Quote:
Originally Posted by RicLouRiv
Ah, maybe it's because the inequality holds for any s, it must hold for the min.
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Excuse me, I can understand your last reply, but confuse in this one.
More tips please.
When 2^(-b) is the minimize of e^(-sa)U(s), i find that 1-a = 1/2 + e (e means epsilon), so P[u>=a] = P[u>= 1/2 - e] , but not P[u>=1/2+e].