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# 数学代写|随机分析作业代写stochastic analysis代考|Doob’s Decomposition and Martingale Transformation

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## 数学代写|随机分析作业代写stochastic analysis代考|random variables

In this chapter, we call a sequence of random variables $X=\left{X_{n}\right}_{n=0}^{\infty}$ a stochastic process.

Definition 2.4.1 We say that a stochastic process $X=\left{X_{n}\right}_{n=0}^{\infty}$ is $\left(\left{\mathcal{F}{n}\right}{n=0}^{\infty}-\right)$ adapted, if a random variable $X_{n}$ is $\mathcal{F}{n}$-measurable for all $n \in \mathbf{Z}{\geqq 0}$.
It is obvious that martingales are adapted.
Definition 2.4.2 We say that a stochastic process $X=\left{X_{n}\right}_{n=0}^{\infty}$ is $\left(\left{\mathcal{F}{n}\right}{n=0^{-}}^{\infty}\right)$ predictable, if $X_{0}=0$ and a random variable $X_{n}$ is $\mathcal{F}{n-1}$-measurable for all $n \geqq 1$. Theorem 2.4.1 Let $\left{X{n}\right}_{n=0}^{\infty}$ be an adapted process such that $E\left[\left|X_{n}\right|\right]<\infty, n=$ $0,1,2, \ldots$. Then we have the following.
(1) There exist a martingale $\left{M_{n}\right}_{n=0}^{\infty}$ with $M_{0}=0$ and a predictable process $\left{A_{n}\right}_{n=0}^{\infty}$ such that
$$X_{n}=X_{0}+M_{n}+A_{n}, \quad n \geqq 0$$
Moreover, if there is a martingale $\left{\tilde{M}{n}\right}{n=0}^{\infty}$ with $\tilde{M}{0}=0$ and a predictable process $\left{\tilde{A}{n}\right}_{n=0}^{\infty}$ such that $X_{n}=X_{0}+\tilde{M}{n}+\tilde{A}{n}, n \geqq 0$, then $\tilde{M}{n}=M{n}$ a.s. and $\tilde{A}{n}=A{n}$ a.s., $n \geqq 0$.

## 数学代写|随机分析作业代写stochastic analysis代考|well-defined

Proposition 2.4.1 Let $\tau$ be a stopping time and $\left{X_{n}\right}_{n=0}^{\infty}$ be an adapted stochastic process. Then a random variable $1_{{\tau<\infty}} X_{\tau}$ is $\mathcal{F}{\tau}$-measurable. Here $1{{\tau<\infty}} X_{\tau}$ is a random variable given by
$$\left(1_{{\tau<\infty}} X_{\tau}\right)(\omega)=\left{\begin{array}{cl} X_{n}(\omega), & \text { if } \tau(\omega)=n \in \mathbf{Z}{\geqq 0}, \ 0, & \text { if } \tau(\omega)=\infty \end{array}\right.$$ for $\omega \in \Omega$. Proof Our assertion follows from the fact that $1{(\tau<\infty)} X_{\tau}$ is well-defined and that
$$\left{1_{{\tau<\infty}} X_{\tau} \in B\right} \cap{\tau=n}=\left{X_{n} \in B\right} \cap{\tau=n} \in \mathcal{F}{n}, \quad n \in \mathbf{Z}{\geqq 0}$$
for any $B \in \mathcal{B}([-\infty, \infty])$.
In the case that $P(\tau<\infty)=1$ we denote $1_{{\tau<\infty}} X_{\tau}$ by $X_{\tau}$ for simplicity of notation.

For any stochastic processes $X=\left{X_{n}\right}_{n=0}^{\infty}$ and $Y=\left{Y_{n}\right}_{n=0}^{\infty}$, we define a stochastic process $Y \bullet X$ by

(Y \bullet X){0}=0, \text { and }(Y \bullet X){n}=\sum_{k=1}^{n} Y_{k}\left(X_{k}-X_{k-1}\right), \quad n \geqq 1

## 数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|RANDOM VARIABLES

It is obvious that martingales are adapted.
Definition 2.4.2 We say that a stochastic process $X=\left{X_{n}\right}_{n=0}^{\infty}$ is $\left(\left{\mathcal{F}{n}\right}{n=0^{-}}^{\infty}\right)$ predictable, if $X_{0}=0$ and a random variable $X_{n}$ is $\mathcal{F}{n-1}$-measurable for all $n \geqq 1$. Theorem 2.4.1 Let $\left{X{n}\right}_{n=0}^{\infty}$ be an adapted process such that $E\left[\left|X_{n}\right|\right]<\infty, n=$ $0,1,2, \ldots$. Then we have the following.
(1) There exist a martingale $\left{M_{n}\right}_{n=0}^{\infty}$ with $M_{0}=0$ and a predictable process $\left{A_{n}\right}_{n=0}^{\infty}$ such that
$$X_{n}=X_{0}+M_{n}+A_{n}, \quad n \geqq 0$$
Moreover, if there is a martingale $\left{\tilde{M}{n}\right}{n=0}^{\infty}$ with $\tilde{M}{0}=0$ and a predictable process $\left{\tilde{A}{n}\right}_{n=0}^{\infty}$ such that $X_{n}=X_{0}+\tilde{M}{n}+\tilde{A}{n}, n \geqq 0$, then $\tilde{M}{n}=M{n}$ a.s. and $\tilde{A}{n}=A{n}$ a.s., $n \geqq 0$.

## 数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|WELL-DEFINED

$$\left(1_{{\tau<\infty}} X_{\tau}\right)(\omega)=\left{\begin{array}{cl} X_{n}(\omega), & \text { if } \tau(\omega)=n \in \mathbf{Z}{\geqq 0}, \ 0, & \text { if } \tau(\omega)=\infty \end{array}\right.$$ for $\omega \in \Omega$. Proof Our assertion follows from the fact that $1{(\tau<\infty)} X_{\tau}$ is well-defined and that
$$\left{1_{{\tau<\infty}} X_{\tau} \in B\right} \cap{\tau=n}=\left{X_{n} \in B\right} \cap{\tau=n} \in \mathcal{F}{n}, \quad n \in \mathbf{Z}{\geqq 0}$$

(Y \bullet X){0}=0, \text { and }(Y \bullet X){n}=\sum_{k=1}^{n} Y_{k}\left(X_{k}-X_{k-1}\right), \quad n \geqq 1