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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?

Glen_b
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Rasmus Bååth
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2 Answers2

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The distributions, if any, are obtained as integrals of the estimating equations. Let us assume for simplicity that the scale parameter is known, and the trimming parameters, if any, are fixed.

  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.

StasK
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    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 02 '13 at 20:46
  • Yes, it is somewhat different indeed -- I said that I am treating the trimming constant as fixed. Making it data-dependent will complicate things, but I believe it would eventually lead to the similar conclusion that some of the data points are "impossible" under the distribution implied by the "likelihood". – StasK Apr 02 '13 at 23:56
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Special cases like the median aside, I don't think that trimmed means are generally ML; if they were, they'd already be some form of M-estimator. However, if you take a distribution which is normal in the middle with, say, exponential tails - the distribution corresponding to a Huber M-estimator - then for a particular level of trimming, the trimmed mean would be expected to be highly efficient.

Glen_b
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