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It is elementary probability theory that the sample mean of an i.i.d. sample follows normal distribution, if the background distribution is normal. But what about the trimmed mean? Is there any result on its distribution for an i.i.d. sample of size $n$? (For normal or general population distribution.)

I'm aware that asymptotic (i.e. $n\rightarrow\infty$) results are available, but I'm now interested in the finite sample (i.e. fixed $n$) sampling distribution.

My actual aim is to come up with the sampling distribution for asymmetric trimming (that is, when only the smallest or largest observations are dropped), but I got stuck at the first step of usual (symmetric) trimming...

Background: I aimed to come up with a very-very-very simple model for publication bias. You take a background distribution of $\mathcal{N}\left(0,\sigma^2\right)$ (the drug is not effective). You take a sample of size $n$ (trials with equal number of participants performed), their average (metaanalysis) is also normal with zero mean. But - and now comes the important part - you drop the minimum observation: the worst trial is not published. What will be the distribution of your metaanalysis in this case? This is simply asymmetrically trimmed mean.

What I have done so far: It is quite easy to do through simulation. Say:

library( lattice )
library( reshape )

x <- matrix( rnorm( 100000, 0, 1 ), nc = 10 )
densityplot( ~value, groups = X2, data = melt( sapply( 10:1, function( i ) { 
apply( x, 1, function( y ) { mean( head( sort( y, decreasing = TRUE ), i ) ) } ) } ) ),
plot.points = FALSE,
panel = function( ... ) {
panel.densityplot( ... )
panel.abline( v = 0 )
} )

But is there any analytical solution? The problem doesn't seem to be exotic, yet, I wasn't able to find anything. My only idea is to use the results for the distribution of order statistics (summing them, taking their non-independence into account), but it seems exceedingly complicated, perhaps there is an easier way...

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    $\begingroup$ For posterity: this question has a Math.SE counterpart. $\endgroup$ – shadowtalker Dec 5 '15 at 22:59
  • $\begingroup$ Are you interested only when the underlying distribution is normal, or others? Which? Certainly some approximations can be found, I hope, will try ... $\endgroup$ – kjetil b halvorsen Jan 16 '19 at 20:47

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