# Is every L-estimator an M-estimator?

This is a follow-up to this question: Do M-estimators and L-estimators overlap?. In particular, the answers to that question suggest that there are L-estimators which are not M-estimators, but do not give such an example.

More concretely, suppose I have an L-estimator $\ell(x_1,...,x_n)$ which is a linear combination of order-statistics of $(x_1,...,x_n)$. Can I always write

$\ell(x_1,...,x_n) = \operatorname{argmin}_{\theta} \sum_{i=1}^n \rho(x_i,\theta)$ for some function $\rho$?

A classic example would be a trimmed mean.

For concreteness consider a 25% trimmed mean, where we average the middle half of the data.

That's an L-estimator, but not an M-estimator. It can in a sense be approximated* by a Huber-type M-estimator but they're not the same.

* (perhaps 'analogy' would be a better term than 'approximation' -- they might not always be very close together -- if the distribution is quite skew for example. In symmetric cases they're very alike.)

While many L-estimators can also be M-estimators, there are a great many that aren't.