#11




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? 
#12




Re: Problem 1.9
Ah, maybe it's because the inequality holds for any s, it must hold for the min.

#13




Re: Problem 1.9
You got it. :)
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#14




Re: Problem 1.9
Quote:
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]. 
#15




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) 
#16




Re: Problem 1.9
That's correct. But it doesn't get us anywhere.

#17




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. 
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