整理隨機過程的連續性、微分、積分和Brownian Motion

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Coursera Stochastic Processes 課程筆記, 共十篇:

根據之前上的 Stochastic processes 課程, 針對以下幾點整理出各自的定義、充分或充要條件:
 - 隨機過程的連續性 (Stochastic continuity)
 - 隨機過程的微分 (Stochastic differentiability)
 - 隨機過程的積分 (Stochasitc Integral)
 - Brownian Motion

每次過段時間都忘記, 查找起來也麻煩, 因此整理一篇

隨機過程的連續性 [Week 6.4]

[Continuity in the probability sense Def]:

$$X_t\xrightarrow[t\rightarrow t_0]{p}X_{t_0} \Longleftrightarrow \mathcal{P}(|X_t-X_{t_0}|>\varepsilon)\xrightarrow[t\rightarrow t_0]{}0, \qquad \forall\varepsilon>0$$ 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]:
$$X_t\xrightarrow[t\rightarrow t_0]{L^2}X_{t_0} \Longleftrightarrow \mathbb{E}(X_t-X_{t_0})^2\xrightarrow[t\rightarrow t_0]{}0$$

[Proposition, Continuity of Stochastic Process]: (充分條件)
$X_t$ stochastic process and $\mathbb{E}X_t=0$
 1. If $K(t,s)$ is continuous at $(t_0, t_0)$, then $X_t$ is continuous in the m.sq. sense at $t=t_0$
 2. If $X_t$ is continuous in the m.sq. sense at $t=t_0,s_0$, then $K(t,s)$ is continuous at $(t_0,s_0)$

💡 有關不同的收斂種類, e.g. in mean-squared sense, in probability sense, in distribution sense and almost surely, 請參考 [Week 6 Self Study]

隨機過程的微分 [Week 6.3]

[Stochastic Derivative Def]:
We say that a random process $X_t$ is differentiable at $t=t_0$ if the following limit converges in m.sq. sense to some random variable $\eta$
$$\frac{X_{t+h}-X_t}{h} \xrightarrow[h\rightarrow0]{L^2}\eta:=X_{t_0}'$$ equivalently we can write as follows
$$\mathbb{E}\left[\left(\frac{X_{t+h}-X_t}{h}-\eta\right)^2\right] \xrightarrow[h\rightarrow0]{}0$$

[Proposition, Differentiability of Stochastic Process]: (充要條件)
Let $X_t$ is a stochastic process and $\mathbb{E}X_t^2<\infty$. Then $X_t$ is differentiable at $t=t_0$ if and only if

$$\begin{align} \exists\frac{dm(t)}{dt}\text{, at }t=t_0 \\ \exists\frac{\partial}{\partial t\partial s}K(t,s)\text{, at }(t_0,t_0) \\ \end{align}$$

💡 mean and covariance functions 在 $t_0$ 都是 differentiable

隨機過程的積分 [Week7]

四種 types 的 stochastic integration, where $X_t,H_t$ are stochastic processes, $f(t)\in L^2([a,b])$ is a deterministic function, $W_t$ is a Brownian motion, and $a,b\in\mathbb{R}$

$$\begin{align} \int_a^b X_t dt \\ \int_a^b f(t)dW_t \\ \int_a^b X_t dW_t \\ \int_a^b X_t dH_t \\ \end{align}$$

💡 在 stochastic integral 跟以往的 integral 很不一樣的一點是, 我們發現 integral 的結果仍然是一個 random variable 閱讀以下內容時可以看出這一點

只列出最簡單的積分形式, 其餘請參考Week7筆記

[Stochastic Integrals of Simplest Type $∫ X_t dt$]:
Given a Stochastic process $X_t:\Omega\times\mathbb{R}_+\rightarrow\mathbb{R}$
Given partition $\Delta:a=t_0\leq t_1\leq ...\leq t_n=b$, $|\Delta|=\max\{t_k-t_{k-1}\},k=1,...,n$
If the following expectation converges to some value $A$ in mean square sense:

$$\mathbb{E}\left[ \left( A - \sum_{k=1}^n X_{t_{k-1}}(t_k-t_{k-1}) \right)^2 \right] \xrightarrow[|\Delta|\longrightarrow0]{} 0$$ Then we denote(define) the converged value $A$ as:
$$\int_a^b X_t dt$$

[Existence of the Simplest Type of Stochastic Integral]: (充分條件)

$X_t:\mathbb{E}[X_t^2]<\infty$, if
 1. $m(t)$ continuous
 2. $K(t,s)$ continuous
Then the stochastic integral exists
$$\int_a^b X_tdt<\infty$$ 有上述定理好處如下:
We assume integral is taken in an bounded interval $[a,b]$
 1. Expectation of stochastic integral:
  $$\mathbb{E}\left[ \int X_tdt \right] = \int \mathbb{E}[X_t]dt$$

 2. Expectation of squared stochastic integral:
  $$\mathbb{E}\left[ \left(\int X_t dt\right)^2 \right] = \int\int\mathbb{E}[X_tX_s]dtds$$
 3. Variance of stochastic integral:
  $$Var\left[ \int X_t dt \right] = \int_a^b\int_a^b K(t,s) dtds$$

總結隨機過程的連續、微分、積分的條件

Differentiability 要關注 $m(t)$ and $K(t,s)$ 的微分性 (充要條件)
而 continuity 則是, 若$m(t)=0$, 且 $K(t_0,t0)$ 連續, 則 $X{t_0}$ 連續 (充分條件)
另外對於 simplest type of stochastic integral $\int X_tdx$ 來說, 關注的是 $m(t)$ and $K(t,s)$ 的連續性, 連續的話則 $\int X_tdx$ 存在 (充分條件)

Brownian Motion [Week4.6]

Brownian Motion 又稱 Wiener process
[Brownian Motion Def1]:
 We say that $B_t$ is Brownian motion if and only if
 $B_t$ is a Gaussin process with $m(t)=0$ and $K(t,s)=\min(t,s)$

[Brownian Motion Def2]:
 We say that $B_t$ is Brownian motion if and only if
 $(0)$ $B_0=0$ almost surely (a.s.)
 $(1)$ $B_t$ is independent increaments, i.e.
  $\forall t_0<t_1<...<t_n$, we have $B_{t_1}-B_{t_0}, ..., B_{t_n}-B_{t_{n-1}}$ are independent
 $(2)$ $B_t-B_s\sim\mathcal{N}(0,t-s)$, $\forall t>s\geq0$

由 Def2 (2) 知
$$B_{t+\Delta t}-B_t\sim\mathcal{N}(0,\Delta t)$$ 寫成 $dB_t$ 的離散近似形式變成
$$B_{t+\Delta t}-B_t=\sqrt{\Delta t}\cdot \varepsilon$$, where $\varepsilon\sim\mathcal{N(0,1)}$
注意到 $dB_t/dt$ 照隨機微分的定義為:
$$\lim_{h\rightarrow0}\frac{B_{t+h}-B_t}{h}$$ 但上式 limit 不存在
因為在 $t_0$ 上面提到的 Stochasitc differentiable 的充要條件為 mean and covariance functions 在 $t_0$ 都是 differentiable
而我們知道 Brownian motion 的 $K(t,s)=\min(t,s)$ 是不連續的, 所以微分不存在