i'm confused about the hypothesies of the following statement:
if $\log f(x,\theta)$ is twice differentiable respect $\theta$, and under certain regularity conditions, then the Fisher information may also be written as $$\mathbb{E} \big[ \big(\frac{\partial}{\partial \theta}\log f(X,\theta)\big)^2 \big]=-\mathbb{E}\big[\frac{\partial^2}{\partial\theta^2}\log f(X,\theta) \big]$$
where $X:\Omega\rightarrow\mathbb{R}$ is a r.v. and $(\Omega,\mathcal{F},\mathbb{P})$ a probability space
which are these regularity conditions?
surely $\log f(x,\theta)$ must be twice differentiable and both the function in the equality must be integrable
but taking the demonstration:
$$\mathbb{E}\big[\frac{\partial^2}{\partial\theta^2}\log f(X,\theta) \big]=\int_{\mathbb{R}}f(x,\theta)* \frac{\partial^2}{\partial\theta^2}\log f(x,\theta) dx= ... \\ ... = \int_{\mathbb{R}}\frac{\partial^2}{\partial \theta^2}f(x,\theta)dx - \int_{\mathbb{R}}\frac{\big( \frac{\partial}{\partial \theta}f(x,\theta) \big)^2}{f(x,\theta)}dx$$
now $$\int_{\mathbb{R}}\frac{\big( \frac{\partial}{\partial \theta}f(x,\theta) \big)^2}{f(x,\theta)}dx = \int_{\mathbb{R}}\big( \frac{ \frac{\partial}{\partial \theta}f(x,\theta)}{f(x,\theta)}\big)^2*f(x,\theta)dx=\mathbb{E} \big[ \big(\frac{\partial}{\partial \theta}\log f(X,\theta)\big)^2 \big]$$
the thesis is obtained having $$\int_{\mathbb{R}}\frac{\partial^2}{\partial \theta^2}f(x,\theta)dx=0$$
to prove the previous equality i think is sufficent that $$\frac{\partial}{\partial \theta}f(x,\theta)\space \mathrm{and} \space f(x,\theta)$$
both satisfy the theorem of exchange of integral and derivative. Do you agree? moreover there are some way to relax the hypothesis?
