# Asymptotic normality of (M) estimators for convex $rho$.

The text from Robust Estimation of Location Parameter (1964), Huber, P.:

It will be assumed thad $$\rho$$ is continuous convex real-valued function of a real variable $$t$$. tending to $$+\infty$$, as $$t > \rightarrow \pm \infty$$.

Let $$x_1, x_2, ..., x_n$$ be independent identically distributed random variables with common distribution function $$F$$. Let $$\left[T_n(x)\right]$$ bet the set of all those $$\xi$$ for which $$Q(\xi) = \sum_{i=1}^n \rho(x_i-\xi)$$ reaches it's minimum $$Q_{\inf}$$. Obviously $$\left[T_n(x)\right]$$ is invariant under translations: $$\left[T_n(x+c)\right] = \left[T_n(x)\right]+c$$. By $$T_n(x)$$ we shall denote any representation of the set valued function $$(x_1,x_2,...,x_n) \rightarrow \left[T_n(x)\right]$$ by a single valued function $$(x_1,x_2,...,x_n) \rightarrow T_n(x) \xi \left[T_n(x)\right]$$, e.g., $$T_n(x) -$$ midpoint of $$\left[T_n(x)\right]$$.

Why it's obvious that $$\left[T_n(x)\right]$$ is invariant under translation? How does $$T_n(x)\xi \left[T_n(x)\right]$$ translates to midpoint of $$\left[T_n(x)\right]$$, and if the set is not ordered, how the middle point be of any use?

Lemma 1. $$Q(\xi)$$ is a convex function of $$\xi$$, and $$\left[T_n(x)\right]$$ is convex, non-empty and compact. If $$\rho$$ is strictly convex, then $$\left[T_n(x)\right]$$ is reduced to a single point.

PROOF. Strict convexity of Q follows immediately from strict convexity of $$\rho$$. The sets $$\{\xi| Q(\xi) \le Q_{\inf} + m^{-1}\}$$ form a decreasing sequency of non empty convex compact sets as $$m\to\infty$$ , hence their intersection $$\left[T_n(x)\right]$$ is non-empty convex compact. If $$Q$$ is convex, and $$\xi', \xi''$$ are two distinct points from $$\left[T_n(x)\right]$$, then we would have $$Q_{\inf} = Q(\frac{1}{2}\xi' + \frac{1}{2}\xi'') < \frac{1}{2}Q(\xi') + \frac{1}{2}Q(\xi'') = Q_{\inf}$$, which is a contradiction.

I don't get the last expression $$Q_{\inf} = Q(\frac{1}{2}\xi') + Q(\frac{1}{2}\xi'')$$. If for example $$Q(\xi') =Q(\xi'') = 0$$, does it follow that $$Q(\frac{1}{2}\xi') + Q(\frac{1}{2}\xi'')=0$$? Why $$Q(\frac{1}{2}\xi') + Q(\frac{1}{2}\xi'')$$ is less than $$Q(\frac{1}{2}\xi' + \frac{1}{2}\xi'')$$?