Re: Problem 1.9
When 2^(b) equal to the minimize of e^(sa)U(s), i try to assume that 1a=1/2e, so that a = 1/2 + e, P[u>=a]=P[u>=1/2+e]
According to (b), P[u>=1/2+e] = P[u>=a] <= (e^(sa)U(s))^N for any s , even if the minimize of e^(sa)U(s) when s = ln(a / (1a)).
Hence P[u>=1/2+e] <= 2^(bN)
