Sample medians as absolute-criterion M-estimators

The IID data context is as in [F4], and the candidate set is a metric space by [F3]. Each order statistic has the finite min-max representation $x_{(j)}={\min }_{S\subseteq \{1,\dots ,n\},\mathrm{\mid }S\mathrm{\mid }=j}{\max }_{i\in S}{x}_i$: every j-element subset contains a value at least $x_{(j)}$, while the subset of the first j ordered entries has maximum $x_{(j)}$. Coordinate projections are continuous. For finite vectors $a,b$, the maxima differ by at most ${\max }_i\mathrm{\mid }{a}_i-b_i\mathrm{\mid }$: each $a_i\le b_i+\delta$ for $\delta ={\max }_i\mathrm{\mid }{a}_i-b_i\mathrm{\mid }$, and the reverse inequality follows by swapping $a,b$; minima satisfy the same bound by negating the vectors. Thus finite maxima and minima, each order statistic, and the stated middle-point selection are continuous, hence Borel measurable under [A1]. The selection is an estimator sequence by [F2].

Put $Q(\theta )={\sum }_i\mathrm{\mid }{x}_i-\theta \mathrm{\mid }$. For $n=2k+1$ and $\theta <x_{(k+1)}$, let $\delta =x_{(k+1)}-\theta >0$. Moving to $x_{(k+1)}$ decreases each of the k+1 distances for indices $i\ge k+1$ by exactly $\delta$; each of the other k distances can increase by at most $\delta$ by the triangle inequality. Therefore $Q(x_{(k+1)})-Q(\theta )\le k\delta -(k+1)\delta =-\delta <0$. For $\theta >x_{(k+1)}$, the k+1 distances for indices $i\le k+1$ decrease by $\delta =\theta -x_{(k+1)}$, while each of the other k distances can increase by at most $\delta$, giving the same strict inequality. Thus the odd-sample median uniquely minimizes $Q$.

For $n=2k$ and $\theta \in [x_{(k)},x_{(k+1)}]$, the first k observations lie at or below $\theta$ and the last k lie at or above it, so $Q(\theta )=k\theta -{\sum }_{i=1}^k{x}_{(i)}+{\sum }_{i=k+1}^{2k}{x}_{(i)}-k\theta$ is constant on this interval. If $\theta <x_{(k)}$, moving to $x_{(k)}$ decreases the k+1 distances for $i\ge k$ by $\delta =x_{(k)}-\theta$, while the other k-1 distances can increase by at most $\delta$; hence $Q(x_{(k)})-Q(\theta )\le (k-1)\delta -(k+1)\delta =-2\delta <0$. If $\theta >x_{(k+1)}$, moving to $x_{(k+1)}$ decreases the k+1 distances for $i\le k+1$ by $\delta =\theta -x_{(k+1)}$, while the other k-1 distances can increase by at most $\delta$, so $Q(x_{(k+1)})-Q(\theta )\le -2\delta <0$. Therefore the interval is exactly the minimizing set of $Q$ and the maximizing set of $P_n{m}_{\theta }$ by [F1].