Write $y_i=|x_i|$. If $n$ is odd, the <sample median> $t=T_n$ equals one observation, with equally many $y_i$ below and above it; the corresponding terms $\operatorname{sign}(y_i/t-1)$ cancel and the median term is zero. If $n$ is even, the conventional median lies strictly between the two middle observations when the $y_i$ are distinct, so exactly half the terms are $-1$ and half are $+1$. In either case
$$
\frac1n\sum_{i=1}^n\psi(x_i/t)=0,
\qquad
\psi(u)=\operatorname{sign}(|u|-1).
$$
Hence $T_n$ is a <Scale M-estimator>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 223 / 3 / a / Solution

{c}

A scale M-estimator solves an equation of the form
$$
\sum_{i=1}^n\psi(x_i/t)=0
$$
for a positive scale $t$.


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 Incoming links: Scale M-estimator