Proposition 2: medians on a constrained feasible set

C1 corrected target

• Limit on a general convex domain: Proposition 2’s coordinatewise medians → minimizers of expected coordinatewise absolute distance over the feasible set. For an ideal point $V$ and feasible set $X$, the target is

$$\operatorname{argmin}_{x\in X}\;\mathbb{E}\!\left[\sum_i |x_i-V_i|\right].$$

The result covers Model A and the coordinatewise Model B rule used in the proof. On a product domain, this is exactly the feasible coordinatewise median set. On a coupled domain, separate coordinate medians can be infeasible.

The source is Proposition 2 and its Appendix C.5 proof. C1 requires a bounded, closed, convex feasible set; it does not require a product domain. The constrained target applies under C1 without adding that geometric restriction.

Why the coordinatewise median can be infeasible

Take $$X=\{x\in\mathbb{R}^3:x_i\geq0,\;1\leq x_1+x_2+x_3\leq2\}.$$ Draw the ideal point from an equal mixture of uniform distributions on balls of radius $0<r<0.05$ centered at $(1,0.05,0.05)$, $(0.05,1,0.05)$, and $(0.05,0.05,1)$. These balls lie inside $X$, and the mixture has a bounded measurable density. The decomposable utility $f_v(x)=-\sum_i(x_i-v_i)^2$ has the unique ideal point $v$, so this example satisfies C1–C3.

Two thirds of each coordinate’s marginal mass is near $0.05$. Its unique median lies strictly between $0.05$ and $0.10$, so the coordinatewise median vector has sum below $0.30$ and lies outside $X$. Algorithm 1 projects each iterate into the closed set $X$ and therefore cannot converge to that vector. Minimizing expected absolute distance over $X$ gives a feasible target.