Empirical likelihood

Remarks

The factor $n$ makes the unconstrained maximum equal to $1$: without the mean constraint, $\prod w_i$ is largest at $w_i = 1/n$. Feasible weights exist exactly when $\mu$ lies in the convex hull of $X_1, \dots, X_n$. When $\mu$ is in the interior of the hull, the maximiser is $w_i = \frac{1}{n(1 + \lambda^{T}(X_i - \mu))}$, where $\lambda \in \R^d$ solves $\sum_i \frac{X_i - \mu}{1 + \lambda^{T}(X_i - \mu)} = 0$. Owen's theorem: if $\mu_0 = E[X_1]$ and the covariance matrix of $X_1$ is finite and nonsingular, then $-2 \ln \mathcal{R}(\mu_0)$ converges in distribution to the chi-squared distribution with $d$ degrees of freedom. So $\set{\mu : -2\ln\mathcal{R}(\mu) \le c}$, with $c$ the $(1-\alpha)$-quantile of $\chi^2_d$, is an asymptotic confidence region of level $1 - \alpha$.

Example: $\mu = \bar X$. The weights $w_i = 1/n$ satisfy the constraint, so $\mathcal{R}(\bar X) = 1$, the largest possible value.

Non-example: data $(1, 2, 3)$ in $\R$ and $\mu = 5$. No weighted average of $1, 2, 3$ equals $5$, because $5$ is outside the convex hull $[1, 3]$, so $\mathcal{R}(5) = 0$.