Stochastic Processes Week 6 Ergodicity, differentiability, continuity


Coursera Stochastic Processes 課程筆記, 共十篇:


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 XndX

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:=XY 亦為在 [1,1] 上均勻分布,亦即 XY 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 XnpX

[Properties]:

自己的想法: XnX 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
Xtptc (is const.)
則表示一定會 (必要條件)
EXttcVar(Xt)t0

[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 XnLrX.

Convergence in the r-th mean, for r1, implies convergence in probability (by Markov’s inequality). Furthermore, if r>s1, 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

XnLrX

then

limnE[|Xn|r]=E[|X|r]

[Relation btw Stochastic Convergences]:

補上課程針對以上四個 convergences 的定義


Week 6.1: Notion of ergodicity. Examples

Motivated by LLN (Law of Large Number)
[LLN Thm]:
ξ1,ξ2, - i.i.d. and Eξ1<, then
1NNn=1ξnpNEξ1
 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:=1TTt=1XtpTc


 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 dc

[Example 1]:
Xt=ξN(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
1TTt=1Xt=ξc
 不存在一個 constant for T, 所以 non-ergodic

[Example 2]:
Xt stochastic process defined as:
Xt=εt+acosπt6
 where a0 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)=acosπt6const, 所以不是 stationary. 考慮 ergodicity:
1TTt=1XtN(aTTt=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
1TTt=1XtN(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=1TTt=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)T0C(T)T0

[Stolz-Cesaro Thm]:
  For an,bnR, bn is strictly increasing and unbounded, we have:
limnanan1bnbn1=qanbnnq

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
  1TT1r=0γ(r)T0
  Then Xt is ergodic
 2. Sufficient condition: If
  γ(r)r0
  Then Xt is ergodic
[Proof 1]:
 考慮 C(T)
C(T)=Cov(XT,1TTt=1Xt)=1TTt=1Cov(XT,Xt)=1TTt=1γ(Tt)=1TT1r=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=cEMt=c
 則 variance converge to 0 as T 表示
VarMT=E[(MTc)2]T0
 則
MTL2cMTpc
 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:=n1r=0γ(r)bn:=n


 則
anan1bnbn1=γ(n1)1n0=qanbn=1nn1r=0γ(r)n0=qXt ergodic

[Example 1]:
Nt is Poisson process with λ. 給定 p>0, 並定義
Xt:=Nt+pNt
 則我們可得知 Xt is ergodic

You can use the independent increment property of Poisson Process to get this formula.
γ(ts)=Cov(Nt+pNt,Ns+pNs)
If |st|>p, then Nt+pNt is independent of Ns+pNs, so γ(ts)=0
If t<s<t+p, then Nt is independent of Ns+pNs, so
Cov(Nt+pNt,Ns+pNs)=Cov(Nt+p,Ns+pNs)=Cov(Nt+p,Ns+p)Cov(Nt+p,Ns)=Cov(Nt+p,Ns+pNt+p+Nt+p)Cov(Nt+pNs+Ns,Ns)=Var(Nt+p)Var(Ns)=λ(t+ps)


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+hXthL2h0η:=Xt0
 equivalently we can write as follows
E[(Xt+hXthη)2]h00

[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=t0tsK(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(XtXs,XsX0)+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(ts), 計算 partial derivatives:
2K(t,s)ts|(t0,t0)=2γ(ts)ts=t(γ(ts))|(t0,t0)=γ(ts)|(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)=eα|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]:
Xtptt0Xt0P(|XtXt0|>ε)tt00,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]:
XtL2tt0Xt0E(XtXt0)2tt00

[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(XtXt0)2=EX2t2EXtXt0+EX2t0=K(t,t)2K(t,t0)+K(t0,t0)tt00


[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[(XtXt0)Xs]|E(XtXt0)2EX2sby assumption tt00

 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.