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Assume a context with $N$ approximately normal distributions where a lower mean implies a 'better' distribution and a high variance or high standard deviation implies a 'better' distribution as well. Unfortunately, it is unknown which of the two weighs heavier and by how much. Does there exist a sound way to go from here to compare the distributions?

Simply calling the distribution with the minimal coefficient of variation $c_v = \sigma/\mu$ the 'best' assumes equal weights, right?

I have seen something on $\mu-\gamma\sigma$ optimization, but I am not sure when one would use this outside of finance.

Bla Blaat
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  • Your question is, basically, "What function $f$ of two variables $(\mu, \sigma)$ which is decreasing in $\mu$ and increasing in $\sigma$ should I use, if I know nothing else?" And the answer, of course, is that no answer can possibly exist. You need a LOT more information... – 5xum Jul 06 '15 at 09:28

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Without any knowledge about the weighting, the only thing you can say is the following:

Distribution $F$ is better than distribution $F'$ if and only $F$ has a weakly lower mean and a weakly higher variance (one of them strictly) than $F'$.

Martin
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  • Based on this, you can eliminate all "dominated" distributions, which may give you a smaller (or even singleton) set of non-dominated distributions. – Martin Jul 06 '15 at 09:39