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Old 12-06-2005, 03:31 PM
BruceZ BruceZ is offline
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Join Date: Sep 2002
Posts: 1,636
Default Re: RoR -- What % loss before double?

I corrected an error in my computation of the mean fractional loss. Since P(f) is actually the probability of losing f or more, we need the density function -P'(f) to compute the expected value of f. This has the effect of multiplying the computed mean by -ln(ror), which for a ror of 0.1% makes the mean fractional loss 14.4%, which is closer to the median value of 10%. The changes are marked in red.
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