(More complete) proof the Fisher information is additive

Question

For independent, identically distributed variables it is well known that the Fisher information is additive, i.e. \begin{align} \mathcal{I}_n(\theta)&=\left<{\left({\frac{\partial}{\partial\theta}\log f(X_1, X_2, \dots, X_n)}\right)^2}\right>\\ &=n\left<{\left({\frac{\partial}{\partial\theta}\log f(X_1)}\right)^2}\right>\\ &=n\mathcal{I}_1(\theta). \end{align} These lines are not meant as a proof but a statement. When I try to work this out, I know I should use that the variables are independent, but I don't know how. If I try it for two variables I get something like this: \begin{align} \left<\left(\frac{\partial}{\partial\theta}\log f(X_1, X_2)\right)^2\right>=&\left<\left(\frac{\partial}{\partial\theta}\log f(X_1)f( X_2)\right)^2\right>\\ =&\left<\left(\frac{\partial}{\partial\theta}\log f(X_1)+\log f(X_2)\right)^2\right>\\ =&\mathcal I_1(\theta)+\mathcal I_2(\theta)+2\left<\left(\frac{\partial}{\partial\theta}\log f(X_1)\right)\left(\frac{\partial}{\partial\theta}\log f(X_2)\right)\right> \end{align} So the last term should be zero. If this term factors, $$\left<\left(\frac{\partial}{\partial\theta}\log f(X_1)\right)\left(\frac{\partial}{\partial\theta}\log f(X_2)\right)\right>\overset{?}{=} \left<\frac{\partial}{\partial\theta}\log f(X_1)\right>\left<\frac{\partial}{\partial\theta}\log f(X_2)\right>, $$ we are done because we evaluate at the true parameter value and the variance of the score is zero. But how do we know this is true? Because the variables are independent, we only know $\left<X_1X_2\right>=\left<X_1\right>\left<X_2\right>$.

Answer 1

One has to utilize the regularity conditions which ensure that the family is stable meaning the gradient and hessian of the likelihood function are uniformly bounded in a nbd of $\theta$ by integrable functions enabling the commutation of differentiation wrt $\theta$ with integration.

When expectation is applied on the product term, it yields owing to independence $$\mathbf E_\theta(\partial_\theta\ln f(X_1\mid \theta)\partial_\theta\ln f(X_2\mid \theta))=\mathbf E_\theta(\partial_\theta\ln f(X_1\mid \theta))\mathbf E_\theta(\partial_\theta\ln f(X_2\mid \theta))=0.$$

Answer 2

\begin{align} \mathbb E\left[\frac{\partial}{\partial\theta}\log f(X_1) \frac{\partial}{\partial\theta}\log f(X_2)\right] &= \mathbb E\left[\frac{\partial}{\partial\theta}\log f(X_1)\right] \mathbb E\left[\frac{\partial}{\partial\theta}\log f(X_2)\right]\\ &\hskip 5cm {\text{(independence)}}\\ &=\mathbb E\left[\frac{\partial}{\partial\theta}\log f(X_1)\right]^2\\ &\hskip 5cm {\text{(identical distribution)}}\\ &= 0\\ &\hskip 5cm {\text{(score identity)}} \end{align}