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Let's say you have a distribution that is either symmetric or positively skewed (and defined over 0-1). Call it F.

Then, you find the distribution of the minimum of n>1 draws from F. Call it Fmin.

Clearly the mean of Fmin is less than the mean of F.

Now, find the distribution of the maximum of the same number n draws from Fmin. Call it Fminimax.

The mean of Fminimax is of course greater than the mean of Fmin. However, the increase from Fmin to Fminimax is less in magnitude than the original decrease from F to Fmin. Meaning, that the "round trip" is not complete, and the minimax operation ends up lowering the mean.

What I meant to ask, was, does this single operation lower the mean for any non-negatively skewed distribution?

I deleted a bunch of stuff from the question that I think I made a mess out of. But, I can always revert the changes if anyone objects.

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    @joriki -- oops! positive. thanks for pointing out.2011-10-13

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The minimax transformation $M$ sends a CDF $F$ to the CDF $M(F)$ defined, for every real number $x$, by $M(F)(x)=(1-(1-F(x))^n)^n$. Let $x_n$ denote the unique root in $(0,1)$ of the equation $x=(1-(1-x)^n)^n$. Starting from a non degenerate distribution $F$ and iterating $M$ produces a sequence of distributions which converges to the Dirac mass at $x_F$, where $x_F$ solves the equation $F(x_F)=x_n$. In particular the sequence of the means converges to $x_F$. If the mean $m_F$ of $F$ is smaller than $x_F$, the whole sequence of the means cannot be nonincreasing.

Hence the result you explain can only be true because of some special property that every neutral or positively skewed distribution would share. Either every such $F$ is such that $m_F\geqslant x_F$, or every such $F$ such that $m_F yields a distribution $M(F)$ which is not positively skewed. Unfortunately, the former seems false and the latter unlikely, but maybe I am missing something.

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    It seems to me there is still a step needed. This shows that the means *could* be nonincreasing, (for $F(m_F) \geq \frac{1}{2}$) but not that they definitively are... (I'm not complaining, just checking to make sure I understand)2011-10-14