【随机过程】1



2019年08月03日    Author:Guofei

文章归类: 0x44_随机过程    文章编号: 471

版权声明:本文作者是郭飞。转载随意,但需要标明原文链接,并通知本人
原文链接:https://www.guofei.site/2019/08/03/stochastic_process1.html


这篇是 coursera.com 上的 Stochastic processes by National Research University Higher School of Economics 的学习笔记

week1

我们有3种工具

  • probability theory
  • mathematical statistics
  • stochastic processes

an example

在讲 mathematical statistics 的时候,举了个经典例子: 想估计池塘里多少鱼,第一次捞上来一批,打上标签,捞q次,统计每次捞上来的比例。
假设:

  • 总共N条鱼,这是需要我们估计的量
  • 第一次捞上来M条鱼,并打标
  • 第k次捞上来$n_k$条鱼,其中打标的有$m_k$条

计算得知,发生的概率为$P (marked=m_k) = \dfrac{C_M^m C_{N-M}^{n-m}}{C_N^n}$
我们使用最大似然估计 $\max\limits_N \sum\limits_{k=1}^q \ln P (marked=m_k)$ 可以估计出N

定义:概率测度

\(\{ \Omega,F,P \}\) ,其中

  • $\Omega$ is sample space
  • $F$ is sigma algebra,满足
    • $\Omega\in F$
    • $A\in F\Longrightarrow \Omega - A \in F$
    • $A_1,A_2,…,A_n \in F \Longrightarrow \cup A_i \in F$
  • $P$ is probability measure
    • $P{ \Omega }=1$
    • $A_i \in F,P(\cup A_i)=\sum P(A_i)$

以扔硬币n次为例,$\Omega$ 中有$2^n$个元素,$F$中有$2^{2^n}$个元素

Random variable is a function $\xi: \Omega \to \mathbb R$ such that $\forall B \in \mathbb {B(R)}:\xi^{-1}(B) \in F$

Let us consider the following example. An agent flips a coin 2 times. In this model, \(\Omega={(h,h);(h,t);(t,h);(t,t)}\), where t means tails and h means heads. σ-algebra $\mathbb F$ contains all possible combinations of those 4 elements in $\Omega$ (by the way, the number of elements in $\mathbb F$ is exactly $2^4$ ).

Let, $\xi(h;h)=1, \xi(h;t)=2, \xi(t;h)=3, \xi(t;t)=4$. (Note that \(P\{\xi=k\}=14,k=\{1,2,3,4\}\) and 0 otherwise.) Clearly, if we apply $\xi^{-1}$ to both sides of those equations, we obtain $\xi^{-1}(1)=(h;h), \xi^{-1}(2)=(h;t), \xi^{-1}(3)=(t;h), \xi^{-1}(4)=(t;t)$. Therefore, $\forall B \in \mathbb {B(R)}: \xi^{-1}(B) \subset F$.

定义:随机过程

$X:T\times \Omega \to \mathbb R$

https://www.coursera.org/learn/stochasticprocesses/lecture/mIqkz/week-1-5-trajectories-and-finite-dimensional-distributions

counting process

$N_t=\arg\max\limits_k (\xi_k\leq t)$

convolution

两个分布X, Y的convolution,实际上就是 $X+Y$ 的分布
求出来就是:
$F_X \ast F_Y = F_{X+Y}(x) = \int_R F_X(x-y)dF(y)$
$P_{X+Y}(x)=\int_R P_X(x-y)P_Y(y)dy$

we define $F^{n\ast}=F\ast …\ast F$
properties:

  • $F^{n\ast}(x)\leq F^n(x)$ if $F(0)=0$
    这是因为 \(\{\xi_1+...+\xi_n\leq x\}\subset\{\xi_1\leq x,...,\xi_n\leq x\}\)
  • $F^{n\ast}(x) \geq F^{(n+1)\ast}(x)$ if $F(0)=0$

Renewal process

$S_0=0,S_n=S_{n-1}+\xi_n,$
$\xi_1,\xi_2,…,i.i.d>0$, as
$P(\xi_i>0)=1$ (equals to $F(0)=0$)

properties:

  • define $u(t)=\sum\limits_{n=1}^\infty F^{n\ast}$
    then $u(t)<\infty$
  • define \(N_t = ???\)
    then $EN_t=u(t)$, because $EN_t=??$
    这一段没太明白

laplace transform

先去另一篇blog复习一下 laplace transform

我们研究一下 $\mathscr L_f(s) = \mathscr L[f(s)]=\int_0^{+\infty}f(t)e^{-st}dt$,其中$f$是概率密度函数
properties:

  • $\mathscr L_f(s) = E[e^{-s\xi}]$
  • $\mathscr L_{f_1\ast f_2}(s) = \mathscr L_{f_1}(s) \ast \mathscr L_{f_2}(s)$
  • 如果$F(0)=0$,那么$\mathscr L_f(s) = \dfrac{\mathscr L_f(s)}{s}$(证法是分部积分法)

week 1.8 week 1.9 有几个漂亮的定理,可以再复习一下


另一个:

教材板块

  1. Poisson Processes
  2. Finite-state Markov Chains
  3. Renewal Processes
  4. Countable-state Markov Chains
  5. Markov Processes with countable state spaces
  6. Random Walks

coursra

https://www.coursera.org/learn/stochasticprocesses

这个课一直在堆公式

交大课程


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