Quote:
Originally Posted by MaciekLeks
How do you know that ![\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") ? 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. |
I don't think the condition right