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 kongweihan 09-29-2015 02:14 AM

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)?

 MaciekLeks 03-21-2016 08:29 AM

Re: Problem 1.9

Quote:
 Originally Posted by kongweihan (Post 12075) 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)?

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.

 waleed 05-12-2016 03:18 AM

Re: Problem 1.9

Quote:
 Originally Posted by MaciekLeks (Post 12298) 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.
I don't think the condition right

 svend 09-17-2016 11:43 AM

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

 k_sze 03-03-2017 08:53 AM

Re: Problem 1.9

Quote:
 Originally Posted by kongweihan (Post 12075) Also, in (c), why do we want to minimize with respect to s and use that in (d)?
I also have trouble understanding (c).

Actually I don't even know how to tackle it. I think I'll need a lot of hand-holding through this one because my math got really rusty since I left school (I'm 34).

 k_sze 03-03-2017 09:25 AM

Re: Problem 1.9

Quote:
 Originally Posted by k_sze (Post 12596) I also have trouble understanding (c). Actually I don't even know how to tackle it. I think I'll need a lot of hand-holding through this one because my math got really rusty since I left school (I'm 34).
Will I need to summon notions such as "moment generating function" for part (c) of this problem?

 Sebasti 03-06-2017 04:38 AM

Re: Problem 1.9

Quote:
 Originally Posted by k_sze (Post 12596) Actually I don't even know how to tackle it. I think I'll need a lot of hand-holding through this one because my math got really rusty since I left school
So I'm not the only one. Gee, thanks ;)

 k_sze 03-15-2017 07:11 AM

Re: Problem 1.9

Quote:
 Originally Posted by k_sze (Post 12597) Will I need to summon notions such as "moment generating function" for part (c) of this problem?
So I did end up using moment generating function. And I think the answer to (c) is , using calculus.

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).

 k_sze 03-16-2017 10:36 AM

Re: Problem 1.9

Quote:
 Originally Posted by k_sze (Post 12606) So I did end up using moment generating function. And I think the answer to (c) is , using calculus. 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).
*sigh*

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

 k_sze 04-01-2017 08:19 AM

Re: Problem 1.9

Quote:
 Originally Posted by k_sze (Post 12607) *sigh* 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
I finally worked out the proof for . Apparently using Jensen's inequality is supposed to be the simple way, but I simply don't get Jensen's inequality, so I used some brute force calculus (first and second derivatives, find the minimum, etc.)

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