Quote:
Originally Posted by RicLouRiv
Ah, maybe it's because the inequality holds for any s, it must hold for the min.

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 1a = 1/2 + e (e means epsilon), so P[u>=a] = P[u>= 1/2  e] , but not P[u>=1/2+e].