Observed Fisher information for the binomial: How is $I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}$ calculated? I am currently studying the textbook In All Likelihood by Yudi Pawitan. Example 2.10 of chapter 2.5 Maximum and curvature of likelihood says the following:

Example 2.10: Based on $x$ from the binomial$(n, \theta)$ the log-likelihood function is
$$\log L(\theta) = x \log \theta + (n - x) \log(1 - \theta).$$
We can first find the score function
$$\begin{align} S(\theta) &\equiv \frac{\partial}{\partial{\theta}} \log L(\theta) \\ &= \frac{x}{\theta} - \frac{n - x}{1 - \theta}, \end{align}$$
giving the MLE $\hat{\theta} = x/n$ and
$$\begin{align} I(\theta) &\equiv -\frac{\partial^2}{\partial{\theta}^2} \log L(\theta) \\ &= \frac{x}{\theta^2} + \frac{n - x}{(1 - \theta)^2}, \end{align}$$
so at the MLE we have the Fisher information
$$I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}. \square$$

How is $I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}$ calculated? I know that the MLE $\hat{\theta}$ is the solution of the score equation $S(\theta) \equiv \frac{\partial}{\partial{\theta}} \log L(\theta) = 0$, and I know that $I(\theta) \equiv - \frac{\partial^2}{\partial{\theta}^2} \log L(\theta)$ is the Fisher information, but I can't figure out precisely what $I(\hat{\theta})$ is, or exactly how $I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}$ was calculated.
 A: Hint: Substitute the value $\hat\theta=x/n$ in $I(\theta) .$ See what happens.
Substitute in place of $\theta$ to get $$\frac{n^2x}{x^2}+ \frac{n^2(n-x) }{(n-x) ^2}.$$ Simplify it.
A: You forgot the expectation operator in the definition of Fisher information: $I(\theta) = E_\theta\left[-\frac{\partial^2}{\partial\theta^2}\log L(\theta)\right]$ instead of what you posted.  After correcting this and noting $E_\theta(x) = n\theta$ given $x \sim B(n, \theta)$, it turns out that
\begin{align}
 & I(\theta) = E_\theta\left[-\frac{\partial^2}{\partial\theta^2}\log L(\theta)\right] \\
=& \frac{1}{\theta^2}E_\theta(x) + \frac{1}{(1 - \theta)^2}E_\theta(n - x) \\
=& \frac{n}{\theta} + \frac{n}{1 - \theta} \\
=& \frac{n}{\theta(1 - \theta)},
\end{align}
matching the goal.

Through the discussion below, it seems that the reference you quoted fails to clearly present the concept of "Fisher information" (and other related sample-level concepts).  I will try to clarify them based on Section 4.3 of Statistical Models by A. C. Davison.
Suppose $N$ i.i.d. observations $x_1, \ldots, x_N$ are drawn from the  distribution $f(x; \theta)$, so that the log-likelihood is
\begin{align}
\ell(\theta) = \log L(\theta) = \sum_{i = 1}^N \log f(x_i; \theta). 
\end{align}
Fisher information (or expected information) is defined by
\begin{align}
I(\theta) = E\left[-\frac{\partial^2\ell(\theta)}{\partial\theta^2}\right]. \tag{1}
\end{align}
While the observed information is defined as:
\begin{align}
J(\theta) = -\frac{\partial^2\ell(\theta)}{\partial\theta^2}
= -\sum_{i = 1}^N \frac{\partial^2 \log f(x_i; \theta)}{\partial\theta^2}. \tag{2}
\end{align}
Note that, if observations are i.i.d., Fisher information only contains parameter $\theta$ (i.e., no observations are included, but the sample size $N$ is still included).
For your example, you have a single observation $x \sim B(n, \theta)$. In this case, your post and my calculation above showed:
\begin{align}
& I(\theta) = \frac{n}{\theta(1 - \theta)}. \tag{3} \\
& J(\theta) = \frac{x}{\theta^2} + \frac{n - x}{(1 - \theta)^2}. \tag{4}
\end{align}
To evaluate "Fisher information evaluated at MLE $\hat{\theta}$", simply plug in $\hat{\theta}$ into $(3)$ (just like evaluating a function's value at a given point) to get $I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}$.  Even though $J(\hat{\theta}) = I(\hat{\theta})$ in this case, that doesn't mean $J(\hat{\theta})$ is conceptually "Fisher information evaluated at MLE $\hat{\theta}$".
