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#1
09-29-2015, 02:14 AM
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Problem 1.9
I'm working through this problem and stuck on (b).
Since ![P[u_n \geq \alpha] \leq e^{-s\alpha}U(s)](/vblatex/img/96f9aa019395ed439da6a24ff79adf1a-1.gif "P[u_n \geq \alpha] \leq e^{-s\alpha}U(s)") , we get![\prod_{n=1}^N P[u_n \geq \alpha] \leq (e^{-s\alpha}U(s))^N](/vblatex/img/edd19e4e1eed52ca220b2ce9df3ead1d-1.gif "\prod_{n=1}^N P[u_n \geq \alpha] \leq (e^{-s\alpha}U(s))^N") We also know ![\prod_{n=1}^N P[u_n \geq \alpha] \leq P[\sum_{n=1}^N u_n \geq N\alpha] = P[u \geq \alpha]](/vblatex/img/9c7c52a7e898bd7fd5267bdbfedfe9dd-1.gif "\prod_{n=1}^N P[u_n \geq \alpha] \leq P[\sum_{n=1}^N u_n \geq N\alpha] = P[u \geq \alpha]") Both terms in the desired inequality is bigger than the common term, so I don't know how these two inequalities can lead to the desired conclusion, what did I miss? Also, in (c), why do we want to minimize with respect to s and use that in (d)? 
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