Coursera Stochastic Processes 課程筆記, 共十篇:
- Week 0: 一些預備知識
- Week 1: Introduction & Renewal processes
- Week 2: Poisson Processes
- Week3: Markov Chains
- Week 4: Gaussian Processes
- Week 5: Stationarity and Linear filters
- Week 6: Ergodicity, differentiability, continuity (本文)
- Week 7: Stochastic integration & Itô formula
- Week 8: Lévy processes
- 整理隨機過程的連續性、微分、積分和Brownian Motion
Self Study: Convergence of random variables
主要參考 wikipedia 的資料筆記 Convergence of random variables
[Converge in distribution Def]:
記做 Xnd→X
Converge in distribution 是最 weak 的, 也就是滿足 converge in distribution 不一定會滿足 converge in probability 或滿足 almost surely 或滿足 converge in mean
[機率論] 兩隨機變數相等表示兩者有相同分布但反之不然, 這篇文章最後給了一個例子:
考慮均勻分布 X 為隨機變數服從均勻分布 U[−1,1] 現在取另一隨機變數 Y:=−X 則 Y 亦為在 [−1,1] 上均勻分布,亦即 X 與 Y 具有同分布。然而
P(X=Y)=0
Converge in distribution 的其他等價定義
[Converge in probability Def]:
記做 Xnp→X
自己的想法: Xn 與 X 的 sample spaces 可以不同, 只要 mapping 到 R 之後相減就好, 所以考量的是 joint distribution
For a random process Xt converges to a constant in probability sense
Xtp→t→∞c (is const.)
則表示一定會 (必要條件)
EXt→t→∞cVar(Xt)→t→∞0
[Almost sure convergence Def]:
記做 Xna.s.→X
可參考: 謝宗翰的隨筆 [機率論] Almost Sure Convergence
自己的想法: 需要 sample space 一樣, 且都是對個別 outcome 去比較的. 把所有這些符合的 outcomes 蒐集起來的集合, 其 probability measure 為 1.
這是一個很強的條件, 幾乎針對每一個 outcome 都要符合.
[Convergence in mean Def]:
記做 XnLr→X.
Convergence in the r-th mean, for r≥1, implies convergence in probability (by Markov’s inequality). Furthermore, if r>s≥1, convergence in r-th mean implies convergence in s-th mean. Hence, convergence in mean square implies convergence in mean.
It is also worth noticing that if
XnLr→X
then
limn→∞E[|Xn|r]=E[|X|r]
[Relation btw Stochastic Convergences]:
- 各種 convergence 的 proofs:
https://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables#propA2
Week 6.1: Notion of ergodicity. Examples
Motivated by LLN (Law of Large Number)
[LLN Thm]:
ξ1,ξ2,… - i.i.d. and Eξ1<∞, then
1N∑Nn=1ξnp→N→∞Eξ1
上式 p→ 表示 convergence in probability
但也滿足 a.s.→ almost sure convergence (Strong LLN)
Ergodicity 嘗試將 LLN 概念延伸到 stochastic process
[Ergodicity Def]:
Xt is a stochastic process, where t=1,2,3,…
Xt is said ergodic if ∃c constant such that
MT:=1TT∑t=1Xtp→T→∞c
where c is some constant. T is called horizon.
And we consider convergence in probability sense.
所以要證明 ergodic 可以 p→, or a.s.→, or m.sq.→, or d→c
[Example 1]:
Xt=ξ∼N(0,1), trajectory 為 constant for all t
m(t)=0,K(t,s)=Varξ=1
所以是 weak stationary, 我們考慮 ergodicity
1T∑Tt=1Xt=ξ≠c
不存在一個 constant for T→∞, 所以 non-ergodic
[Example 2]:
Xt stochastic process defined as:
Xt=εt+acosπt6
where a≠0 and ε1,ε2,… i.i.d. N(0,1)
其 trajectory 為下圖曲線, 並對該曲線每個位置都有 std normal noise
m(t)=acosπt6≠const, 所以不是 stationary. 考慮 ergodicity:
1T∑Tt=1Xt∼N(aT∑Tt=1cosπt6,1T)
互相獨立之 normal distributions 相加仍為 normal, mean and variance 都為相加
variance 收斂到 0, 觀察 mean:
由於 trajectory 是以 12 為一個週期, 所以最多只會有 12 個非 0 的值, 而每一個都小於等於 1
結論是 mean 也收斂到 0 (因為不管哪一個 outcome, 其 trajectory 最後都到 0, 所以收斂的 random variable 為 contant 0)
所以
1T∑Tt=1Xt∼N(0,0) for T→∞
所以 =0 a.s. 因此是 ergodic
Example 1 是 (weak) stationary but non-ergodic
Example 2 是 non-stationary but erogodic
因此 stationary 跟 ergodic 是不同概念
Week 6.2: Ergodicity of wide-sense stationary processes
[Proposition]:
For Xt is a discrete time stochastic process. Define
MT=1TT∑t=1XtC(T)=Cov(XT,MT)
If ∃α such that the covariance function is bounded by α, i.e.
|K(s,t)|≤α;∀s,t
Then
Var(MT)→T→∞0⟺C(T)→T→∞0
[Stolz-Cesaro Thm]:
For an,bn∈R, bn is strictly increasing and unbounded, we have:
limn→∞an−an−1bn−bn−1=q⟹anbn→n→∞q
See wiki for more detials and proof
當 stationary 時, 有兩個 sufficient conditions 滿足則保證 ergodic
[Sufficient Conditions of Ergodicity when WSS]:
Xt is weakly stationary and γ(⋅) is its auto-covariance function with |γ(⋅)|<∞.
1. Sufficient condition: If
1T∑T−1r=0γ(r)→T→∞0
Then Xt is ergodic
2. Sufficient condition: If
γ(r)→r→∞0
Then Xt is ergodic
[Proof 1]:
考慮 C(T)
C(T)=Cov(XT,1TT∑t=1Xt)=1TT∑t=1Cov(XT,Xt)=1TT∑t=1γ(T−t)=1TT−1∑r=0γ(r)⟶0 (by assumption)
由 proposition 知道 Var(MT)→0, as T→∞
又因為已知 Xt is weakly stationary, 所以 ∃c constant, s.t.
EXt=c⇒EMt=c
則 variance converge to 0 as T→∞ 表示
VarMT=E[(MT−c)2]→T→∞0
則
MTL2→c⟹MTp→c
根據定義 Xt is ergodic. Q.E.D.
[Proof 2]:
我們利用 proof 1 的結果和 Stolz-Cesaro thm, 定義
an:=n−1∑r=0γ(r)bn:=n
則
an−an−1bn−bn−1=γ(n−1)1→n→∞0=q⟹anbn=1nn−1∑r=0γ(r)→n→∞0=q⟹Xt ergodic
[Example 1]:
Nt is Poisson process with λ. 給定 p>0, 並定義
Xt:=Nt+p−Nt
則我們可得知 Xt is ergodic
You can use the independent increment property of Poisson Process to get this formula.
γ(t−s)=Cov(Nt+p−Nt,Ns+p−Ns)
If |s−t|>p, then Nt+p−Nt is independent of Ns+p−Ns, so γ(t−s)=0
If t<s<t+p, then Nt is independent of Ns+p−Ns, so
Cov(Nt+p−Nt,Ns+p−Ns)=Cov(Nt+p,Ns+p−Ns)=Cov(Nt+p,Ns+p)−Cov(Nt+p,Ns)=Cov(Nt+p,Ns+p−Nt+p+Nt+p)−Cov(Nt+p−Ns+Ns,Ns)=Var(Nt+p)−Var(Ns)=λ(t+p−s)
For s<t<s+p. you can deal with it similarly.
[Example 2]:
A,B 都是 r.v.s 有如下圖的 expectation and uncorrelated relation
ω 是任意給定的 fixed value
則我們發現 Xt is weakly stationary and ergodic
Week 6.3: Definition of a stochastic derivative
[Stochastic Derivative Def]:
We say that a random process Xt is differentiable at t=t0
if the following limit converges in m.sq. sense to some random variable η
Xt+h−XthL2→h→0η:=X′t0
equivalently we can write as follows
E[(Xt+h−Xth−η)2]→h→00
[Proposition, Differentiability of Stochastic Process]:
Let Xt is a stochastic process and EX2t<∞.
Then Xt is differentiable at t=t0 if and only if
∃dm(t)dt, at t=t0∃∂∂t∂sK(t,s), at (t0,t0)
[Brownian Motion is NOT differentiable at any time t]:
[Differentiability of Independent Increments]:
Xt is independent increments and X0=0 a.s. 則我們知道
K(t,s)=Var(Xmin(t,s))
因此大部分的 stochastic process with independent increments 都不是可微的.
我們寫一下 covariance function 的推導:
K(t,s)=Cov(Xt,Xs)(for t>s) =Cov(Xt−Xs,Xs−X0)+Cov(Xs,Xs)=0+Var(Xs)(consider t>s and t<s) =Var(Xmin(t,s))
[Differentiability of Weakly Stationary]:
Xt is weakly stationary. 所以 m(t) is constant ⇒ differentiable at all t.
我們知道 K(t,s)=γ(t−s), 計算 partial derivatives:
∂2K(t,s)∂t∂s|(t0,t0)=∂2γ(t−s)∂t∂s=∂∂t(−γ′(t−s))|(t0,t0)=−γ″(t−s)|(t0,t0)=−γ″(0)
所以 Xt is differentiable (at any time t) if and only if −γ″(0) 存在
[Example 1]:
承上 weakly stationary differentiable. 如果 γ(r)=e−α|r|
則 Xt is not differentiable, 因為 −γ″(0) 不存在 (γ′(0) 就不存在了). γ(t) 如下:
[Example 2]:
承上 weakly stationary differentiable. 如果 γ(r)=cos(αr)
則 Xt is differentiable
Week 6.4: Continuity in the mean-squared sense
[Continuity in the probability sense Def]:
Xtp→t→t0Xt0⟺P(|Xt−Xt0|>ε)→t→t00,∀ε>0
Converges in mean-squared sense 確保了 converges in probability, 且 in m.sq. sense 比較容易確認, 因此以下著重在 in m.sq. sense
[Continuity in the mean-squared sense Def]:
XtL2→t→t0Xt0⟺E(Xt−Xt0)2→t→t00
[Proposition, Continuity of Stochastic Process]:
Xt stochastic process and EXt=0
1. If K(t,s) is continuous at (t0,t0), then Xt is continuous in the m.sq. sense at t=t0
2. If Xt is continuous in the m.sq. sense at t=t0,s0, then K(t,s) is continuous at (t0,s0)
[Proof 1]:
proved by definition of m.sq. sense
E(Xt−Xt0)2=EX2t−2EXtXt0+EX2t0=K(t,t)−2K(t,t0)+K(t0,t0)→t→t00
[Proof 2]:
K(t,s)−K(t0,s0)=(K(t,s)−K(t0,s))+(K(t0,s)−K(t0,s0))
使用 EXt=0 and 科西不等式, 考慮第一項 K(t,s)−K(t0,s)
|K(t,s)−K(t0,s)|=|E[(Xt−Xt0)Xs]|≤√E(Xt−Xt0)2⋅√EX2sby assumption →t→t00
第二項 K(t0,s)−K(t0,s0) 也一樣. Q.E.D.
Differentiability 要關注 m(t) and K(t,s) 的微分性, 而 continuity 只關注 K(t,s) 的連續性
另外對於 simplest type of stochastic integral ∫Xtdx 來說, 關注的是 m(t) and K(t,s) 的連續性
[Corollary]:
Covariance function K(t,s) is continuous at (t0,s0),∀t0,s0 if and only if
K(t,s) is continuous at diagonal, i.e. (t0,t0),∀t0
[Proof]:
只需證明 ⟸
K(t,s) is continuous at (t0,t0), ∀t0. 由 proposition 1 知道 Xt is continuous in m.sq. sense ∀t
因此 Xt is continuous in the m.sq. sense at t=t0,s0 for all points
由 proposition 2 知道 K(t,s) is continuous at (t0,s0),∀t0,s0
Q.E.D.