How do we deduce this fisher information relation? Given a RS $X_{1},X_{2},\ldots,X_{n}$ whose distribution is well known (unless its parameters), how do we prove the following Fischer Information relationship
\begin{align*}
I_{F}(\theta) =\textbf{E}\left[\left(\frac{\partial\ln f(\textbf{X}|\theta)}{\partial\theta}\right)^{2}\right] = -\textbf{E}\left[\frac{\partial^{2}\ln f(\textbf{X}|\theta)}{\partial\theta^{2}}\right]
\end{align*}
MY APPROACH
To begin with, I started with the observation that
\begin{align*}
&\frac{\partial^{2}\ln f(\textbf{X}|\theta)}{\partial\theta^{2}} = \frac{\partial}{\partial\theta}\left[\frac{\partial\ln f(\textbf{X}|\theta)}{\partial\theta}\right] = \frac{\partial}{\partial\theta}\left[\frac{1}{f(\textbf{X}|\theta)}\frac{\displaystyle\partial f(\textbf{X}|\theta)}{\displaystyle\partial \theta}\right] = \ldots
\end{align*}
But I do not know how to proceed from here. Can someone help me out? Thanks in advance!
 A: Following your way, calling $f(\textbf{X}|\theta)=f$, for notational simplicity:
$$\begin{align}\frac{\partial^2 \ln f}{\partial\theta^2}&=\frac{\partial}{\partial\theta}\left(\frac{\partial\ln f}{\partial\theta}\right)=\frac{\partial}{\partial\theta}\left(\frac{1}{f}\frac{\partial f}{\partial\theta}\right)\\&=-\frac{1}{f^2}\left(\frac{\partial f}{\partial\theta}\right)^2+\frac{1}{f}\frac{\partial^2 f}{\partial\theta^2}\end{align}$$
Note that the first summand, call it $\mathcal{F}$, can be manipulated as:
$$\mathcal{F}=-\left(\frac{1}{f}\frac{\partial f}{\partial\theta}\right)^2=-\left(\frac{\partial\ln f}{\partial\theta}\right)^2$$
because inner term is the same with the rightmost term we found in the first line.
For the second summand, we have:
$$E\left[\frac{1}{f}\frac{\partial^2 f}{\partial\theta^2}\right]=\int \frac{\partial^2 f}{\partial\theta^2}dX=\frac{\partial^2}{\partial\theta^2}\left(\int fdX\right)=\frac{\partial^2 f}{\partial\theta^2}(1)=0$$
So, only the first summand remains non-zero, which is what we need to find. Note that. in order for this equation to be true, the differentiation by $\theta$ needs to be drawn out of the integral via assumed regularity conditions.
