For a Fisher Information matrix $I(\theta)$ of multiple variables, is it true that $I(\theta) = nI_1(\theta)$? For a Fisher Information matrix $I(\theta)$ of multiple variables, is it true that $I(\theta) = nI_1(\theta)$? That is, if $\theta = (\theta_1, \ldots, \theta_k)$, will it be the case that the fisher information matrix of multiple parameters for an entire dataset will just be $n$ times the fisher information matrix for the first data point, assuming the data is iid?
Update: As a concrete example, consider a sequence of random variables $y_1, \ldots, y_n$ such that $y_i = \beta_0 + \beta_1 x_i + \epsilon_i$ where $\epsilon_i$ is assumed to be i.i.d. $N(0,\sigma^2)$, with $\sigma^2$ known. Additionally, assume that $n$ is even. I am trying to find the Fisher Information Matrix for $\beta = (\beta_1, \beta_2)$. I know that the log-likelihood for one observation is:
$$
l(\beta_0, \beta_1) \propto -\frac{1}{2\sigma^2}(y-\beta_0-\beta_1x)^2
$$
Hence, we have that the observed information matrix (before the expectation), is:
$\frac{\partial^2 l}{\partial \beta_0^2}= \frac{-1}{\sigma^2}$, $\frac{\partial^2 l}{\partial \beta_1^2}= \frac{-x^2}{\sigma^2}$, and $\frac{\partial^2 l}{\partial \beta_0 \partial \beta_1} = \frac{-x}{\sigma^2}$. 
Thus the information matrix for a single observation is:
$$
-E\left(\frac{\partial^2 l}{\partial \beta^2}\right) =\frac{1}{\sigma^2}\left( \begin{array}{ccc}
1 & x \\
x & x^2  \end{array}\right)
$$
and the information matrix for n pairs of observations $(x_i, y_i)$ is given by:
$$
I(\beta_0, \beta_1) =\frac{1}{\sigma^2}\left( \begin{array}{ccc}
n & \sum_{i=1}^{n}x_i \\
\sum_{i=1}^{n}x_i  & \sum_{i=1}^{n}x_i^2   \end{array}\right)
$$
Above, because the fisher information is additive, all we did to move from the single observation to the multiple observation case was just to add entry by entry. HOWEVER, I know that in general if $Y_1, \ldots, Y_n$ are iid, then $I(\theta) = nI_1(\theta)$. My question is, why is it NOT the case above we could have just multiplied each entry by $n$, and instead had to add? 
 A: Since the wikipedia article https://en.wikipedia.org/wiki/Fisher_information  do not contain a proof, I will write one here. Let $X_1, X_2, \dotsc, X_n$ be independent random variables with density function $f(x;\theta)$ (which might in addition depend on known covariates, so this covers more than the iid case). Then the loglikelihood function is 
$$
   \ell(\theta) = \sum_i \log f(X_i;\theta)
$$
and the score function is 
$$
   s(\theta) = \frac{\partial \ell(\theta)}{\partial \theta}= \sum_i\frac{\partial}{\partial\theta}\log f(X;\theta)
$$
The Fisher information matrix then can be written
$$ \DeclareMathOperator{\E}{\mathbb{E}}
I(\theta) = \E\left[\sum_i \left( \frac{\partial}{\partial\theta}\log f(X_i;\theta) \right)\left( \frac{\partial}{\partial\theta}\log f(X_i;\theta) \right)^T \mid \theta\right]
$$
and now the result follows by moving the summation sign outside the expectation operator, which shows that $I(\theta)=\sum_i I_i(\theta)$ where $I_i(\theta)$ is the Fisher information from variable $X_i$. In the iid case that becomes $I(\theta)=n I_1(\theta)$.
A: Let $X$ be a random variable with probability density function
$f(x;\theta)$.
Assume that the observations $x_1,\ldots,x_n$ are independent realizations
of $X$.
Let us prove that the Fisher matrix is:
\begin{align}
I(\theta) = n I_1(\theta)
\end{align}
where $I_1(\theta)$ is the Fisher matrix for one single observation:
\begin{align}
I_1(\theta)_{jk} = 
\mathbb{E}\left[\left(\frac{\partial \log(f(X_1 ; \theta))}{\partial \theta_j}\right)
\left(\frac{\partial \log(f(X_1 ; \theta))}{\partial \theta_k}\right)\right]
\end{align}
for any $j,k=1,\ldots, m$ and any $\theta \in \mathbb{R}^m$.
Since the observations are independent and have the same PDF, the log-likelihood is:
\begin{align}
\ell(\theta) = \sum_{i=1}^n \log(f(x_i;\theta))
\end{align}
for any $\theta\in\mathbb{R}^m$.
Let $s$ be the score, defined as the gradient of the log-likelihood:
\begin{align}
s(\theta) = \frac{\partial \ell(\theta)}{\partial \theta}
\end{align}
for any $\theta\in\mathbb{R}^m$.
Indeed, for any $j,k=1,\ldots, m$, the equation $\mathbb{E}(s(\theta)) = 0$ implies:
\begin{align*}
I(\theta)_{jk}
&= \textbf{Cov}(s(\theta)_j, s(\theta_k)) \\
&= \textbf{Cov}\left(\frac{\partial \ell(\theta)}{\partial \theta_j}, 
\frac{\partial \ell(\theta)}{\partial \theta_k}\right).
\end{align*}
The independence of the realizations implies:
\begin{align*}
I(\theta)_{jk}
&= \textbf{Cov}\left(\frac{\partial}{\partial \theta_j} \sum_{i_1=1}^n \log(f(X_{i_1};\theta)), 
\frac{\partial}{\partial \theta_k} \sum_{i_2=1}^n \log(f(X_{i_2};\theta))\right) \\
&= \textbf{Cov}\left(\sum_{i_1=1}^n \frac{\partial}{\partial \theta_j} \log(f(X_{i_1};\theta)), 
\sum_{i_2=1}^n \frac{\partial}{\partial \theta_k} \log(f(X_{i_2};\theta))\right) \\
&= \sum_{i_1=1}^n \sum_{i_2=1}^n \textbf{Cov}\left(\frac{\partial}{\partial \theta_j} \log(f(X_{i_1};\theta)), 
\frac{\partial}{\partial \theta_k} \log(f(X_{i_2};\theta))\right),
\end{align*}
for any $j,k=1,\ldots, m$, where the last equation uses the linearity
properties of the covariance.
However, the observations are independent, therefore,
$$
\textbf{Cov}\left(\frac{\partial}{\partial \theta_j} \log(f(X_{i_1};\theta)), 
\frac{\partial}{\partial \theta_k} \log(f(X_{i_2};\theta))\right) 
= 0 
$$
if $i_1 \neq i_2$.
Hence,
\begin{align*}
I(\theta)_{jk}
&= \sum_{i=1}^n \textbf{Cov}\left(\frac{\partial}{\partial \theta_j} \log(f(X_i;\theta)), 
\frac{\partial}{\partial \theta_k} \log(f(X_i;\theta))\right).
\end{align*}
Moreover, since all the observations have the same distribution,
\begin{align*}
&\textbf{Cov}\left(\frac{\partial}{\partial \theta_j} \log(f(X_i;\theta)), 
\frac{\partial}{\partial \theta_k} \log(f(X_i;\theta))\right) \\
&= \textbf{Cov}\left(\frac{\partial}{\partial \theta_j} \log(f(X_1;\theta)), 
\frac{\partial}{\partial \theta_k} \log(f(X_1;\theta))\right) \\
&= I_1(\theta)
\end{align*}
for any $i=1,\ldots, n$.
Hence,
\begin{align*}
I(\theta)_{jk}
&= \sum_{i=1}^n I_1(\theta)
\end{align*}
which concludes the proof.
A: The proof by kjetil b halvorsen is incorrect, you missed the cross-terms. The correct equation for the $I(\theta)$ should read something like this:
$$
  I(\theta) = \mathbb{E}\left[\left(\sum\limits_i\frac{\partial}{\partial\theta}\log f(X_i;\theta)\right)\left(\sum\limits_j\frac{\partial}{\partial\theta}\log f(X_j;\theta)\right)^T | \theta\right]
$$
