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## 数学代写|随机微积分代写Stochastic calculus代考|Kolmogorov’s continuity criterion

There are elements of Lévy’s construction that admit interesting generalizations, perhaps the most important of which is Kolmogorov’s continuity criterion.

Remember that when dealing with an uncountable number of random variables, in general one can measure only events that depend on countably many of them. As a consequence, one can change individual members on sets of measure 0 without effecting the probability of events that can be measured. With this in mind, one says that two families of random variables are versions of one another if one family can be obtained from the other by changing individual random variables on sets of measure 0 .

In the proof of the following theorem, we use the fact that if $Q$ is a closed cube in $\mathbb{R}^N$ and, for each vertex $\mathbf{v}$ of $Q, a_{\mathbf{v}}$ is an element of a vector space $E$, then there is a unique function $f: Q \longrightarrow E$, known as the multilinear extention of the $a_{\mathbf{v}}$ ‘s, such that $f(\mathbf{v})=a_{\mathbf{v}}$ for each vertex $\mathbf{v}$ and $f$ is an affine function of each coordinate. For example, if $Q=[0,1]^2$, then
$$f\left(x_1, x_2\right)=\left(1-x_1\right)\left(1-x_2\right) a_{(0,0)}+\left(1-x_1\right) x_2 a_{(0,1)}+x_1\left(1-x_2\right) a_{(1,0)}+x_1 x_2 a_{(1,1)} .$$
The general case can be proved by translation, scaling, and induction on $N$.
Theorem 2.1.2. Suppose that $\left{X(\mathbf{x}): \mathbf{x} \in[0, R]^N\right}$ is a family of random variables taking values in a Banach space $E$, and assume that, for some $p \in[1, \infty), C<\infty$, and $r \in(0,1]$
$$\mathbb{E}\left[|X(\mathbf{y})-X(\mathbf{x})|_E^p\right]^{\frac{1}{p}} \leq C|\mathbf{y}-\mathbf{x}|^{\frac{N}{p}+r} \quad \text { for all } \mathbf{x}, \mathbf{y} \in[0, R]^N .$$
Then there exists a version $\left{\widetilde{X}(\mathbf{x}): \mathbf{x} \in[0, R]^N\right}$ of $\left{X(\mathbf{x}): \mathbf{x} \in[0, R]^N\right}$ such that $\mathbf{x} \in[0, R]^N \longmapsto \widetilde{X}(\mathbf{x})(\omega) \in E$ is continuous for all $\omega \in \Omega$. In fact, for each $\alpha \in[0, r)$, there is a $K<\infty$, depending only on $N, p, r$, and $\alpha$, such that
$$\mathbb{E}\left[\sup _{\substack{\mathbf{x}, \mathbf{y} \in[0, R]^N \ \mathbf{x} \neq \mathbf{y}}}\left(\frac{|\widetilde{X}(\mathbf{y})-\widetilde{X}(\mathbf{x})|_E}{|\mathbf{y}-\mathbf{x}|^\alpha}\right)^p\right]^{\frac{1}{p}} \leq K C R^{\frac{n}{p}+r-\alpha}$$

## 数学代写|随机微积分代写Stochastic calculus代考|Brownian motion and Wiener measure

For various reasons, it has become common to give a more flexible description of what is meant by a Brownian motion. Given a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, a filtration (i.e., a non-decreasing family) $\left{\mathcal{F}t: t \geq 0\right}$ of sub- $\sigma$-algebras, and a family ${B(t): t \geq 0}$ of $\mathbb{R}^N$-valued random variables, one says that the triple $\left(B(t), \mathcal{F}_t, \mathbb{P}\right)$ is an $\mathbb{R}^N$-valued Brownian motion if (i) $\mathbb{P}$-almost surely, $B(0)=\mathbf{0}$ and $t \rightsquigarrow B(t)$ is continuous. (ii) For each $s \geq 0, B(s)$ is $\mathcal{F}_s$-measurable, and, for $t>s, B(t)-B(s)$ is independent of $\mathcal{F}_s$ and has distribution $\gamma{0,(t-s) \mathbf{I}}$.

It should be clear that ${B(t): t \geq 0}$ is a Brownian motion in the sense that (2.1.1) holds if and only if $\left(B(t), \mathcal{F}t, \mathbb{P}\right)$ is a Brownian motion in the preceding sense with $\mathcal{F}_t=\sigma({B(\tau): \in[0, t]})$, the $\sigma$-algebra generated by the path up until time $t$. However, there are many times when the natural choice of $\mathcal{F}_t$ ‘s are larger than this choice. For example, if $N \geq 2$ and $\left(B(t), \mathcal{F}_t, \mathbb{P}\right)$ is an $\mathbb{R}^N$-valued Brownian motion, then, for each $\mathbf{e} \in \mathbb{S}^{N-1},\left((\mathbf{e}, B(t)){\mathbb{R}^N}, \mathcal{F}t, \mathbb{P}\right)$ will be an $\mathbb{R}$-valued Brownian motion even though $$\mathcal{F}_t \neq \sigma\left(\left{(\mathbf{e}, B(\tau)){\mathbb{R}^N}: \tau \in[0, t]\right}\right) .$$
Another way to think about Brownian motion is in terms of Gaussian families. Suppose that ${X(t): t \geq 0}$ is a family of centered, Gaussian, $\mathbb{R}^N$-valued random variables on $(\Omega, \mathcal{F}, \mathbb{P})$, and assume that the span of $\left{(\boldsymbol{\xi}, X(t)){\mathbb{R}^N}: t \geq 0 \& \boldsymbol{\xi} \in \mathbb{R}^N\right}$ is a centered Gaussian family. If $\mathcal{F}_t$ is the $\mathbb{P}$-completion of the $\sigma$-algebra generated by ${X(\tau): \tau \in[0, t]}$, then there is a Brownian motion $\left(B(t), \mathcal{F}_t, \mathbb{P}\right)$ such that $X(t)=B(t)$ (a.s., $\left.\mathbb{P}\right)$ for each $t \geq 0$ if and only if $\mathbb{E}^{\mathbb{P}}\left[(\boldsymbol{\xi}, X(s)){\mathbb{R}^N}(\boldsymbol{\eta}, X(t)){\mathbb{R}^N}\right]=s \wedge t(\boldsymbol{\xi}, \boldsymbol{\eta}){\mathbb{R}^N}$ for all $s, t \in[0, \infty)$ and $\boldsymbol{\xi}, \boldsymbol{\eta} \in \mathbb{R}^N$.

# 随机微积分代考

## 数学代写|随机微积分代写Stochastic calculus代考|Kolmogorov’s continuity criterion

Lévy 的构造中有一些元素可以进行有趣的概括，其中最重要的可能是 Kolmogorov 的连续性准则。

$$f\left(x_1, x_2\right)=\left(1-x_1\right)\left(1-x_2\right) a_{(0,0)}+\left(1-x_1\right) x_2 a_{(0,1)}+x_1\left(1-x_2\right) a_{(1,0)}+x_1 x_2 a_{(1,1)} .$$

$$\mathbb{E}\left[|X(\mathbf{y})-X(\mathbf{x})|E^p\right]^{\frac{1}{p}} \leq C|\mathbf{y}-\mathbf{x}|^{\frac{N}{p}+r} \quad \text { for all } \mathbf{x}, \mathbf{y} \in[0, R]^N .$$ 然后存在一个版本 \left 的分隔符缺失或无法识别 的\1eft 的分隔符缺失或无法识别 有人都是连续的 $\omega \in \Omega$. 事实上，对于每个 $\alpha \in[0, r)$ ，有一个 $K<\infty$ ，仅取决于 $N, p, r$ ，和 $\alpha$, 这样 $$\mathbb{E}\left[\sup {\mathbf{x}, \mathbf{y} \in[0, R]^N \mathbf{x} \neq \mathbf{y}}\left(\frac{|\widetilde{X}(\mathbf{y})-\widetilde{X}(\mathbf{x})|_E}{|\mathbf{y}-\mathbf{x}|^\alpha}\right)^p\right]^{\frac{1}{p}} \leq K C R^{\frac{n}{p}+r-\alpha}$$

## 数学代写|随机微积分代写Stochastic calculus代考|Brownian motion and Wiener measure

\eft 的分隔符缺失或无法识别

$\backslash$ left 的分隔符缺失或无法识别 是一个中心高斯族。如果 $\mathcal{F}_t$ 是个 $\mathbb{P}$ – 完成 $\sigma$ – 代数由 $X(\tau): \tau \in[0, t]$, 那么有一个布朗运动 $(B(t), \mathcal{F}, \mathbb{P})$ 这样 $X(t)=B(t)$ (作为， $\mathbb{P})$ 对于每个 $t \geq 0$ 当且仅当 $\mathbb{E}^{\mathbb{P}}\left[(\boldsymbol{\xi}, X(s)) \mathbb{R}^N(\boldsymbol{\eta}, X(t)) \mathbb{R}^N\right]=s \wedge t(\boldsymbol{\xi}, \boldsymbol{\eta}) \mathbb{R}^N$ 对所有人s, $t \in[0, \infty)$ 和 $\boldsymbol{\xi}, \boldsymbol{\eta} \in \mathbb{R}^N$.

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## MATLAB代写

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