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# 数学代写|随机分析作业代写stochastic analysis代考|Stopping Time

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## 数学代写|随机分析作业代写stochastic analysis代考|stopping time

From now on, we mainly consider the case that $\mathbf{T}=\mathbf{Z}{\geq 0}={0,1,2, \ldots}$ in this chapter. Also, we fix a filtration $\left{\mathcal{F}{n}\right}_{n=0}^{\infty}=\left{\mathcal{F}{n}\right}{n \in \mathbf{Z}_{\geqq 0}}$.

Definition 2.3.1 We say that $\sigma$ is an $\left(\left{\mathcal{F}{n}\right}{n=0^{-}}^{\infty}\right)$ stopping time, if $\sigma$ is a mapping from $\Omega$ into $\mathbf{Z}{\geqq 0} \cup{\infty}$ such that $${\sigma=n} \in \mathcal{F}{n}, \quad n=0,1, \ldots$$
Proposition 2.3.1 (1) Suppose that $\tau$ is a mapping from $\Omega$ into $\mathbf{Z}{\geq 0} \cup{\infty}$. Then $\tau$ is a stopping time, if and only if $${\tau \leqq n} \in \mathcal{F}{n}, \quad n=0,1, \ldots$$
(2) Let $m \in \mathbf{Z}{\geqq 0} \cup{\infty}$ and let $\tau \equiv m$, i.e., $\tau(\omega)=m$, $\omega \in \Omega$. Then $\tau$ is a stopping time. (3) If $\sigma$ and $\tau$ are stopping times, $\sigma \wedge \tau, \sigma \vee \tau$, and $\sigma+\tau$ are stopping times. Here $\sigma \wedge \tau$ is a mapping defined by $(\sigma \wedge \tau)(\omega)=\sigma(\omega) \wedge \tau(\omega), \omega \in \Omega$. The definitions of $\sigma \vee \tau$ and $\sigma+\tau$ are similar. Proof We have Assertion (1) from the definition of filtration and the fact that $${\tau \leqq n}=\bigcup{k=0}^{n}{\tau=k} \text { and }{\tau=n}={\tau \leqq n} \backslash{\tau \leqq n-1}$$
We have Assertions (2) and (3) from the definition of stopping times, Assertion
(1) and the fact that
$${\sigma \wedge \tau \leqq n}={\sigma \leqq n} \cup{\tau \leqq n}, \quad{\sigma \vee \tau \leqq n}={\sigma \leqq n} \cap{\tau \leqq n}$$
and
$${\sigma+\tau=n}=\bigcup_{k=0}^{n}({\sigma=k} \cap{\tau=n-k})$$
We define a subset $\mathcal{F}{\tau}$ of $\mathcal{F}$ for each stopping time $\tau$ by $$\mathcal{F}{\tau}=\left{A \in \mathcal{F} ; A \cap{\tau=n} \in \mathcal{F}{n} \text { for any } n \in \mathbf{Z}{\geqq 0}\right}$$
We leave the proof of the following proposition to the reader as an exercise.

## 数学代写|随机分析作业代写stochastic analysis代考|Let τ be a stopping time

Proposition 2.3.2 Let $\tau$ be a stopping time.
(1) Let $A \in \mathcal{F}$. Then $A \in \mathcal{F}{\tau}$, if and only if $A \cap{\tau \leqq n} \in \mathcal{F}{n}$ for any $n \in \mathbf{Z}{\geqq 0}$. (2) $\mathcal{F}{\tau}$ is a sub- $\sigma$-algebra.
(3) $\tau$ is $\mathcal{F}{\tau}$-measurable. (4) Let $m \in \mathbf{Z}{\geqq 0}$. If $\tau \equiv m$, then $\mathcal{F}{\tau}=\mathcal{F}{m}$.
Proposition 2.3.3 Let $\sigma$ and $\tau$ be stopping times.
(1) If $A \in \mathcal{F}{\sigma}$, then $A \cap{\sigma \leqq \tau} \in \mathcal{F}{\tau}$ and $A \cap{\sigma \leqq \tau} \in \mathcal{F}_{\sigma \wedge \tau}$.

(2) If $\sigma(\omega) \leqq \tau(\omega)$ for all $\omega \in \Omega$, then $\mathcal{F}{\sigma} \subset \mathcal{F}{\tau}$.
(3) $\mathcal{F}{\sigma} \cap \mathcal{F}{\tau}=\mathcal{F}{\sigma \wedge \tau}$. (4) ${\tau<\sigma},{\tau=\sigma},{\tau>\sigma} \in \mathcal{F}{\sigma \wedge \tau}$.
(5) If $A \in \mathcal{F}{\sigma}$, then $A \cap{\sigma=\tau} \in \mathcal{F}{\sigma \wedge \tau}$.
Proof (1) For any $n \in \mathbf{Z}{\geqq 0}$ we see that $$(A \cap{\sigma \leqq \tau}) \cap{\tau=n}=(A \cap{\sigma \leqq n}) \cap{\tau=n} \in \mathcal{F}{n}$$
So we see that $A \cap{\sigma \leqq \tau} \in \mathcal{F}{\tau}$. Then by this result we see that $$A \cap{\sigma \leqq \tau}=A \cap{\sigma \leqq(\sigma \wedge \tau)} \in \mathcal{F}{\sigma \wedge \tau} .$$
This implies Assertion (1).
Assertion (2) follows from Assertion (1), since ${\sigma \leqq \tau}=\Omega$.
Suppose that $A \in \mathcal{F}{\sigma} \cap \mathcal{F}{\tau}$. Then by Assertion (1) we see that $A=(A \cap{\sigma \leqq$ $\tau}) \cup(A \cap{\tau \leqq \sigma}) \in \mathcal{F}{\sigma \wedge \tau}$. So $\mathcal{F}{\sigma} \cap \mathcal{F}{\tau} \subset \mathcal{F}{\sigma \wedge \tau}$. Also, by Assertion (2) we see that $\mathcal{F}{\sigma \wedge \tau} \subset \mathcal{F}{\sigma} \cap \mathcal{F}_{\tau}$. So we have Assertion (3).

Letting $A=\Omega$ in Assertion (1), we see that ${\sigma \leqq \tau},{\tau \leqq \sigma} \in \mathcal{F}_{\sigma \wedge \tau}$. This implies Assertion (4).
Assertion (5) follows from Assertions (1) and (4).

## 数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|STOPPING TIME

(2) Let $m \in \mathbf{Z}{\geqq 0} \cup{\infty}$ and let $\tau \equiv m$, i.e., $\tau(\omega)=m$, $\omega \in \Omega$. Then $\tau$ is a stopping time. (3) If $\sigma$ and $\tau$ are stopping times, $\sigma \wedge \tau, \sigma \vee \tau$, and $\sigma+\tau$ are stopping times. Here $\sigma \wedge \tau$ is a mapping defined by $(\sigma \wedge \tau)(\omega)=\sigma(\omega) \wedge \tau(\omega), \omega \in \Omega$. The definitions of $\sigma \vee \tau$ and $\sigma+\tau$ are similar. Proof We have Assertion (1) from the definition of filtration and the fact that $${\tau \leqq n}=\bigcup{k=0}^{n}{\tau=k} \text { and }{\tau=n}={\tau \leqq n} \backslash{\tau \leqq n-1}$$
We have Assertions (2) and (3) from the definition of stopping times, Assertion
(1) and the fact that
$${\sigma \wedge \tau \leqq n}={\sigma \leqq n} \cup{\tau \leqq n}, \quad{\sigma \vee \tau \leqq n}={\sigma \leqq n} \cap{\tau \leqq n}$$
and
$${\sigma+\tau=n}=\bigcup_{k=0}^{n}({\sigma=k} \cap{\tau=n-k})$$
We define a subset $\mathcal{F}{\tau}$ of $\mathcal{F}$ for each stopping time $\tau$ by $$\mathcal{F}{\tau}=\left{A \in \mathcal{F} ; A \cap{\tau=n} \in \mathcal{F}{n} \text { for any } n \in \mathbf{Z}{\geqq 0}\right}$$

## 数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|LET Τ BE A STOPPING TIME

(1) Let $A \in \mathcal{F}$. Then $A \in \mathcal{F}{\tau}$, if and only if $A \cap{\tau \leqq n} \in \mathcal{F}{n}$ for any $n \in \mathbf{Z}{\geqq 0}$. (2) $\mathcal{F}{\tau}$ is a sub- $\sigma$-algebra.
(3) $\tau$ is $\mathcal{F}{\tau}$-measurable. (4) Let $m \in \mathbf{Z}{\geqq 0}$. If $\tau \equiv m$, then $\mathcal{F}{\tau}=\mathcal{F}{m}$.
Proposition 2.3.3 Let $\sigma$ and $\tau$ be stopping times.
(1) If $A \in \mathcal{F}{\sigma}$, then $A \cap{\sigma \leqq \tau} \in \mathcal{F}{\tau}$ and $A \cap{\sigma \leqq \tau} \in \mathcal{F}_{\sigma \wedge \tau}$.

(2) If $\sigma(\omega) \leqq \tau(\omega)$ for all $\omega \in \Omega$, then $\mathcal{F}{\sigma} \subset \mathcal{F}{\tau}$.
(3) $\mathcal{F}{\sigma} \cap \mathcal{F}{\tau}=\mathcal{F}{\sigma \wedge \tau}$. (4) ${\tau<\sigma},{\tau=\sigma},{\tau>\sigma} \in \mathcal{F}{\sigma \wedge \tau}$.
(5) If $A \in \mathcal{F}{\sigma}$, then $A \cap{\sigma=\tau} \in \mathcal{F}{\sigma \wedge \tau}$.
Proof (1) For any $n \in \mathbf{Z}{\geqq 0}$ we see that $$(A \cap{\sigma \leqq \tau}) \cap{\tau=n}=(A \cap{\sigma \leqq n}) \cap{\tau=n} \in \mathcal{F}{n}$$
So we see that $A \cap{\sigma \leqq \tau} \in \mathcal{F}{\tau}$. Then by this result we see that $$A \cap{\sigma \leqq \tau}=A \cap{\sigma \leqq(\sigma \wedge \tau)} \in \mathcal{F}{\sigma \wedge \tau} .$$

Suppose that $A \in \mathcal{F}{\sigma} \cap \mathcal{F}{\tau}$. Then by Assertion (1) we see that $A=(A \cap{\sigma \leqq$ $\tau}) \cup(A \cap{\tau \leqq \sigma}) \in \mathcal{F}{\sigma \wedge \tau}$. So $\mathcal{F}{\sigma} \cap \mathcal{F}{\tau} \subset \mathcal{F}{\sigma \wedge \tau}$. Also, by Assertion (2) we see that $\mathcal{F}{\sigma \wedge \tau} \subset \mathcal{F}{\sigma} \cap \mathcal{F}_{\tau}$. 所以我们有断言