Quote:
Originally Posted by kongweihan
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)? |
How do you know that
? I think that is a problem in your proof that you assumed that the joint probability works with Problem 1.9(b) inequality.
To proof (b) I went this way:
1. I used Markov Inequality
![\mathbb{P}[u\geq\alpha]\leq\frac{\mathbb{E}_{u}[u]}{\alpha}](/vblatex/img/9a0652a5d0910e27b7fcc63df1e9d783-1.gif "\mathbb{P}[u\geq\alpha]\leq\frac{\mathbb{E}_{u}[u]}{\alpha}")
2. Problem 1.9(a) gave me this:
![\mathbb{P}[t\geq\alpha]=\mathbb{P}[e^{sNt}\geq e^{sN\alpha}]\leq\frac{\mathbb{E}[e^{sNt}]}{e^{sN\alpha}}](/vblatex/img/2da8ba404c75e00204c519f8fe0a7179-1.gif "\mathbb{P}[t\geq\alpha]=\mathbb{P}[e^{sNt}\geq e^{sN\alpha}]\leq\frac{\mathbb{E}[e^{sNt}]}{e^{sN\alpha}}")
, hence
![\mathbb{P}[u\geq\alpha]\leq\frac{\mathbb{E}_{u}[e^{sNu}]}{e^{sN\alpha}}](/vblatex/img/4fe8b061a70335ba1b6f210c354166ea-1.gif "\mathbb{P}[u\geq\alpha]\leq\frac{\mathbb{E}_{u}[e^{sNu}]}{e^{sN\alpha}}")
Using this the rest of the proof is quite nice to carry out.