= Solution
{c}
For the <independent> <statistical sample>, write the <likelihood function> and <log-likelihood function> as
$$
L_n(\theta)=\prod_{i=1}^nf(X_i;\theta),\qquad
\ell_n(\theta)=\sum_{i=1}^n\log f(X_i;\theta).
$$
The expected <Fisher information> in the whole <statistical sample> is
$$
\boxed{I_n(\theta)=\mathbb E_\theta\!\left[\left\{\frac{\partial\ell_n(\theta)}{\partial\theta}\right\}^2\right]
=nI_1(\theta),\qquad
I_1(\theta)=\int\left\{\partial_\theta\log f(x;\theta)\right\}^2 f(x;\theta)\,dx.}
$$
Under the regularity conditions that permit <differentiation under the integral sign>, the <score function> has <expected value> zero and
$$
I_n(\theta)=-\mathbb E_\theta\ell_n''(\theta).
$$
For an interior <statistical parameter>, assuming <identifiability> and $0<I_1(\theta)<\infty$, the regular <asymptotic normality of a maximum likelihood estimator> states
$$
\boxed{\sqrt n(\widehat\theta_n-\theta)\xrightarrow dN\!\left(0,I_1(\theta)^{-1}\right).}
$$
Thus the large-sample <normal approximation> is $\widehat\theta_n\mathrel{\dot\sim}N(\theta,I_n(\theta)^{-1})$. The inverse information is the leading sampling <variance> approximation; the displayed convergence specifies its precise asymptotic meaning.
Back to article page
