# For what distribution is a trimmed mean the maximum likelihood estimator?

The sample mean is the maximum likelihood estimator of $\mu$ for a normal distribution $\text{Normal}(\mu,\sigma)$. The sample median is the maximum likelihood estimator of $m$ for a Laplace distribution $\text{Laplace}(m,s)$ (also called the double exponential distribution).

Does a distribution exist with a location parameter that the trimmed sample mean is the maximum likelihood estimator for?

1. For the sample mean, the estimating equation is $${\rm E}(x-\mu)=0.$$ Imagining that this is the derivative of the log-likelihood, with an awful lot of abuse of notation and loss of rigor, we have $$\frac{{\rm d}\ln l(\mu;x)}{{\rm d}\mu} = x-\mu, \quad \ln l(\mu;x) = a (x-\mu)^2, \quad l(\mu;x) \propto \exp[ a(x-\mu)^2],$$ where the $a$ parameter (integration constant) has to be negative to ensure that it integrates to something meaningful.
2. For the sample median, the estimating equation is $${\rm E \, sign}(x-\mu)=0.$$ Integrate this to get $$l(\mu;x) \propto \exp[ a|x-\mu| ],$$ where again we would have to choose $a$ to be negative to make sense.
3. For the trimmed mean, the estimating equation is $${\rm E}\rho(x,\mu,c) = 0, \quad \rho(x,\mu,c) = \left\{ \begin{array}{ll} x-\mu, & |x-\mu|\le c, \\ 0, & |x-\mu|>c. \end{array} \right.$$ Let's see what it integrates to: $$l(\mu;x, c) = \left\{ \begin{array}{ll} \exp[ a(x-\mu)^2], & |x-\mu|\le c, \\ b, & |x-\mu|>c. \end{array} \right.$$ Looks like a censored normal in the center, but look at the tails: they are improper if $b>0$. So to get a proper distribution, we have to set $b=0$. But then we have a logical inconsistency: this distribution would have to give a zero pdf to some actual data in the trimmed tails. This is self-contradictory, and shows some undesirable side effects of trimming.
Sometimes, it is beneficial to establish "likelihoodity" of a method to show its asymptotic normality, and efficiency for a narrow class of distributions. In general, asymptotic normality of the trimmed mean can follow from the theory of $M$-estimates.
• Is that really the estimating equation of a trimmed mean? In your equation $c$ seems to be a constant that "discards" data that is $c$ away from the mean while in the usual version of a trimmed mean you define what proportion of the data points should be discarded from the tails from the data. Aren't these two different things or am I missing something? – Rasmus Bååth Apr 2 '13 at 20:46