1
3
2012
0

Test TeX

Let me start with a test for TeX:

[tex]$-{\hbar^2 \over 2m}$[\tex]

Category: 未分类 | Tags:
3
5
2011
29

Lecture Notes for Stochastic Differential Equations --- A parapharse from Professor Xue's Lectures (Lecture 2)

Introduction

In this lecture, we reviewed the definition of the conditional probability with respect to a sub-$\sigma$-field. And then gave several special examples to compute the conditional probability. After mentioning the conditional probability with respect to a random variable, and some function measurability, an easier way to check a conditional probability is given. Finally, we introduce the so-called $\pi$-$\lambda$ theorem and given an application of it.

General Case

Consider the probability measure space $(\Omega, \mathcal F, \mathbb P)$. Event $A \in \mathcal F$ and $\mathcal G$ is a sub-$\sigma$-field. The conditional probability $\mathbb P\{A|\mathcal G\}$satiesfies

  1. $\mathbb P\{A|\mathcal G\}$ is $\mathcal G$-measurable
  2. $\int_G\mathbb P\{A|\mathcal G\}dP = \mathbb P\{A\cap G\}, \forall G \in \mathcal G$.

In the sense of a.e. equal, the random variable is uniquely determined.

These two requirements may seemingly being surprising at the first sight, yet they really have their probabilistic interpretations.

The first entry tells us that by the knowledge of $\mathcal G$, we could define $\mathbb P\{A|\mathcal G\}$.

The second entry is understood while determing the price of a gamble. If the expected gain is zero, then the price of entering a gamble must be $\mathbb P\{A|\mathcal G\}$. To see this, if $A$ happens, the gain is $1 - \mathbb P\{A|\mathcal G\}$, and if $A$ does not happen, the gain will be $- \mathbb P\{A|\mathcal G\}$. Together, the gain will be

$$(1 - \mathbb P\{A|\mathcal G\})\mathbf 1_A - \mathbb P\{A|\mathcal G\} \mathbf 1_{A^c} = \mathbf 1_A - \mathbb P\{A|\mathcal G\}$$

which will lead to the result, if we integral over $G$.

Ex. What is $\mathbb P\{A|\mathcal G\}$ if $A \in \mathcal G$?

Solution. This is Example 33.3. The key is $\mathbf 1_A$.

Ex. What is $\mathbb P\{A|\mathcal G\}$ if $\mathcal G = \{\emptyset, \Omega\}$?

Solution. This is Example 33.4. The key is $\mathbb P\{A\}$.

Ex. Suppose $A$ is independent of $\mathcal G$, i.e., $\mathbb P\{A\cap G\} = \mathbb P\{A\} \mathbb P\{G\}, \forall G\in\mathcal G$, then what is $\mathbb P\{A|\mathcal G\}$?

Solution. Intuitively it is $\mathbb P\{A\}$.

Ex. Suppose $\Omega = \mathbb R^2$, $\omega = (x, y)$, $\mathcal F = \mathcal B(\mathbb R^2)$, and

$$\mathbb P\{A\} = \int_A f(x, y)\,dx dy, \quad \forall A \in \mathcal F.$$

Here, $f(x, y)$ is Borel-measurable functon, $f: \mathbb R^2 \to \mathbb R$, and $f(x, y) \ge 0$ almost everywhere, and $\iint_{\mathbb R^2} f(x, y) \,dxdy =1$. Countable additivity is trivial to verify. It follows that $\mathbb P$ is a probability measure. Suppose $\mathcal G$ is a sub-$\sigma$-field generated by the form $E\times \mathbb R^1 = \{(x, y): x\in E\}, \forall E\in \mathcal B(\mathbb R^1)$. Let $A$ be $\mathbb R^1\times F = \{(x,y): y\in F\}, \forall F \in \mathcal B(\mathbb R^1)$. What is $\mathbb P\{A|\mathcal G\}$?

Solution. One claims that

$$\varphi(x, y) = \frac{\int_Ff(x,t)d\,t}{\int_{\mathbb R^1}f(x,t)\,dt}$$

is a version of $\mathbb P\{A|\mathcal G\}$. To verify this, one should verify second property of the definition. Since a general element in $\mathcal G$ takes the form $E \times \mathbb R^1$, it is essential to prove that

$$\int_{E \times \mathbb R^1} \varphi(x, y)\,dP(x, y) = \mathbb P\{A\cap(E\times\mathbb R^1)\}.$$

Notice that the right hand side of the equation is exactly $\mathbb P\{E \times F\}$. To evaluate the value on the left hand side, one must resort to the Fubini's Theorem and a previous Theorem.

It is also possible to define a conditionial probability with respect to random variable $X$. This could be done by define $\sigma(X) = \{X^{-1}(B), B\in \mathcal B(\mathbb R^1)\}$. And define $\mathbb P\{A|X\} = \mathbb P\{A|\sigma(X)\}$. 

 

Category: Lecture notes | Tags:
3
4
2011
0

Lecture Notes for Stochastic Differential Equations --- A parapharse from Professor Xue's Lectures

In this lecture note, we are going to discuss 7 topics, which is outlined as follows:

  1. Conditioned Expectation
  2. Discrete-Time Martingale
  3. Continuous-Time Martingale
  4. Stochastic Integral
  5. Strong Solution of SDE
  6. Weak Solution of SDE (optional)
  7. Applications (optional)

References

  1. Steven E. Shreve and Ioannis Karatres, Brownian Motion and Stochastic Calculus
  2. Rick Durret, Probability: Theory and Examples
  3. Patrick Billingslay: Probability and Measure

Topic 1: Conditioned Probability, Distribution and Expectations

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability measure space, and $X$ be a random variable. Actually, $X$ is a functional from $\Omega$ to $\overline{\mathbb R}$ satisfying that for all $B \in \mathcal B(\mathbb R^1)$,

$$X^{-1}(B) \in \mathcal F.$$

This means every pre-image of a Borel set in the real number is in the $\sigma$-algebra.

Mathematical expectation is defined by

$$\mathbb E X = \int_\Omega X dP = \int_\Omega X \mu(dX).$$

where $\mu(B)$ represents the probability of the preimage of Borel set $B \in \mathcal B(\mathbb R^1)$, i.e.,

$$\mu(B) = \mathbb P \{X^{-1}(B)\}.$$

Ex: Prove that $\mu$ is a measure on $(\mathbb R^1, \mathcal B(\mathbb R^1))$.

Proof. Let us write the definition of a measure. The following is copied from Wikipedia measure (mathematics). If $\mu$ is a measure, three conditions must hold, i.e.,

Let $\Sigma$ be a $\sigma$-algebra over $X$, a function $\mu$ from $\Sigma$ to the extended real number line is called a measure, if it satisfies the following properties:

  • Non-negativity
  • Null empty set
  • Countable additivity

First of all, $\mu$ is a set function from $\mathcal B(\mathbb R^1)$ to $[0, 1]$. Non-negativity follows from the non-negativity of  $\mathbb P$. To see the null empty set property, we check that if $B = \emptyset$implies that $X^{-1}(B) = \emptyset$. By the null empty set property of $\mathbb P$, we arrive at the null empty set property of $\mu$. And finally, for countable additivity, suppose $\{B_i\}_{i=1}^\infty$ are disjoint sets in $\mathcal B(\mathbb R^1)$, then by the definition of $\mu$, we have

$$\mu\left(\bigcup_{i=1}^\infty B_i\right) = \mathbb P\left\{X^{-1}\left(\bigcup_{i=1}^\infty B_i \right)\right\}.$$

By the operation of set, and the disjointness of $\{B_i\}_{i=1}^\infty$ we have

 $$\mathbb P\left\{X^{-1}\left(\bigcup_{i=1}^\infty B_i \right)\right\} = \mathbb P\left\{ \bigcup_{i=1}^\infty X^{-1}(B_i)\right\}.$$

and finally by the countable additivity, we have

$$\mathbb P\left\{ \bigcup_{i=1}^\infty X^{-1}(B_i)\right\} = \sum_{i=1}^\infty \mathbb P\{X^{-1}(B_i)\} = \sum_{i=1}^\infty \mu(B_i)$$

which arrive at the conclusion that satisfy countable additivity.

1.1 Sub-$\sigma$-field and Information

In this subsection, an heuristic example is provided to explain the meaning of a sub-$\sigma$-field. In general cases, a sub-$\sigma$-field could be approximatedly understood as information.

Example: [Toss of a coin 3 times]

All the possible out come constructs the sample space $\Omega$, which takes the form

$$\Omega = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}.$$

After first toss, the sample space could be divided into two parts, as \Omega = \Omega_1 \cup \Omega_2, where

$\Omega_1 = \left\{HHH, HHT, HTH, HTT\right\}, and $\Omega_2 = \Omega - \Omega_1$.

We can consider the corresponding $\sigma$-algebra: $\mathcal G_1 = \{\emptyset, \Omega, \Omega_1, \Omega_2\}$, which stands for the ``information'' after the first toss. When a sample $\omega$ is given, whose first experiment is a head, we can tell that $\omega$is not in $\emptyset$, $\omega$is in $\Omega$, $\omega$ is in $\Omega_{11}$ and $\omega$is not in $\Omega_{12}$. And look it in another angle, we see that, this $\sigma$-algebra contains all the possible situations for different ``first toss'' cases.

It is quite easy to generalize to the ``information'' $\sigma$-field after second toss $\mathcal G_2 = \sigma\left(\{\Omega_{11}, \Omega_{12}, \Omega_{21}, \Omega_{22}\}\right)$.

Generally speaking, if $\mathcal G$is a sub-$\sigma$-field of  $\mathcal F$, the information of $\mathcal G$ is understood as

for all $A \in \mathcal G$, one know whether $\omega \in A$ or not. In other word, the indicator function  is well-defined $\mathbf 1_A(\omega)$.

1.2 Conditional Probability

In this subsection, a theoretical treatment of conditional probability is concerned. As we know in the elementary probability theory, the nature definition for conditional probability is govened by the following equation

$$\mathbb P \{ A \vert B\} = \frac{\mathbb P\{A\cap B\}}{\mathbb P\{B\}}.$$

Therefore, it is natural to raise the question how to define $\mathbb P\{A|\mathcal G\}$, where $\mathcal G$is a sub-$\sigma$-field?

Note: The lecture notes follows majorly from reference book 3---Patrick Billingsley's Probability and Measure, Section 33.

Sometimes, $\mathcal G$ is quite complicated. Thus, instead, we consider the simple case when $\mathcal G$ is generated by some disjoint $\{B_i\}_{i=1}^\infty$ instead, where $\Omega = \bigcup_{i=1}^\infty B_i$. Therefore, we have $\mathcal G = \sigma(\{B_1, B_2, \ldots, B_n, \ldots\})$.

Then $\mathbb P\{A|\mathcal G\}$can be defined pathwisely, by

$$\mathbb P\{A|\mathcal G\}_\omega = \mathbb P\{A|B_i\} = \sum_{i=1}^\infty \mathbf 1_{B_i} \frac{\mathbb P\{A\cap B_i\}}{\mathbb P\{B_i\}}$$

for all sample $\omega$ in sample space $\Omega$.

Before we goto see the formal definition, we examine an example, which comes from the problem of predicting the telephone call probability.

Example: [Apply Simple Case to Computing Conditional Probability]

Consider a Poisson process $\{N_t, t\ge 0\}$ on measure space $(\Omega, \mathcal F, \mathbb P)$. Let $\mathcal G = \sigma(\{N_t\})$ and $A = \{\omega\in \Omega| N_s(\omega) = 0, s<t\}$. Compute $\mathbb P\{A|\mathcal G\}$.

Solution. Recall some of the knowledge of Poisson process now. By Wikipedia Poisson Process, we have

  • $N_0=0$
  • Independent increments
  • Stationary increments
  • No counted occurrences are simultaneous

The result of this defintion is that $N_t \sim \mathrm{Pois}(\alpha)$ where $\alpha$ is the intensity.

Note: If $\alpha$ is a constant, this is the case of homogeneous Poisson process, which is also named as Lévy processes.

It follows that

$$ \mathbb P\left\{ N_t - N_s = n \right\} = e^{-\alpha(t-s)} \frac{(t-s)^n}{n!}$$

for all  $n=0,1,2,\ldots$ and $0 \le s < t < +\infty$.

Another explaination is also need for $\mathcal G = \sigma(\{N_t\})$. This is a sigma field

Now, let $B_i = \{\omega: N_t(\omega) = i\}, i=0, 1, 2, \ldots$. Obviously, the union of all these set is the sample space $\Omega$. Moreover, they are obviously disjoint. Then by the computation formula in the simple case, we have

$$\mathbb P\{A|\mathcal G\}_\omega = \mathbb P\{A|B_i\} = \frac{\mathbb P\{A\cap B_i\}}{\mathbb P\{B_i\}}, \quad \omega \in B_i.$$

 To compute $\mathbb P\{A\cap B_i\}$ and $\mathbb P\{B_i\}$, we have

$$\begin{split} \mathbb P\{A\cap B_i\} &= \mathbb P\{N_0=0, N_s=0, N_t=i\} \\ &= \mathbb P\{N_0=0, N_s=0\}\mathbb P\{N_s=0, N_t=i\}\\ &= e^{-\alpha t}\frac{\alpha (t-s)^i}{i!}\end{split}$$

and

$$\mathbb P\{B_i\}=\mathbb P\{N_0=0, N_t = i\}=e^{-\alpha t} \frac{(\alpha t)^i}{i!}$$

which gives rise to the final result

$$\mathbb P\{A|\mathcal G\}=\left(1 - \frac{s}t\right)^{N_t(\omega)}.$$

Ex. Prove that $\mathbb P\{A|\mathcal G\}$is $\mathcal G$-measurable;

Proof. Recall the definition of $\mathcal G$-measurable. If a random variable $X:\Omega\to\overline{\mathbb R}$ is $\mathcal G$-measurable, then for all $B \in \mathcal B(\mathbb R)$, we have its pre-image $X^{-1}(B) \in \mathcal G$. Or equivalently, $\{\omega\in\Omega|X(\omega) \in B\} \in \mathcal G$. In this case, we are going to prove that $\{\omega\in\Omega|\mathbb P\{A|\mathcal G\}_\omega \in B\} \in \mathcal G$. It reduced to the problem that $\{\omega\in\Omega|\mathbb P\{A|\mathcal G\}_\omega \in B\}$ could be written as some union of $B_i$?

Since for all $\omega \in B_i$, $\mathbb P\{A|\mathcal G\}$ is a constant.

Ex. Prove that $\int_{B_i}\mathbb P\{A|\mathcal G\} dP = \int_{B_i} \mathbf 1_A dP$ holds.

Proof. Note that the left hand side equals to $\int_{B_i}\frac{\mathbb P\{A\cap B_i\}}{\mathbb P\{B_i\}} dP which is identical with the right hand side.

Now we go further to the general cases for $\mathcal G$. Suppose, $(\Omega, \mathcal F, \mathbb P)$ is a probability measure space, $\mathcal G$ is a sub-$\sigma$-field, event $A\in \mathcal F$. Then, we claim the conditional probability of $A$ given $\mathcal G$ is a random variable satisfying (1) $\mathbb P\{A|\mathcal G\}$is $\mathcal G$-measurable; (2) $\int_{B_i}\mathbb P\{A|\mathcal G\} dP = \int_{B_i} \mathbf 1_A dP$ . The random variable exists and unique, by Radon-Nikodym Theorem.

Let $G \in \mathcal G$ and $\nu(G) = \mathbb P\{A\cap G\}$

Ex. Prove that $\nu$ is a measure on $(\Omega, \mathcal G)$.

Proof. $\nu$ is a function that $\nu: \mathcal G \to [0, 1]$. Non-negative property and emptyset property are trivial. Countable additivity follows from the countable additivity of probability measure $\mathbb P$.

Ex. Prove that $\nu$ is absolute continuous with respect to $\mathbb P$ on $(\Omega, \mathcal G)$.

Proof. By Wikipedia, absolute continuity, we know that if $\nu$ is absolute continuous with respect to $\mathbb P$ on $(\Omega, \mathcal G)$, this means that for all $A\in \mathcal G$ and $\mathbb P\{A\} = 0$ implies that $\nu(A) = 0$. This is obvious, since $0\le \mathbb P\{A \cap G\} = \nu(A) = \mathbb P\{A \cap G\} \le \mathbb P\{A\} = 0$.$0\le \mathbb P\{A \cap G\} = \nu(A) = \mathbb P\{A \cap G\} \le \mathbb P\{A\} = 0$..$0\le \mathbb P\{A \cap G\} = \nu(A) = \mathbb P\{A \cap G\} \le \mathbb P\{A\} = 0$ 

Ex. If  $f$ and $g$ are $\mathcal G$-measurable, and $\int_GfdP = \int_GgdP, \forall g \in \mathcal G$, then $f=g$ almost surely.

Proof.

Category: Lecture notes | Tags:
3
2
2011
0

Hello is-programmer!

To day, I'd like to begin my journey of blogging my PhD. study life in Fudan university, as a member of Computational Mathematics direction.

For a student of computational mathematics, one has to find his/her academic background. In Fudan University, a lot of professors are entitled with computational PhD. As my current boss, Professor Weiguo Gao, he is interested in electronic structure calculations, and numerical problems in network and control flows. As Professor Yangfeng Su, the computations in Very-Large-Scale-Integrationi (VLSI). As Professor Jungong Xue puts his effort in stochastic problems. All of them have a deep understanding in numerical linear algebra and related algorithms, as well as having a basic understanding of the background they are going to deal with. Last month, due to the application of CSC Program, I have got a chance to work with Carlos Javier Garcia-Cervera. And I got the invitation letter. The research is entitled with ``First principle calculations of Continuous Solvent''. I think it is a very good starting point to open my career of doing research.

So far as I know, computational physics, chemistry, or even biology is a hot research field all over the world. Take computational physics an example, people are eager to know what solid behaves. A number of questions could be raised. What is their permittivity? How to compute their heat conductivity? Which color will the look like? How to predict the magnetic behavior of a iron? Of course, there are a lot of work had been done and the theory is also well established.

What can I do? Maybe I should go to do a search of articles that already exist about the continuous solvent. In period one, I think the very first paper I should read is written by Professor Weiguo Gao and Professor Zhipan Liu. The second step is to guess out what Professor Carlos would like to view the problem. The third step is to read some of the publication of Professor Weinan E to see what was going on. In period two, I would like to do a review on the numerical methods that exists for continuous solvents.

Ah, I need to make friends with Wei Wang and Weile Jia from Beijing USTC. It is pretty strange that I can not find any information about them on Google. Nah, forget it.

Ah, I also want to have a talk with Snoozy Liu tomorrow. When will she go to the office? I have to take care of her. And next, I am going to discuss with her future, which is quite an impossible problem at the first sight, yet we have to get an answer. Yes, just as the Social Network tells us, the further is unpredictable. Yes, she is on duty tomorrow, but is not sure when she will arrive. Therefore, we can have a talk maybe around noon. In my opinion, I should respect her idea of her life. As a boyfriend, I should give her suggestion on what is worthy to be done or what is not. But the decision right is her. So the next question is that what is my idea about her career? She wants to be a what? As a faculty? As a staff working in the bank? Yes, probably, she only got two answers listed above. And she might agree that she wants to become a faculty more than a staff, I bet. To become a Professor, one has to have something that he/she is not replaceable. How to demonstrate that you are the exclusive one? The factors we can control: according to your career path, according to the paper you published. My conclusion will be that it is a good experience of study over sea. But is it worthy to quit from Fudan to apply for the US school? I should listen to her motivation! If the motivation is right, then, I should give a supportive attitude and will encourage her to fulfill her career. Yes! A wise solution exists that she could prepare the English tests first to consider more on her career path.

The real work we are going to do now is sure, to get a firm basic of knowledge for the future. She might agree to take a TOEFL and a GRE test this year. Usually it will cost her 3 months to prepare TOEFL and half a year of preparing GRE. Here comes the question, whether it is a good idea to go to France? Learning a completely new language is surely a time-consuming task. It has already been too full for her to get the mathematics right and the English good! So, I don't think so. Maybe its a good idea to postpone studying French to the first year of graduate school. I should ask Yingxing Zhao or Zhenjie Ren for advices. Therefore, preparing TOEFL test will be the Task No. 1. Preparing GRE will be the Task No. 2. Preparing French will be the Task No. 3. And I will not enourage her for getting a job now, since she had already decided to get a PhD in Fudan.

OK, back to my career, I should now do three tasks in the next 24 hours.

First one, finish the C code for B variation of GMRES.

Two, get matrices from FEMLAB.

Three, do all experiments.

Have a nice night~!

Category: Ideas | Tags:

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