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JGWeissman comments on The Optimizer's Curse and How to Beat It - Less Wrong
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In any group there's going to be random noise, and if you choose an extreme value, chances are that value was inflated by noise. In Bayesian, given that something has the highest value, it probably had positive noise, not just positive signal. So the correction is to correct out the expected positive noise you get from explicitly choosing the highest value. Naturally, this correction is greater for when the noise is bigger.
So imagine choosing between black boxes. Each black box has some number of gold coins in it, and also two numbers written on it. The first number, A, on the box is like the estimated expected value, and the second number, B, is like the variance. What happened is that someone rolled two distinct dice with B sides, subtracted die 1 from die 2, and added that to the number of gold coins in the box.
So if you see a box with 40, 3 written on it, you know that it has an expected value of 40 gold coins, but might have as few as 37 or as many as 43.
Now comes the problem: I put 10 boxes in front of you, and tell you to choose the one with the most gold coins. The first box is 50, 1 - a very low-variance box. But the last 9 boxes are all high-uncertainty, all with B=20. The expected values printed on them are as follows [I generated the boxes honestly] : 53, 52, 37, 60, 44, 36, 56, 45, 54. Ooh, one of those boxes has a 60 on it! Pick that one!
Okay, don't pick that one. Think about it - there are 9 boxes with high variance, and the one you picked probably has unusually large noise. To be special among 9 proposals with high variance, it probably has noise at the 80th+ percentile. What's the 80th percentile of noise for 1d20 - 1d20? I bet it's larger than 10. You're better off just going with the 50, 1 box.
And it's a good thing you applied that correction, because I generated the boxes by typing "RandomInteger[20,9] - RandomInteger[20,9] + 45" into Wolfram alpha - they each 45 coins each.
So this illustrates that what beating the optimizer's curse really is is a sort of "correction for multiple comparisons." If you have a lot of noisy boxes, some of them will look large even when they're not, even larger than non-noisy boxes.
That is a good example of how the optimizer's curse causes an overestimate of the maximum expected value, and even reliably causes a wrong choice to be associated with the maximum expected value. But how do I apply the correction mathematically, so I can know for which expected values on the high uncertainty boxes I should expect their best of them to be better or worse than the low uncertainty box? Even better, how can I deal with situations where the uncertainties of the expected values are not so conveniently categorized (and whose actual values aren't conveniently uniform)?
Oh - I learned how, by the way. You start with some prior over how you expect the actual coins to be distributed, and then you convolute in the noise distribution of each box to get the combined distribution for each box. Then, given where the number on the outside of each box falls on the combined distribution, you can assign how much of that you expect to be signal and how much you expect to be noise by distributing improbability equally between signal and noise. Then you subtract out the expected noise.
I'm not sure. It's probably in the paper.