\mu)^{2})}{E\sigma^{4}}=\frac{n}{\gamma}+\frac{3}{\sigma^{4}}E(\sum(x_{i}-\mu)^{2})$
$$E(\sum(x_{i}-\mu)^{2})=?$$
${\displaystyle \operatorname {E} [X]=\int _{\mathbb {R} }xf(x)\,dx.}$ But what is f(x) here ? Could it possibly be the function itself ${\displaystyle \operatorname {E} [X]=\int _{\mathbb {R} }x∑(x_i−μ)2\,dx.}$
From wiki, we know that Fisher's information is:
$$(
 \begin{matrix}
  \frac{1}{\sigma^{2}} & 0  \\
  0 & \frac{1}{2\sigma^{4}}  
 \end{matrix})
$$
But I need a number, what is that matrix supposed to mean?
What is $I(\sigma^{2})$ for a normal distribution with $\mu$ - known and $\sigma^{2}$- unknown?

REPLY [4 votes]: Fisher information is only a matrix when we have 2 or more unknown parameters. Since $\mu$ is known, we will get a single number.
For a random sample $X_1, \dots, X_n$, the Fisher information for the sample can be defined as 
$$I_X(\theta)=-n \operatorname{E}\left[\frac{\partial^2\ln f(X|\theta)}{\partial \theta^2}\right]$$
where here $\theta = \sigma^2$. So we don't need to consider the individual $X_i$.