A little encouragement

November 8, 2013 Leave a comment

If you know who is David Aldous, you will laugh really hard after reading this extract from the report of a referee who rejected his paper. Even big name’s papers can be rejected, so don’t be discouraged young mathematicians

http://www.stat.berkeley.edu/~aldous/least_interesting_paper_ever.txt

Advertisements
Categories: Uncategorized

A useful identity in convergence of random series

July 23, 2012 Leave a comment

Let X_1, X_2,\ldots, be independent and let S_{m,n} = X_{m+1} + \cdots + X_n. Then for any a>0

P(\max_{m<j\leq n} |S_{m,j}| > 2a)\min_{m<k\leq n} P(|S_{k,n}|\leq a) \leq P(|S_{m,n}|>a).

The identity can be used to prove the following results:

1) P. Levy: Let X_1, X_2,\ldots be independent and let S_n = X_1+\cdots+X_n. If \lim_{n\rightarrow \infty} S_n exists in probability, then it also exists almost surely.

2) Let  X_1, X_2,\ldots  be i.i.d. and S_n = X_1+\cdots+X_n. If \lim_{n\rightarrow \infty} S_n/n \rightarrow 0 in probability, then (\max_{1\leq m\leq n} S_m)/n \rightarrow 0 in probability.

Categories: Probability Theory

An application of Fourier analysis (in Vietnamese),

Author: Prof. Ngo Bao Chau

Categories: Uncategorized

Lecture 3: The outer measure generated by a measure

September 17, 2011 Leave a comment

I’ll post lecture 1 and 2 soon.

In lecture 1 and 2, I defined a measure \mu on a semiring S and an outer measure on the power set of X. Our goal is that we want to define a measurable set E\subseteq X. As we recall, a measurable set E\subseteq X satifies the following condition: \mu(A) =\mu(A\cap E) + \mu(A\cap E^c) for any A\subseteq X. Notice that \mu here is an outer measure.

In this lecture, a measure \mu is extended to an outer measure \mu^*. As proved before, the collection \Lambda of measurable sets with respect to this outer measure is a \sigma-algebra. We will show that S\subseteq\Lambda. Thus, semirings are a basic collections of sets which a measure theory can be built on.

Let a measure space (X,S,\mu) be fixed. The outer measure \mu^* is defined as follows:

\mu^*(A) = \inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } A\subseteq \bigcup_{i=1}^\infty A_i \}.

Theorem: \mu^* is an outer measure.

Proof:

i) Let A_i=\emptyset for i\geq 1. Then \mu^*(\bigcup_{i=1}^\infty A_i)=\mu^*(\emptyset) \leq \sum_{i=1}^\infty \mu(\emptyset)=0. Thus \mu*(\emptyset)=0.

ii) Let A,B\in X and A\subseteq B. We consider a sequence \{A_i\} of sets in S such that B\subseteq \bigcup_{i=1}^\infty A_i. Then clearly, A\subseteq\bigcup_{i=1}^\infty A_i. Thus, \mu^*(A)\leq\sum_{i=1}^\infty \mu(A_i) Further, we have \mu^*(B)=\inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } B\subseteq \bigcup_{i=1}^\infty A_i \}. This follows that \mu^*(A)\leq \mu^*(B). Hence, \mu^* is monotone.

iii) Let \{A_i\} be a sequence of pairwise disjoint sets in X. If \sum_{i=1}^\infty \mu^*(A_i)=\infty. Then clearly, \mu^*(\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i).

Thus we can assume \sum_{i=1}^\infty \mu^*(A_i)<\infty. Since each \mu^*(A_i) <\infty, there exists a sequence \{A_i\}^{n} of sets in S such that A_i\subseteq \bigcup_{n=1}^\infty A_i^n and \sum_{n=1}^\infty \mu(A_i^n)\le \mu^* (A_i)+2^{-i}\epsilon. Hence, \bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty \bigcup_{n=1}^\infty A_i^n. Furthermore,

\mu^* (\bigcup_{i=1}^\infty A_i)\leq \sum_{i=1}^\infty\sum_{n=1}^\infty\mu(A_i^n) 0, we have \mu^* (\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i) . This proves the \sigma-subadditivity of \mu^*. \square

The next theorem will show that \mu^* is indeed an extension of \mu from S to \mathcal{P}(X).

Theorem: If E\in S, then \mu^*(E)=\mu(E).

Proof:

Let A_1=E and A_i=\emptyset for i>1. Then E\subseteq \bigcup_{i=1}^\infty A_i. Thus, \mu^*(E)\leq \mu(\bigcup_{i=1}^\infty A_i)=\mu(E).

Consider a sequence \{A_i\} of sets in S satisfying A\subseteq \bigcup_{i=1}^\infty A_i. Since \mu is a measure on semiring S. It is \sigma-subadditive on S. Therefore, we have \mu(A)\leq\sum_{i=1}^\infty \mu(A_i). Thus, \mu(A)\leq\mu^*(A). \square

The outer measure generated by a measure have some nice characteristics which can be shown by the following theorem.

Theorem: Let E\subseteq X. The followings are equivalent:

(1) E is \mu^*-measurable.

(2) \mu(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c) with any A\in S and \mu(A)<\infty.

(3) \mu(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c) with any A\in S and \mu(A)<\infty.

(4) \mu^*(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c) with any A\subseteq X.

Proof:

Categories: Uncategorized

Polynomial in a Field of Characteristic p

June 30, 2011 1 comment

I’ll go ahead and it post it here as well:
Hey Ken, I’m trying to check if this proof is correct(it may not be the only way, but its one I thought of)

To Prove: Let F be a field with charF=p. Let q(x)=x^p-x-c where c\in F. Show if q is reducible, then q has a root in F (actually all).

This is actually an iff in the book, but this is the nontrivial direction.

My idea is the following. Let u be a root of q. Let f,g be nonunit irreducible factors of q. Now it is easy to see the set of all roots is the set \{u+i:i=1,\ldots,p\}. Thus, Z(f)=u+I, Z(g)=u+J for some finite sets I,J\subset \mathbb{Z}.

Notice this means \prod_{i\in I}u+i, \prod_{i\in J} u+i are both in F. Let f_1=\prod_{i\in I}u+i and f_2=\prod_{i\in J} u+i. Then u is a root for both f_1-\prod_{i\in I}{x+i} and f_2-\prod_{i\in J}x+i which are in F[x]. But since the degrees of these are |I|,|J|; respectively, the only way this is possible is if |I|=|J|. But we could do this for any irreducible factor so each irreducible factor must have the same degree.

Hence, if q=q_1\cdots q_k with each q_i irreducible, then p=deg(q)=k\cdot n for some n\geq 1. Since p is prime either k or n must be 1. If n=1 then each root is in F. If k=1 then q is irreducible. Since p is reducible the result follows.

After writing this out carefully it feels completely correct (of course I left out some easy details).

Categories: Algebra questions

Complex #34: sneaky use of Rouche’s theorem

June 15, 2011 Leave a comment

Problem #34 is: Prove that there does not exist a polynomial of the form p(z)= z^n+a_{n-1}z^{n-1}+\ldots+a_0 such that |p (z)| < 1 for all z such that |z|=1.

Proof: Suppose that there does exist such a polynomial. Note that |z^n|>|-p(z)| for |z|=1. By Rouche’s theorem, z^n-p(z)=-a_{n-1}z^{n-1}-\ldots-a_0 has the same number of zeros (with multiplicities) as z^n inside the unit disk. But that is impossible, since z^n has n zeros and z^n-p(z) has at most n-1 zeros. QED.

Categories: Complex questions

Complex #27: integrating over a long rectangle

June 15, 2011 Leave a comment

Problem 27: Show that \int_{\gamma} e^{iz}e^{-z^2}dz has the same value on every straight line path \gamma (oriented in the +x direction) parallel to the real axis.

Proof: Consider the rectangle R=[-M,M]\times [0,Y], where Y is some given number and M is a very large number. Since the integrand is entire, the integral over the boundary of R is zero by the Cauchy Integral Theorem, and as M\to\infty, the horizontal pieces correspond to \int_{\gamma} e^{iz}e^{-z^2}dz with \gamma being the line Im (z)=0 or the line Im(z)=Y, with one being negative and the other positive. Thus, the result is proved if we can show that the contributions from the vertical pieces go to zero as M goes to \infty.

Observe that the absolute value of the integrand at z=x+iy is e^{-x^2+y^2-y}, which on each vertical boundary piece is e^{-M^2+y^2-y} for 0\le y\le Y. Integrating this over y will give a value bounded by a constant times e^{-M^2}, which clearly goes to zero as M\to\infty.

QED.

Categories: Complex questions