http://probability101.wordpress.com

I’ve decided that I will almost surely start another blog where I will keep track of my adventures in probability (hehe notice the horrible attempt at humor).  I will still use this blog to post math related things as well as other things going on in my life. However, this new blog will be strictly about probability.  When I get around to actually creating the blog (which will be done with WordPress because of the LaTex support) I will be sure to provide a link.  The first few post will be me documenting what I have learned from the book mentioned at the beginning of this post.  Eventually I hope to discuss material covered in courses I’ll be taking as well as “reviews” of articles/papers I am sure to read.

To start things off, I will assume that you are familiar with Hölder’s Inequality, if not the follow the link and at least read everything before the “contents” section.  So this is a common version of the inequality, however there is another inequality that can be derived from this (which one of my teachers call the Generalized Hölder’s Inequality).  Its statement is as follows:

Suppose that and that as well as . Then , and if we choose such that , we have that

proof: First, if then we have that and .  Now if we integrate and take r-th roots we have

Seeing how we got that takes a little thinking, but it shouldn’t be too difficult to convince yourself that this is correct.

Now assume that .  Noticing that , we can use Hölder’s to obtain

.  This can be rewritten as .  Then taking r-th roots gives us . ♦♦♦

Of all the things we will need to accomplish out goal, the above theorem was the most difficult to prove.  Next we will prove what has been labeled “The world’s second most obvious inequality”.  Don’t worry, this is almost as obvious as the “The world’s most obvious inequality” .

Let , then for and , we have

proof: It is almost insulting to write this out since it more or less proves itself.  You just do the obvious thing and everything works itself out.  However, for the sake of completeness I will provide the proof.

. ♦♦♦

Now we finally have all the necessary tools to accomplish our goal.

If for some then for all .  Additionally,

proof: By using we will show that .  Since , we can choose such that

Thus it follows that .  So the first half of the theorem is proved.

Now since in is fixed, if we must have that .  Hence

For the reverse direction we will make use of Chebyshev’s Inequality   Before doing so however, if then we are done, so suppose that .  Now choose such that .  Let , so .  Hence

Now since was arbitrary we have that .  Thus we have the desired equality in . ♦♦♦

B1: Prove that any positive rational number can be written as the ratio of factorials of primes (not necessarily distinct).

For example

That wasn’t the example they gave on the exam but it gets the point across.

SPOILER ALERT:

I won’t provide a proof but provide a hint that will more or less give it away.

The first thing to notice is that you can reduce to showing this for integers since any rational is just a ratio or integers.  Then from here you see that you can reduce it further to the case of only showing that it in fact holds for primes.  At this point you just show how you can use the knowlege about a primes’ representations (as the desired ratio) to construct it for any integer and thus any rational.  The proof is just a basic induction (on the primes if you want or the integers) and your base cases are the numbers 1 (not prime but must be shown for completeness) and 2.

In the next post I hope to post all the problems from the exam and then solve some of then in the proceeding post.  Though I’ll have solutions, or hints for most of the problems, I actually didn’t think of most of them during the examination.  I guess at this point I’ll tell you which ones I solved, well submitted solutions for at least.  A1, A2, B1, and B5.

Some sad news, I knew the answers to; B2 but didn’t know how to show that my lower bound was in fact correct, B4 but forgot everything I learned in Fourier Series so got stuck on an integral, and A4 and once again I was headed in the right direction but gave up too early.

Regardless I feel pretty good about my performance.  The other team members seemed to all have done at least as good, so we should hopefully place just as high as last year (13 I believe).

1. NYU
2. Cornell
3. Columbia
4. Carnegie Mellon
5. Duke
6. Wisconsin-Madison

This boils down to 5 math programs and 1 in operations research.  Part of me wants to also apply to UF but as things are right now I don’t think that I will.  After doing a little preliminary number crunching I realized that I will easily exceed $600 for the whole process.

Assume that in a metric space you are given 2 closed disjoint sets and .  Prove that if one of them is compact then with and .

proof: Without loss of generality (WLOG) assume that is compact.  Then define as .  Otherwise, define it to be the infimum of the distance from the point in to a point in the set .  Now is continuous and thus attains its infimum.  But this must be positive, because if it were not we would have contradicted the fact that the sets were disjoint and closed.

This proof works, but I would also like to find one that relies on a set being compact if any open cover has a finite subcover.  Sure it might be more work than necessary, but I feel that it should be good exercise.

Given two disjoint closed sets and in a metric space , prove that there exists disjoint open sets and such that and .

In my proof I made the following mistake: Let for , . Then .

I should have stopped at this time, since I remember talking about something very similar to this in my analysis class.  It turns out that to gaurantee you need that at least one of to be compact.

Here is the problem statement (generalized).  Consider a function , and ,  such that for any sequence .  Prove that the set of points where is not zero is countable.

Proof:

Suppose for the sake of contradiction that the set is not countable, thus uncountable.  Define .  It should be obvious that . So one of the must be uncountable.  Thus if we choose points from this set (one of the uncountable sets) we can make as large as we want. Hence we must have that the set of points wehre is at most countable.

Not that in fact we can ‘weaken’ the assumption on so that it reads .

If is integrable over , then

Well here is a proof of a simplified version.  I only assume that and

proof: By “the world’s most obvious integral inequality” So obviously

For the reverse direction let then on some set Therefore To get the result take then .

The only part that bothers me is making sure that is measurable, but that shoudn’t be too difficult to fix.