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这是华盛顿大学 随机分析课程的代写成功案例。
Math735课程简介
Course description
Stochastic analysis is a term that refers to stochastic integration and stochastic differential equations and related themes. Here is a list of topics we expect to cover. The amount of time devoted to the fundamentals in the beginning will depend on the level of background that the audience possesses.
Foundations of probability theory, especially conditional expectation
Generalities about stochastic processes, Brownian motion, Poisson process
Martingales
Stochastic integral with respect to Brownian motion (quick overview of the Math 635 stochastic integral)
Predictable processes and stochastic integral with respect to cadlag martingales and semimartingales
Itô’s formula
Stochastic differential equations
Local time for Brownian motion, Girsanov’s theorem
White noise integrals and a stochastic partial differential equation
Prerequisites
Prerequisites
This course has flexible prerequisites. The ideal background would be one or two semesters of graduate measure-theoretic probability theory, such as our 733 or 733-734. An essential prerequisite is a certain degree of mathematical maturity, so familiarity with advanced probability is not absolutely necessary. The course will rely on modern integration theory (measure theory covered in Math 629 and 721) and advanced probability, and we can cover some of these points quickly in the beginning.
Evaluation
Course grades will be based on take-home work and a possible in-class exam. Homework will be posted on Learn@UW. You can also see your score record on Learn@UW.
Lecture notes
The course is based on lecture notes written by the instructor, available on Learn@UW. No textbook purchase is required.
Math735 Stochastic Analysis HELP（EXAM HELP， ONLINE TUTOR）
Exercise 3 Let $\left(M_t(\omega)\right)_{t \in T}$ be a martingale with respect to the filtration $\mathbb{F}=\left(\mathcal{F}_t\right)$ with $M_0(\omega)=0$. Here $T$ could be either $\mathbb{R}^{+}$or $\mathbb{N}$.
Define a family of random times $\tau_x: x \in \mathbb{R}$
$$
\tau_x(\omega)= \begin{cases}\inf \left{s: M_s \geq x\right} & \text { for } x \geq 0 \ \inf \left{s: M_s \leq x\right} & \text { for } x<0\end{cases}
$$
Show that $\tau_x$ is a stopping time.
Solution 3 Assume that $T=\mathbb{N}$. Then if $x \geq 0$
$$
\left{\tau_x \leq t\right}=\left{\inf \left{s: M_s \geq x\right} \leq t\right}=\left{M_s \geq x \text { for some } s \leq t\right}=\bigcup_{s=1}^t\left{M_s \geq x\right}
$$
where $\left{M_s \geq x\right} \in \mathcal{F}s \subset \mathcal{F}_t$, so $\left{\tau_x \leq t\right} \in \mathcal{F}_t$. Thus $\tau_x$ is a stopping time. Similarly if $x<0$, then $$ \left{\tau_x \leq t\right}=\left{\inf \left{s: M_s \leq x\right} \leq t\right}=\left{M_s \leq x \text { for some } s \leq t\right}=\bigcup{s=1}^t\left{M_s \leq x\right} \in \mathcal{F}_t
$$
Exercise 4 Let
$$
M_t(\omega)=\sum_{s=1}^t X_s(\omega)
$$
be a binary random walk where $t \in \mathbb{N}$ and $\left(X_s: s \in \mathbb{N}\right)$ are i.i.d. random variables with
$$
P\left(X_s= \pm 1\right)=P\left(X_s= \pm 1 \mid \mathcal{F}{s-1}\right)=1 / 2 $$ $X_s$ is $F_s$ measurable and $P$-independent from $F{s-1}$.
Show that $\left(M_t\right){t \in \mathbb{N}}$ and $\left(M_t^2-t\right){t \in \mathbb{N}}$ are $\mathbb{F}$-martingales.
Consider the stopping time $\sigma(\omega)=\min \left(\tau_a, \tau_b\right)$ where $a<0<b \in \mathbb{N}$, and the stopped martingales $\left(M_{t \wedge \sigma}\right){t \in \mathbb{N}}$ and $\left(M{t \wedge \sigma}^2-t \wedge \sigma\right){t \in \mathbb{N}}$. Show that Doob’s martingale convergence theorem applies and $$ \lim {t \rightarrow \infty} M_{t \wedge \sigma}(\omega)=M_\sigma(\omega)
$$
exists $P$-almost surely.
Consider now $\left(M_{t \wedge \sigma}^2-t \wedge \sigma\right)$. Use the martingale property together with the reverse Fatou lemma to show that $E(\sigma)<\infty$ which implies $P(\sigma<\infty)=1$.
For $a<0<b \in \mathbb{N}$, compute $P\left(\tau_a<\tau_b\right)$.
Solution $4 \quad\left(M_t\right.$ and $M_t^2-t$ are martingales:) Clearly $M_t$ and $M_t^2-t$ are bounded by $t$ and $t^2-t$, so they are integrable. They are also martingales since if $s<t$, then
$$
E\left(M_t \mid \mathcal{F}s\right)=E\left(M{t-1}+X_t \mid \mathcal{F}s\right)=M_s+E\left(X_t \mid \mathcal{F}_s\right)=M_s $$ by induction and independence. Similarly $$ \begin{aligned} E\left(M_t^2-t \mid \mathcal{F}_s\right) & =E\left(M{t-1}^2+2 M_{t-1} X_t+X_t^2-t \mid F_s\right)=M_s^2-s+E\left(2 M_{t-1} X_t+X_t^2-1 \mid F_s\right) \
& =M_s^2-s+2 E\left(M_{t-1} \mid F_s\right) E\left(X_t \mid F_s\right)=M_s^2-s
\end{aligned}
$$
(Limit of the stopped martingale:) We have to check that
$$
\sup {t \geq 0} E\left(M{t \wedge \sigma}^{-}\right)<\infty \text {. }
$$
But this is clear since $M_{t \wedge \sigma}^{-} \leq-a$. Hence the limit
$$
\lim {t \rightarrow \infty} M{t \wedge \sigma}
$$
exists almost surely. Similarly we see that the limit
$$
\lim {t \rightarrow \infty} M{t \wedge \sigma}^2-t \wedge \sigma
$$
exists almost surely because $\left(M_{t \wedge \sigma}^2-t \wedge \sigma\right)^{+} \leq a^2+b^2$.
(We have $E(\sigma)<\infty$ 🙂 We know that $M_{t \wedge \sigma}^2-(t \wedge \sigma)$ is a martingale that almost surely converges to $M_\sigma^2-\sigma$. Thus by reverse Fatou lemma
$$
E\left(M_\sigma^2-\sigma\right) \geq \limsup {t \rightarrow \infty} E\left(M{t \wedge \sigma}^2-(t \wedge \sigma)\right)=0
$$
In particular then
$$
E(\sigma) \leq E\left(M_\sigma^2\right)<\infty
$$
(The probability $P\left(\tau_a<\tau_b\right)$ 🙂 By Doob’s optional stopping theorem we have
$$
\begin{aligned}
& E\left(M_\sigma\right)=E\left(M_0\right)=0=P\left(\tau_a<\tau_b\right) a+\left(1-P\left(\tau_a<\tau_b\right)\right) b \
& \text { so } P\left(\tau_a<\tau_b\right)=\frac{b}{b-a}
\end{aligned}
$$
Exercise 5 Let $M_t(\omega)=B_t(\omega), t \in \mathbb{R}^{+}$, a Brownian motion which is assumed to be $\mathbb{F}$-adapted, and such that for all $0<s<t$ the increment $\left(B_t-B_s\right)$ is $P$-independent from the $\sigma$-algebra $\mathcal{F}_s$.
Note that since by assumption the Brownian motion is $\mathbb{F}$-adapted, it follows that $\mathcal{F}_t^B=\sigma\left(B_s: 0 \leq s \leq t\right) \subset \mathcal{F}_t$, which could be strictly bigger.
Show that $B_t, M_t=B_t^2-t$ and $Z_t^a=\exp \left(a B_t-\frac{1}{2} a^2 t\right)$ are $\mathbb{F}$-martingales.
Let $\sigma(\omega)=\min \left(\tau_a(\omega), \tau_b(\omega)\right)$, for $a<0<b \in \mathbb{R}$. We will see in the lectures that the Doob martingale convergence theorem applies also to continuous martingales in continuous time. By following the same line of proof as in the random walk case check that $P(\sigma<\infty)=1$.
Let $a<0<b \in \mathbb{R}$. Compute $P\left(\tau_a<\tau_b\right)$.
Solution $5 \quad\left(B_t, B_t^2-t\right.$ and $Z_t^a=\exp \left(a B_t-\frac{1}{2} a^2 t\right)$ are martingales: $)$ Because $B_t$ has Gaussian distribution, it is integrable. Moreover
$$
E\left(B_t \mid \mathcal{F}s\right)=E\left(B_t-B_s+B_s \mid \mathcal{F}_s\right)=E\left(B_t-B_s\right)+B_s=B_s $$ We also have $$ E\left(B_t^2-t \mid \mathcal{F}_s\right)=E\left(\left(B_t-B_s\right)^2+2 B_t B_s-B_s^2-t \mid \mathcal{F}_s\right)=E\left(\left(B_t-B_s\right)^2\right)+2 B_s^2-B_s^2-t=B_s^2-s $$ and $$ E\left(e^{a B_t-\frac{1}{2} a^2 t} \mid \mathcal{F}_s\right)=E\left(e^{a\left(B_t-B_s\right)} e^{a B_s-\frac{1}{2} a^2 t} \mid \mathcal{F}_s\right)=E\left(e^{a\left(B_t-B_s\right)}\right) e^{a B_s-\frac{1}{2} a^2 t}=e^{a B_s-\frac{1}{2} a^2 s}, $$ because for a Gaussian random variable $X$ with mean 0 and variance $\sigma^2$, we have $$ \begin{aligned} E\left(e^{a X}\right) & =\int{-\infty}^{\infty} e^{a x} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2} \frac{x^2}{\sigma^2}} d x \
& =\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2} \frac{x^2-2 a \sigma^2 x}{\sigma^2}} d x \
& =\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2} \frac{\left(x-a \sigma^2\right)^2-a^2 \sigma^4}{\sigma^2}} d x \
& =e^{\frac{1}{2} a^2 \sigma^2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2} \frac{x^2}{\sigma^2}} d x=e^{\frac{1}{2} a^2 \sigma^2} .
\end{aligned}
$$
(We have $P(\sigma<\infty)=1$ 🙂 By Doob’s martingale convergence theorem, the martingale $B_{t \wedge \sigma}^2-(t \wedge \sigma)$ converges to $B_\sigma^2-\sigma$ almost surely. The reverse Fatou lemma gives
$$
E\left(B_\sigma^2-\sigma\right) \geq \limsup {t \rightarrow \infty} E\left(B{t \wedge \sigma}^2-(t \wedge \sigma)\right)=0
$$
and hence $E(\sigma) \leq E\left(B_\sigma^2\right)<\infty$.
(Compute $P\left(\tau_a<\tau_b\right)$ 🙂 Like in the previous exercise,
$$
\begin{aligned}
& \qquad E\left(M_\sigma\right)=E\left(M_0\right)=0=a P\left(\tau_a<\tau_b\right)+b\left(1-P\left(\tau_a<\tau_b\right)\right) \text {, } \
& \text { so } P\left(\tau_a<\tau_b\right)=\frac{b}{b-a} .
\end{aligned}
$$
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