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. 
Re: Problem 1.9
Quote:

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
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). 
Re: Problem 1.9
Quote:

Re: Problem 1.9
Quote:

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). 
Re: Problem 1.9
Quote:
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
Quote:

All times are GMT 7. The time now is 06:34 AM. 
Powered by vBulletin® Version 3.8.3
Copyright ©2000  2021, Jelsoft Enterprises Ltd.
The contents of this forum are to be used ONLY by readers of the Learning From Data book by Yaser S. AbuMostafa, Malik MagdonIsmail, and HsuanTien Lin, and participants in the Learning From Data MOOC by Yaser S. AbuMostafa. No part of these contents is to be communicated or made accessible to ANY other person or entity.