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If p is bimodal, then p = 2*p1 is not possible. Either p = 0 or p = 1. The values in between do not occur. Then the only thing to quibble over is G. But if you

• india 16y ago
  No. The event being bimodal only implies that the probability is centered around two values, not that p =1 or p = 0 for those values. For example, take a biased coin that turns up head once in a million. This is bimodal with p(head) = 10^-6. PG's argument appears to be that the error in your estimate of p is almost always larger than the value of p and one should invest as long as 2*p is also within the estimation error.
  • jessriedel 16y ago
    > PG's argument appears to be that the error in your estimate of p is almost always larger than the value of p and one should invest as long as 2*p is also within the estimation error. I don't think this is correct under canonical Bayesian reasoning. All uncertainties (whether because of ignorance or "objective uncertainties") can be bundled into your probability assessment. There is no "error" on your probability. I'm not expert though. I'm only....80% sure.