统计代写|概率论代写Probability theory代考|STAT4528

统计代写|概率论代写Probability theory代考|Discrete Martingale Representation Theorem and the CRR Model

By virtue of the stochastic integral, we have transformed a martingale $X$ via a gambling strategy $H$ into a new martingale $H \cdot X$. Let us change the perspective and ask: For fixed $X$, which are the martingales $Y$ (with $Y_0=0$ ) that can be obtained as discrete stochastic integrals of $X$ with a suitable gambling strategy $H=H(Y)$ ? Possibly all martingales $Y$ ? This is not the case, in general, as the example below indicates. However, we will see that all martingales can be represented as stochastic integrals if the increments $X_{n+1}-X_n$ can take only two values (given $X_1, \ldots, X_n$ ). In this case, we give a representation theorem and use it to discuss the fair price for a European call option in the stock market model of CoxRoss-Rubinstein. This model is rather simple and describes an idealized market (no transaction costs, fractional numbers of stocks tradeable and so on). For extensive literature on stochastic aspects of mathematical finance, we refer to the textbooks $[9,42,48,57,86,102,121]$ or $[160]$.

Example $9.41$ Consider the very simple martingale $X=\left(X_n\right)_{n=0,1}$ with only two time points. Let $X_0-0$ almost surely and $\mathbf{P}\left[X_1–1\right]-\mathbf{P}\left[X_1-0\right]-\mathbf{P}\left[X_1-\right.$ $1]=\frac{1}{3}$. Let $Y_0=0$. Further, let $Y_1=2$ if $X_1=1$ and $Y_1=-1$ otherwise. Then $Y$ is manifestly a $\sigma(X)$-martingale. However, there is no number $H_1$ such that $H_1 X_1=Y_1$

Let $T \in \mathbb{N}$ be a fixed time. If $\left(Y_n\right)_{n=0,1, \ldots, T}$ is an $\mathbb{F}$-martingale, then $Y_n=$ $\mathbf{E}\left[Y_T \mid \mathcal{F}_n\right]$ for all $n \leq T$. An $\mathbb{F}$-martingale $Y$ is thus determined uniquely by the terminal values $Y_T$ (and vice versa). Let $X$ be a martingale. As $(H \cdot X)$ is a martingale, the representation problem for martingales is thus reduced to the problem of representing an integrable random variable $V:=Y_T$ as $v_0+(H \cdot X)_T$, where $v_0=\mathbf{E}\left\lceil Y_T\right]$

We saw that, in general, this is not possible if the differences $X_{n+1}-X_n$ take three (or more) different values. Hence we now consider the case where only two values are possible. Here, at each time step, a system of two linear equations with two unknowns has to be solved. In the case where $X_{n+1}-X_n$ takes three values, the system has three equations and is thus overdetermined.

统计代写|概率论代写Probability theory代考|Doob Decomposition and Square Variation

Let $X=\left(X_n\right){n \in \mathbb{N}_0}$ be an adapted process with $\mathbf{E}\left[\left|X_n\right|\right]<\infty$ for all $n \in \mathbb{N}_0$. We will decompose $X$ into a sum consisting of a martingale and a predictable process. To this end, for $n \in \mathbb{N}_0$, define $$M_n:=X_0+\sum{k=1}^n\left(X_k-\mathbf{E}\left[X_k \mid \mathcal{F}{k-1}\right]\right)$$ and $$A_n:=\sum{k=1}^n\left(\mathbf{E}\left[X_k \mid \mathcal{F}{k-1}\right]-X{k-1}\right)$$
Evidently, $X_n=M_n+A_n$. By construction, $A$ is predictable with $A_0=0$, and $M$ is a martingale since
$$\mathbf{E}\left[M_n-M_{n-1} \mid \mathcal{F}{n-1}\right]=\mathbf{E}\left[X_n-\mathbf{E}\left[X_n \mid \mathcal{F}{n-1}\right] \mid \mathcal{F}_{n-1}\right]=0$$

Theorem 10.1 (Doob decomposition) Let $X=\left(X_n\right)_{n \in \mathbb{N}_0}$ be an adapted integrable process. Then there exists a unique decomposition $X=M+A$, where $A$ is predictable with $A_0=0$ and $M$ is a martingale. This representation of $X$ is called the Doob decomposition. $X$ is a submartingale if and only if $A$ is monotone increasing.

Proof We only have to show uniqueness of the decomposition. Hence, let $X=$ $M+A=M^{\prime}+A^{\prime}$ be two such decompositions. Then $M-M^{\prime}=A^{\prime}-A$ is a predictable martingale; hence (see Exercise 9.2.2) $M_n-M_n^{\prime}=M_0-M_0^{\prime}=0$ for all $n \in \mathbb{N}_0$.

Example 10.2 Let $I=\mathbb{N}0$ or $I={0, \ldots, N}$. Let $\left(X_n\right){n \in I}$ be a square integrable $\mathbb{F}$-martingale (that is, $\mathbf{E}\left[X_n^2\right]<\infty$ for all $\left.n \in I\right)$. By Theorem $9.35, Y:=\left(X_n^2\right){n \in I}$ is a submartingale. Let $Y=M+A$ be the Doob decomposition of $Y$. Then $\left(X_n^2-\right.$ $\left.A_n\right){n \in I}$ is a martingale. Furthermore, $\mathbf{E}\left[X_{i-1} X_i \mid \mathcal{F}{i-1}\right]=X{i-1} \mathbf{E}\left[X_i \mid \mathcal{F}{i-1}\right]=$ $X{i-1}^2$; hence (as in (10.1))
\begin{aligned} A_n &=\sum_{i=1}^n\left(\mathbf{E}\left[X_i^2 \mid \mathcal{F}{i-1}\right]-X{i-1}^2\right) \ &=\sum_{i=1}^n\left(\mathbf{E}\left[\left(X_i-X_{i-1}\right)^2 \mid \mathcal{F}{i-1}\right]-2 X{i-1}^2+2 \mathbf{E}\left[X_{i-1} X_i \mid \mathcal{F}{i-1}\right]\right) \ &=\sum{i=1}^n \mathbf{E}\left[\left(X_i-X_{i-1}\right)^2 \mid \mathcal{F}{i-1}\right] . \quad \diamond \end{aligned} Definition 10.3 Let $\left(X_n\right){n \in I}$ be a square integrable $\mathbb{F}$-martingale. The unique predictable process $A$ for which $\left(X_n^2-A_n\right){n \in I}$ becomes a martingale is called the square variation process of $X$ and is denoted by $\left(\langle X\rangle_n\right){n \in I}:=A$.

概率论代考

统计代写|概率论代写概率论代考|Doob分解与平方变分

.

$$\mathbf{E}\left[M_n-M_{n-1} \mid \mathcal{F}{n-1}\right]=\mathbf{E}\left[X_n-\mathbf{E}\left[X_n \mid \mathcal{F}{n-1}\right] \mid \mathcal{F}_{n-1}\right]=0$$

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: