# For MLE, why does the information inequality imply identifiability

Let $$X = \langle X_1, \dots, X_n \rangle^{\top}$$ be a finite sample of observation $$X$$ where $$X \sim \mathbb{P}_{\theta_0}$$ with $$\theta_0 \in \Theta$$ and density $$f_X(x; \theta_0)$$. The true parameter $$\theta_0$$ is globally identifiable if

$$\forall \theta \neq \theta_0 \implies \mathbb{P}[f_X(x; \theta) \neq f_X(x; \theta_0)] > 0$$

My lecture notes then say that because of the information inequality for the log likelihood,

$$\mathbb{E}_{\theta_0} [\ell_n(\theta)] \leq \mathbb{E}_{\theta_0}[\ell_{n}(\theta_0)], \qquad \forall \theta \in \Theta \subseteq \mathbb{R}^d.$$

this implies global identification, i.e.

$$\mathbb{E}_{\theta_0} [\ell_n(\theta)] < \mathbb{E}_{\theta_0}[\ell_{n}(\theta_0)], \qquad \forall \theta \neq \theta_0.$$

I don't follow this last step. Why does the inequality become strict? How is this last inequality related to the definition of global identification?

Let $$W$$ and $$U$$ denote two continuous (for simplicity) r.v.s with densities $$f_W(w)$$ and $$f_U(u)$$. The expected value of the r.v. $$Z:=f_W(U)/f_U(U)$$ is $$\begin{eqnarray*} E(Z)&=&\int_{S_U}\frac{f_W(u)}{f_U(u)}rd F_U(u)\\ &=&\int_{S_U}f_W(u)d u\\ &\leqslant&1, \end{eqnarray*}$$ where $$S_U$$ is the support of $$f_U(u)$$. Then, we use Jensen's inequality to get $$E[\log(f_W(U)/f_U(U))]\leqslant\log(E[f_W(U)/f_U(U)])\leqslant\log(1)=0$$ Now, take $$\log f_W(U)=L(\theta;U)$$ and $$\log f_U(U)=L(\theta_0;U)$$, so that $$E[L(\theta;U)]-E[L(\theta_0;U)]=E[\log(f_W(U)/f_U(U))]\leqslant 0,$$ which establishes the weak inequality.
Now, under global identification, $$Z$$ is a nondegenerate r.v., so that the inequality is strict.