= Solution
Write $f_\theta(x)$ for the joint density and assume regularity permits differentiation under the <integral>. Since
$$
\int f_\theta(x)\,dx=1,
$$
one differentiation gives
$$
\mathbb E_\theta S_n(\theta)
=\int \frac{\partial_\theta f_\theta}{f_\theta}f_\theta\,dx
=\partial_\theta\int f_\theta\,dx=0.
$$
Differentiating the score itself,
$$
\partial_\theta S_n
=\frac{\partial_\theta^2f_\theta}{f_\theta}
-\left(\frac{\partial_\theta f_\theta}{f_\theta}\right)^2.
$$
Taking expectation gives
$$
\begin{aligned}
\mathbb E_\theta[\partial_\theta S_n(\theta)]
&=\int\partial_\theta^2f_\theta(x)\,dx
-\mathbb E_\theta[S_n(\theta)^2]\\
&=-I_n(\theta).
\end{aligned}
$$
Thus the <information identity> is
$$
\boxed{I_n(\theta)
=-\mathbb E_\theta\!\left[
\frac{\partial}{\partial\theta}S_n(\theta)
\right]
=-\mathbb E_\theta[\ell_n''(\theta)]}.
$$
Back to article page
