Re: Problem 1.9
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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 2. Problem 1.9(a) gave me this: , hence Using this the rest of the proof is quite nice to carry out. 
Re: Problem 1.9
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Re: Problem 1.9
Here's my take on Problem 1.9, part(b), which is following the same lines as the description of MaciekLeks above.
We have: Since is monotonically increasing in t. Also, is non negative for all t, implying Markov inequality holds: The last line being true since [math]x_n[\math] are independent. From there it directly follows that 
Re: Problem 1.9
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Actually I don't even know how to tackle it. I think I'll need a lot of handholding through this one because my math got really rusty since I left school (I'm 34). 
Re: Problem 1.9
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Re: Problem 1.9
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Re: Problem 1.9
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But now I'm stuck at (d). Directly substituting is probably wrong? Because can be simplified to the point where no logarithm appears (unless I made a really big mistake). 
Re: Problem 1.9
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I did end up getting by substituting for , but only after simplifying all the way down, until there is no more or , otherwise I get two powers of 2 with no obvious way to combine them. So now the remaining hurdle is to prove that . Yay 
Re: Problem 1.9
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Re: Problem 1.9
Quick question on Part D...
I agree that s = ln(a / (1a)) is the value of s that minimizes e^(sa)U(s), where I'm using "a" to represent "alpha." I think that, if you plug this value of s in, you get 2^(b), where I'm using "b" to represent "beta." So, 2^(b) < e^(sa)U(s), by definition of minimization. Then, raising to the power of N, we get: 2^(bN) < (e^(sa)U(s))^N, but this inequality is the wrong way. Any tips? 
Re: Problem 1.9
Ah, maybe it's because the inequality holds for any s, it must hold for the min.

Re: Problem 1.9
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Re: Problem 1.9
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More tips please. When 2^(b) is the minimize of e^(sa)U(s), i find that 1a = 1/2 + e (e means epsilon), so P[u>=a] = P[u>= 1/2  e] , but not P[u>=1/2+e]. 
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) 
Re: Problem 1.9
That's correct. But it doesn't get us anywhere.

Re: Problem 1.9
Ahhhm, That was a reply to kongweihan's OP.
It is correct as shown by the first line: Since... But ends up barking up the wrong tree. There is a nice description on wikipedia of Chernoff bound. It is similar to maciekleks path. 
Re: Problem 1.9
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