#1




Problem 1.9
I'm working through this problem and stuck on (b).
Since , we get We also know 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)? 
#2




Re: Problem 1.9
Quote:
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. 
#3




Re: Problem 1.9
Quote:

#4




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 
#5




Re: Problem 1.9
Quote:
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). 
#6




Re: Problem 1.9
Will I need to summon notions such as "moment generating function" for part (c) of this problem?

#7




Re: Problem 1.9
So I'm not the only one. Gee, thanks

#8




Re: Problem 1.9
Quote:
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). 
#9




Re: Problem 1.9

#10




Re: Problem 1.9

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