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First of all, making a group 9x larger doesn't change the shape of the distribution. The distribution would look like 9 exp(|x|), not exp(9|x|). Secondly, I ca
by yummyfajitas 11y ago
First of all, making a group 9x larger doesn't change the shape of the distribution. The distribution would look like 9 exp(|x|), not exp(9|x|).
Secondly, I calculate a p-value for this test here:
https://news.ycombinator.com/item?id=10484309 https://news.ycombinator.com/item?id=10484309
It's proportional to the smaller of the sample sizes. I'll have a more detailed writeup soon. Also, this test is non-parametric, so you don't actually need to know either C or the distributions f and g - all you need is a certain level of uniform regularity in f and g.
- radioactivist 11y agoA note: The original post is correct in what it says. It is saying if one draws N independent samples from an exp(1) distribution then the minimum is distributed as exp(N). This is true. Because of this, for any finite sample there is a systematic 'bias' by your metric if the sample size differs between the two groups. However, you are correct in that this difference vanishes as N becomes large and further that one can make controlled statements about the expected size of this difference, given the distribution.
- yummyfajitas 11y agoOops - I did misread what he wrote. He was discussing the distribution of the min of the sample, not the distribution itself. Also, there may be a bias in the true p-value, but there is none in the p-value bound based on h(d) and min(N1, N2).