Thread: Problem 1.9
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Old 03-21-2016, 07:29 AM
MaciekLeks MaciekLeks is offline
Join Date: Jan 2016
Location: Katowice, Upper Silesia, Poland
Posts: 17
Default Re: Problem 1.9

Originally Posted by kongweihan View Post
I'm working through this problem and stuck on (b).

Since 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

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]

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 \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}

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}}, hence \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.
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