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Dec 19, 2018 at 11:19 comment added James Martin You can even make do with only $x_1, x_2, x_3$, since $(x_1-x_2)/(x_1-x_3)$ already has a non-trivial distribution which does not depend on $\mu$ and $\sigma$.
Dec 19, 2018 at 10:52 history answered Sebastian Nowozin CC BY-SA 4.0