# 微积分网课代修|极限理论代写Limit Theory代考|MATH6710 Martingales

• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学

## 微积分网课代修|极限理论代写Limit Theory代考|Martingales

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $\mathbb{F}=\left(\mathcal{F}{k}\right){k \geq 0}$ a filtration, that is, a nondecreasing sequence of sub- $\sigma$-fields of $\mathcal{F}$. Set $\mathcal{F}{\infty}:=\sigma\left(\bigcup{k=0}^{\infty} \mathcal{F}{k}\right)$. A sequence $\left(X{k}\right){k \geq 1}$ of random variables on $(\Omega, \mathcal{F}, P)$ is called adapted to $\mathbb{F}$ if $X{k}$ is measurable w.r.t. $\mathcal{F}{k}$ for every $k \in \mathbb{N}$, and a sequence $\left(X{k}\right)_{k \geq 1}$ of integrable random variables adapted to $\mathbb{F}$ is called a martingale difference sequence w.r.t. $\mathbb{F}$, if $E\left(X_{k} \mid \mathcal{F}_{k-1}\right)=0$ for all $k \in \mathbb{N}$.

Let $\left(X_{k}\right){k \geq 1}$ be a martingale difference sequence w.r.t. the filtration $\mathbb{F}$, and let $\left(a{n}\right){n \geq 1}$ be a sequence of positive real numbers. Then $$X{n k}:=\frac{1}{a_{n}} X_{k} \quad \text { for } 1 \leq k \leq n \quad \text { and } \quad \mathcal{F}{n k}:=\mathcal{F}{k} \quad \text { for } 0 \leq k \leq n, n \in \mathbb{N}$$
defines a martingale difference array $\left(X_{n k}\right){1 \leq k \leq n, n \in \mathbb{N}}$ w.r.t. $\left(\mathcal{F}{n k}\right){0 \leq k \leq n, n \in \mathbb{N}}$, and the $\sigma$-fields are nested because $\mathcal{F}{n+1, k}=\mathcal{F}{k}=\mathcal{F}{n k}$ for all $n \in \mathbb{N}$ and $0 \leq k \leq n$. Therefore, Theorem $6.1$ and the sufficient conditions of Sect. $6.3$ can be applied with $\mathcal{G}=\mathcal{F}{\infty}$ and yield stable central limit theorems for the normalized partial sums $a{n}^{-1} \sum_{k=1}^{n} X_{k}$ of $\left(X_{k}\right)_{k \geq 1}$ under appropriate moment conditions. For ease of reference we explicitly formulate here the two sets of sufficient conditions for martingale difference sequences that will be applied later on.

## 微积分网课代修|极限理论代写Limit Theory代考|A Continuous Time Version

We finally present a continuous-time version of Theorem $6.23$ and Corollary $6.24$ for path-continuous (local) martingales. Its proof is obtained by using the associated Dambis-Dubins-Schwarz Brownian motion.

Theorem 6.31 Let $M=\left(M_{t}\right){t \geq 0}$ be a path-continuous local $\mathbb{F}$-martingale, where $\mathbb{F}=\left(\mathcal{F}{t}\right){t \geq 0}$ denotes a right-continuous filtration in $\mathcal{F}$, and let $a:(0, \infty) \rightarrow(0, \infty)$ be a nondecreasing function with $a(t) \rightarrow \infty$ as $t \rightarrow \infty$. Assume for the (continuous) quadratic characteristic $$\frac{\langle M\rangle{t}}{a(t)^{2}} \rightarrow \eta^{2} \quad \text { in probability as } t \rightarrow \infty$$
for some $\mathbb{R}{+}$-valued random variable $\eta$. Then $$\frac{M{t}}{a(t)} \rightarrow N\left(0, \eta^{2}\right) \quad \text { stably as } t \rightarrow \infty$$
and if $P\left(\eta^{2}>0\right)>0$,
$$\frac{M_{t}}{\langle M\rangle_{t}^{1 / 2}} \rightarrow N(0,1) \quad \text { mixing under } P_{\left{\eta^{2}>0\right}} \text { as } t \rightarrow \infty$$
$$\left(M_{t} / 0:=0 .\right)$$

Proof Since $\left\langle M-M_{0}\right\rangle=\langle M\rangle$, we may assume $M_{0}=0$. Let $\left(s_{n}\right){n \geq 1}$ be an arbitrary sequence in $(0, \infty)$ with $s{n} \uparrow \infty$. The assertions reduce to
$$\frac{M_{s_{n}}}{a\left(s_{n}\right)} \rightarrow N\left(0, \eta^{2}\right) \quad \text { stably as } n \rightarrow \infty$$
and
$$\frac{M_{s_{n}}}{\langle M\rangle_{s_{n}}^{1 / 2}} \rightarrow N(0,1) \quad \text { mixing under } P_{\left{\eta^{2}>0\right}} \text { as } n \rightarrow \infty$$

## 微积分网课代修|极限理论代写Limit Theory代考| Martingales

$\mathcal{F} \infty:=\sigma\left(\bigcup k=0^{\infty} \mathcal{F} k\right)$. 序列 $(X k) k \geq 1$ 上的随机变量 $(\Omega, \mathcal{F}, P)$ 称为适应 $\mathbb{F}$ 如

$X n k:=\frac{1}{a_{n}} X_{k} \quad$ for $1 \leq k \leq n \quad$ and $\quad \mathcal{F} n k:=\mathcal{F} k \quad$ for $0 \leq k \leq n, n \in \mathbb{N}$

，以及 $\sigma$-字段嵌套，因为 $\mathcal{F} n+1, k=\mathcal{F} k=\mathcal{F} n k$ 面向所有人 $n \in \mathbb{N}$ 和
$0 \leq k \leq n$. 因此，定理 $6.1$ 以及第三节的充分条件。 $6.3$ 可与 $\mathcal{G}=\mathcal{F} \infty$ 并产生归一化

## 微积分网课代修|极限理论代写Limit Theory代考| A Continuous Time Version

$\frac{\langle M\rangle t}{a(t)^{2}} \rightarrow \eta^{2} \quad$ in probability as $t \rightarrow \infty$

$\frac{M t}{a(t)} \rightarrow N\left(0, \eta^{2}\right) \quad$ stably as $t \rightarrow \infty$

$\backslash$ frac $\left{M_{-}{t}\right}{\backslash$ langle M\rangle_{t $\left.} \wedge{1 / 2}\right} \backslash$ rightarrow $N(0,1)$ lquad \text ${$ mixing under $} P_{-}{\backslash l e f t{\backslash$ leta^ ${2}>0 \backslash$ right $}$
$\left(M_{t} / 0:=0 .\right)$

$$\frac{M_{s_{n}}}{a\left(s_{n}\right)} \rightarrow N\left(0, \eta^{2}\right) \quad \text { stably as } n \rightarrow \infty$$