# 数学代写|随机微积分代写Stochastic calculus代考|FE610

## 数学代写|随机微积分代写Stochastic calculus代考|Some properties and extentions

Given $\boldsymbol{\eta}1, \boldsymbol{\eta}_2 \in P M^2\left(\mathbb{R}^M\right),(3.2 .2)$ plus a simple polarization argument $^6$ shows that $$\left(I{\boldsymbol{\eta}1}(t) I{\boldsymbol{\eta}3}(t)-\int_0^t\left(\boldsymbol{\eta}_1(\tau), \boldsymbol{\eta}_2(\tau)\right){\mathbb{R}^M} d \tau, \mathcal{F}_t, \mathbb{P}\right)$$ is a martingale.
Now suppose that $\eta \in P M^2\left(\mathbb{R}^M\right)$ and that $\zeta$ is a stopping time relative to $\left{\mathcal{F}t: t \geq 0\right}$. Because $(t, \omega) \rightsquigarrow \mathbf{1}{[0, \zeta(\omega))}(t)$ is adapted and left continuous, it, and therefore $(t, \omega) \rightsquigarrow \mathbf{1}{[0, \zeta(\omega))}(t) \eta(t, \omega)$, are progressively measurable. Further, by Hunt’s stopping time theorem, \begin{aligned} &\mathbb{E}^{\mathbb{P}}\left[I{\boldsymbol{\eta}}(t \wedge \zeta) \int_0^t \mathbf{1}{[0, \zeta)}(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R} M}\right] \ &\quad=\mathbb{E}^{\mathbb{P}}\left[I_{\boldsymbol{\eta}}(t \wedge \zeta) \int_0^{t \wedge \zeta} \mathbf{1}{[0, \zeta)}(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R}^M}\right]=\mathbb{E}^{\mathbb{P}}\left[\int_0^{t \wedge \zeta}|\boldsymbol{\eta}(\tau)|^2 d \tau\right], \end{aligned}
and so
$$\mathbb{E}^{\mathbb{P}}\left[\left|I_{\boldsymbol{\eta}}(t \wedge \zeta)-\int_0^t \mathbf{1}{[0, \zeta)}(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R}^M}\right|^2\right]=0 .$$
Hence
$$I_{\boldsymbol{\eta}}(t \wedge \zeta)=\int_0^t \mathbf{1}{[0, \zeta)}(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R}^M} .$$
In particular, if $\zeta_1$ and $\zeta_2$ are a pair of stopping times and $\zeta_1 \leq \zeta_2$, then
$$\int_0^{\imath \wedge \zeta_1} \mathbf{1}{\left[0, \zeta_2\right)}(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R}^M}=\int_0^l \mathbf{1}{\left[0, \zeta_1\right)}(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R}^M} .$$
Similarly, if $\zeta_1 \leq \zeta_2<\infty$ are stopping times, then
\begin{aligned} \int_{\zeta_1}^{\zeta_2}(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R}^M} &:=I{\boldsymbol{\eta}}\left(\zeta_2\right)-I_{\boldsymbol{\eta}}\left(\zeta_1\right) \ &=\int_0^{\infty} \mathbf{1}{\left[\zeta_1, \zeta_2\right)}(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)){\mathbb{R}^M} . \end{aligned}

## 数学代写|随机微积分代写Stochastic calculus代考|Stochastic integral equations

Let $\sigma: \mathbb{R}^N \longrightarrow \operatorname{Hom}\left(\mathbb{R}^M ; \mathbb{R}^N\right)$ and $b: \mathbb{R}^N \longrightarrow \mathbb{R}^N$ be uniformly Lipschitz continuous functions. We can now interpret the construction in $\S 2.2$ in terms of stochastic integrals. For each $n \geq 0, X_n(\cdot, \mathbf{x})$ is given by (3.0.1). Thus
$\mathbb{E}^{\mathcal{W}}\left[\left|X_n(\cdot, \mathbf{x})\right|_{[0, t]}^2\right] \leq 3|\mathbf{x}|^2+12 \int_0^t \mathbb{E}^{\mathcal{W}}\left[\left|\sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|_{\text {H.S. }}^2\right] d \tau$
$+3 t \int_0^t \mathbb{E}^{\mathcal{W}}\left[\left|b\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|^2\right] d \tau$
$\mathbb{E}^{\mathcal{W}}\left\lfloor\left|X_n(t, \mathbf{x})-X_n(s, \mathbf{x})\right|^2\right\rfloor \leq 2 \int_s^t \mathbb{E}^{\mathcal{W}}\left\lfloor\left|\sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|_{\text {H.S. }}^2\right\rfloor d \tau$
$+2 t \int_s^t \mathbb{E}^{\mathcal{W}}\left[\left|b\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|^2\right]$ and
\begin{aligned} &\mathbb{E}^{\mathcal{W}}\left[\left|X_{n+1}(\cdot, \mathbf{x})-X_n(\cdot, \mathbf{x})\right|_{[0, t]}^2\right] \ &\leq 8 \int_0^t \mathbb{E}^{\mathcal{W}}\left[\left|\sigma\left(X_{n+1}\left(\lfloor\tau\rfloor_{n+1}, \mathbf{x}\right)\right)-\sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|_{\text {H.S. }}^2\right] d \tau \ &\quad+2 t \int_0^t \mathbb{E}^{\mathcal{W}}\left[\left|b\left(X_{n+1}\left(\lfloor\tau\rfloor_{n+1}, \mathbf{x}\right)\right)-b\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|^2\right] d \tau \end{aligned}
Given these, one can proceed as in $\S 2.2 .1$ and thereby recover (2.2.2) and (2.2.3). In addition, knowing $(2.2 .3)$, we see that $X(., \mathbf{x})$ is progressively measurable with respect to the filtration $\left{W_t: t \geq 0\right}$ and
$$\lim {n \rightarrow \infty} \mathbb{E}^{\mathcal{W}}\left[|| \int_0^t \sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right) d w(\tau)-\left.\int_0^t \sigma(X(\tau, \mathbf{x})) d w(\tau)\right|^2\right]=0$$ Therefore $X(\cdot, \mathbf{x})$ solves the stochastic integral equation in (3.0.2). In fact, it is the only solution, since if $\widetilde{X}(\cdot, \mathbf{x})$ were a second solution, then \begin{aligned} &\mathbb{E}^{\mathcal{W}}\left[|| \widetilde{X}(t, \mathbf{x})-\left.X(t, \mathbf{x})\right|^2\right] \ &\leq 2 \int_0^t \mathbb{E}^{\mathcal{W}}\left[|\sigma(\widetilde{X}(\tau, \mathbf{x}))-\sigma(X(\tau, \mathbf{x}))|{\text {H.S. }}^2\right] d \tau \ &\quad \quad+2 t \int_0^t \mathbb{E}^{\mathcal{W}}\left[|b(\widetilde{X}(\tau, \mathbf{x}))-b(X(\tau, \mathbf{x}))|^2\right] d \tau \ &\leq 2\left(|\sigma|_{\text {Lip }}^2+t|b|_{\text {Lip }}^2\right) \int_0^t \mathbb{E}^{\mathcal{W}}\left[|\widetilde{X}(\tau, \mathbf{x})-X(\tau, \mathbf{x})|^2\right] d \tau \end{aligned}
which, by Lemma $1.2 .4$, means that $\widetilde{X}(t, \mathbf{x})=X(t, \mathbf{x})($ a.s., $\mathcal{W})$.

# 随机微积分代考

## 数学代写|随机微积分代写Stochastic calculus代考|Some properties and extentions

$$\mathbb{E}^{P^2}\left[\left|I_\eta(t \wedge \zeta)-\int_0^t \mathbf{1}[0, \zeta)(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)) \mathbb{R}^M\right|^2\right]=0$$

$$I_\eta(t \wedge \zeta)=\int_0^t \mathbf{1}[0, \zeta)(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)) \mathbb{R}^M$$

$$\int_0^{\sim \wedge \zeta_1} \mathbf{1}\left[0, \zeta_2\right)(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)) \mathbb{R}^M=\int_0^l \mathbf{1}\left[0, \zeta_1\right)(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)) \mathbb{R}^M$$

$$\int_{\zeta_1}^{\zeta_2}(\boldsymbol{\eta}(\tau), d B(\tau)) \mathbb{R}^M:=\operatorname{I\eta }\left(\zeta_2\right)-I_\eta\left(\zeta_1\right) \quad=\int_0^{\infty} \mathbf{1}\left[\zeta_1, \zeta_2\right)(\tau)(\boldsymbol{\eta}(\tau), d B(\tau)) \mathbb{R}^M$$

## 数学代写|随机微积分代写Stochastic calculus代考|Stochastic integral equations

\begin{aligned} &\mathbb{E}^{\mathcal{W}}\left[\left|X_n(\cdot, \mathbf{x})\right|{[0, t]}^2\right] \leq 3|\mathbf{x}|^2+12 \int_0^t \mathbb{E}^{\mathcal{W}}\left[\left|\sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|{\mathrm{H} . \mathrm{S} .}^2\right] d \tau \ &+3 t \int_0^t \mathbb{E}^{\mathcal{W}}\left[\left|b\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|^2\right] d \tau \ &\left.\mathbb{E}^{\mathcal{W}}\left[\left|X_n(t, \mathbf{x})-X_n(s, \mathbf{x})\right|^2\right] \leq\left. 2 \int_s^t \mathbb{E}^{\mathcal{W}}|| \sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|{\mathrm{H} . \mathrm{S}} ^2\right] d \tau \ &+2 t \int_s^t \mathbb{E}^{\mathcal{W}}\left[\left|b\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|^2\right] \text { 和 } \ &\mathbb{E}^{\mathcal{W}}\left[\left|X{n+1}(\cdot, \mathbf{x})-X_n(\cdot, \mathbf{x})\right|{[0, t]}^2\right] \quad \leq 8 \int_0^t \mathbb{E}^{\mathcal{W}}\left[\left|\sigma\left(X{n+1}\left(\lfloor\tau\rfloor_{n+1}, \mathbf{x}\right)\right)-\sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right)\right|_{\mathrm{H.S.} .}^2\right] d \tau \end{aligned}

$$\lim n \rightarrow \infty \mathbb{E}^{\mathcal{W}}\left[| \int_0^t \sigma\left(X_n\left(\lfloor\tau\rfloor_n, \mathbf{x}\right)\right) d w(\tau)-\left.\int_0^t \sigma(X(\tau, \mathbf{x})) d w(\tau)\right|^2\right]=0$$

$$\mathbb{E}^{\mathcal{W}}\left[| \widetilde{X}(t, \mathbf{x})-\left.X(t, \mathbf{x})\right|^2\right] \quad \leq 2 \int_0^t \mathbb{E}^{\mathcal{W}}\left[|\sigma(\widetilde{X}(\tau, \mathbf{x}))-\sigma(X(\tau, \mathbf{x}))| \text { H.S. }^2\right] d \tau \quad+2 t \int_0^t \mathbb{E}^{\mathcal{W}}$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: