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

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

Thích Học Toán

Một bài toán kinh điển trong luyện thi học sinh giỏi là bài này. Chứng minh rằng nếu $latex {\alpha}&fg=000000$ là số vô tỷ, dãy các số $latex {n\alpha -[n\alpha]}&fg=000000$, phần thập phân của $latex {n\alpha}&fg=000000$ với $latex {n\in \mathbb N}&fg=000000$ biên thiên, trù mật trong đoạn $latex {[0,1]}&fg=000000$. Lời giải dựa trên nguyên lý Dirichlet, còn được gọi là nguyên lý chuồng thỏ hoặc chuồng bồ câu tùy vào khu vực địa lý nơi bạn sinh sống.

Dùng chuỗi Fourier, Hermann Weyl chứng minh định đề mạnh hơn nhiều. Ông chứng minh rằng tập các phần thập phân $latex {n \alpha – [n \alpha]}&fg=000000$ phân bố đều trên đoạn $latex {[0,1]}&fg=000000$. Nếu lấy một đoạn con $latex {[a,b]}&fg=000000$ nằm giữa 0 và 1, xác suất để $latex {n \alpha – [n \alpha]}&fg=000000$ rơi vào trong đoạn này đúng bằng $latex {b-a}&fg=000000$.

Gọi $latex {I_{[a,b]}}&fg=000000$ là hàm đặc trưng của…

View original post 249 more words

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