How to prove alternative form of Fisher information

Question

Assuming the FI regularity conditions hold. The Fisher information matrix $I(\theta;X)$ about $\theta$ based on $X$ is defined as the matrix with elements $$ I_{i,j}(\theta;X)=Cov_{\theta}\Big(\frac{\partial}{\partial\theta_{i}}\log f_{X}(X|\theta),\frac{\partial}{\partial\theta_{j}}\log f_{X}(X|\theta)\Big) $$ However, an alternative form is given in my studybook $$ I_{i,j}=-E_{\theta}(\frac{\partial^{2}}{\partial\theta_{i}\partial\theta_{j}}\log f_{X}(x|\theta)) $$ I am not able to proof the alternative form, since I am not able to find why the following equality holds $$ -\frac{\partial^{2}}{\partial\theta_{i}\partial\theta_{j}}\log f_{X}(x|\theta)=\frac{\partial}{\partial\theta_{i}}\log f_{X}(x|\theta)\frac{\partial}{\partial\theta_{j}}\log f_{X}(x|\theta) $$

Answer 1

We have that:

$-\frac{\partial^{2}}{\partial\theta_{i}\partial\theta_{j}}\log f_{X}(x|\theta)=-\frac{\frac{\partial^{2}}{\partial\theta_{i}\partial\theta_{j}}f_{X}(x|\theta)}{f_{X}(x|\theta)}+\frac{\frac{\partial}{\partial\theta_{i}}f_{X}(x|\theta)\frac{\partial}{\partial\theta_{j}} f_{X}(x|\theta)}{f^{2}_{X}(x|\theta)}$

Now taking the expectation left and right of this relation, we find that:

$-\mathbb{E}_{\theta}[\frac{\partial^{2}}{\partial\theta_{i}\partial\theta_{j}}\log f_{X}(x|\theta)]=\mathbb{E}[(\frac{\partial}{\partial\theta_{i}}\log f_{X}(x|\theta)(\frac{\partial}{\partial\theta_{j}}\log f_{X}(x|\theta))]$

This is because of the fact that:

$-\mathbb{E}_{\theta}[\frac{\partial}{\partial\theta_{i}} \log f_X(x;\theta)]=\frac{\partial}{\partial\theta_{i}}\int f_X(x|\theta)\text{dx}$

So the alternative form follows.