# 随机分析入门|MATH5975 Introduction to Stochastic Analysis代写 unsw代写

In this course, you will learn the basic concepts and techniques of Stochastic Analysis, such as: Brownian motion, martingales, Itˆo stochastic integral, Itˆo’s formula,
stochastic differential equations, equivalent change of a probability measure, integral representation of martingales with respect to a Brownian filtration, relations
to second order partial differential equations, the Feynman-Kac formula, and jump
processes.

Let $\left(X_{n}\right){n \in \mathbb{N}}$ be a submartingale w.r.t. the filtration $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ such that $$\forall n \in \mathbb{N}, \exists k \geq n: \mathbb{E}\left(X{k}\right) \leq \mathbb{E}\left(X_{n}\right)$$
Show that $\left(X_{n}\right)_{n \in \mathbb{N}}$ is a martingale!

We consider the Doob composition $X_{n}=M_{n}+A_{n}$ with $M_{n}$ a martingale and $A_{n} \geq 0$ an increasing predictable process. It follows
$$\mathbb{E}\left[X_{n}\right]=\underbrace{\mathbb{E}\left[M_{n}\right]}{=\mathbb{E}\left[M{n+1}\right]}+\underbrace{\mathbb{E}\left[A_{n}\right]}{\geq \mathbb{E}\left[A{n-1}\right]} \geq \mathbb{E}\left[X_{n-1}\right]$$
Hence, the sequence of $\left(\mathbb{E}\left[X_{n}\right]\right){n \in \mathbb{N}}$ is an increasing sequence of real numbers. On the other hand, by $$\forall n \in \mathbb{N}, \exists k \geq n: \mathbb{E}\left(X{k}\right) \leq \mathbb{E}\left(X_{n}\right)$$
we can a extract a decreasing a subsequence $\left(\mathbb{E}\left[X_{n_{k}}\right]\right){k \in \mathbb{N}}$. Hence, the whole sequence $\left(\mathbb{E}\left[X{n}\right]\right){n \in \mathbb{N}}$ must be constant. By (1), this is only possible if $\mathbb{E}\left[A{n}\right]=0$ for all $n$. Since $A_{n} \geq 0$, it follows $A_{n}=0$. Hence, $X_{n}=M_{n}$ is a martingale.

## MATH5975 COURSE NOTES ：

Exercise ：Let $\tau_{1}(\omega)$ and $\tau_{2}(\omega)$ be stopping times with respect to the filtration $\mathbb{F}=\left(F_{1}: t \in T\right)$ taking values in $T$. Here $T$ could be either $\mathbb{R}^{+}$or $\mathbb{N}$.

Use the definition of stopping time to show that $\sigma(\omega)=\min \left(\tau_{1}(\omega), \tau_{2}(\omega)\right)$ is a F-stopping time.
Solution ：We have
$${\sigma(\omega) \leq t}=\left{\omega: \tau_{1}(\omega) \leq t \text { or } \tau_{2}(\omega) \leq t\right}=\left{\tau_{1}(\omega) \leq t\right} \cup\left{\tau_{2}(\omega) \leq t\right} \in \mathcal{F}_{t},$$
so $\sigma$ is a stopping time.