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Lumifer comments on Median utility rather than mean? - Less Wrong Discussion
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As the Wikipedia says, "If the distribution has finite variance". That's not necessarily a good assumption.
Consider a policy with three possible outcomes: one pony; two ponies; the universe is converted to paperclips. What's the median outcome? One pony. Don't you want a pony?
The median is a robust estimator meaning that it's harder for outliers to screw you up. The price for that, though, is indifference to the outliers which I am not sure is advisable in the utility context.
In fact, "Pascal's mugging" scenarios tend to pop up when you allow for utility distributions with infinite variance.
For Pascal's Muggings I don't think you care that much about variance -- what you want is a gargantuan skew.
Indeed. But the argument about convergence when you get more and more options still applies.
Still -- only is you true underlying distribution has finite variance. Check some plots of, say, a Cauchy distribution -- it doesn't take much of heavy tails to have no defined variance (or mean, for that matter).
Not everything converges to a Gaussian.
You did notice that I mentioned the Cauchy distribution by name and link in the text, right?
And the Cauchy distribution is the worst possible example for defending the use of the mean - because it doesn't have one. Not even, a la St Petersburg paradox, an infinite mean, just no mean at all. But it does have a median, exactly placed in the natural middle.
Your argument works somewhat better with one of the stable distributions with an alpha between 1 and 2. But even there, you need a non-zero beta or else median=mean! The standard deviation is an upper bound on the difference, not necessarily a sharp one.
It would be interesting to analyse the difference between mean and median for stable distributions with non-zero beta; I'll get round to that some day. My best guess is that you could use some fractional moment to bound the difference, instead of (the square root of) the variance.
EDIT: this is indeed the case, you can use Jensen's inequality to show that the q-th root of the q-th absolute value central moment, for 1<q<2, can be substituted as a bound between mean and moment. For q<alpha, this should be finite.
I only brought up Cauchy to show that infinite-variance distributions don't have to be weird and funky. Show a plot of a Cauchy pdf to someone who had, like, one undergrad stats course and she'll say something like "Yes, that's a bell curve" X-/
Actually, there's no need for higher central moments. The mean absolute deviation around the mean (which I would have called the first absolute central moment) bounds the difference between mean and median, and is sharper than the standard deviation.