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Jan 12, 2012 at 23:31 comment added cardinal Can I assume you've already looked through the Johnson, Kotz & Balakrishnan volumes?
Jan 11, 2012 at 15:41 comment added Cyan The square root in my distribution comes from a hyperbola. The x/2 term flattens the positive-x asymptote against the x-axis, and the exponential maps 0 (well, $-\delta$) to 1 (or rather, $1-\epsilon$).
Jan 11, 2012 at 14:45 comment added Steve Huntsman That square root reminds me of the Wigner semicircle distribution. You might try seeing if there is some nice transformation that takes you to it.
Jan 11, 2012 at 13:50 history asked Cyan CC BY-SA 3.0