如果你也在 怎样代写随机分析stochastic analysis这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。随机分析stochastic analysis是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。
随机分析stochastic analysis应用随机微积分的最著名的随机过程是维纳过程（为纪念诺伯特-维纳而命名），它被用来模拟路易-巴切莱特在1900年和阿尔伯特-爱因斯坦在1905年描述的布朗运动以及其他受随机力作用的粒子在空间的物理扩散过程。自20世纪70年代以来，维纳过程被广泛地应用于金融数学和经济学中，以模拟股票价格和债券利率的时间演变。
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数学代写|随机分析作业代写stochastic analysis代考|probability space
Let $\mathbf{T}$ be an ordered set. We only consider the case that $\mathbf{T}$ is a subset of $[-\infty, \infty]$ in this book. In almost all cases, $\mathbf{T}$ is ${0,1,2, \ldots, m}, m \geqq 1$, or $\mathbf{Z}{\geq 0}={0,1,2, \ldots}$, or $[0, T], T>0$, or $[0, \infty)$, although we consider the case that $\mathbf{T}$ is $\mathbf{Z}{\leqq 0}={n \in$ $\mathbf{Z} ; n \leqq 0}$.
Let $(\Omega, \mathcal{F}, P)$ be a probability space.
Definition 2.1.1 We say that $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ is a filtration, if the following two conditions are satisfied.
(1) $\mathcal{F}{t}, t \in \mathbf{T}$ are sub- $\sigma$-algebras. (2) $\mathcal{F}{s} \subset \mathcal{F}{t}$ for any $s, t \in \mathbf{T}$ with $s{t}\right}_{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingale, if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right]=X{s}
$$
Definition 2.1.3 Let $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ submartingale (supermartingale), if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right] \geqq(\leqq) X{s}
$$
In the case that it is obvious what filtration $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ we handle, we call $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingales (sub-, supermartingales) martingales (sub-, supermartingales) for simplicity.
The following is obvious.
数学代写|随机分析作业代写stochastic analysis代考|martingale
Proposition 2.1.1 (1) If $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are martingales, and if $a, b \in \mathbf{R}$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is a martingale.
(2) If $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are submartingales (supermartingales), and if $a, b \geqq 0$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is also a submartingale (supermartingale).
Also, we have the following.
Proposition 2.1.2 (1) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a martingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.
(2) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a submartingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a non-decreasing convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.
Proof In both cases, by Jensen’s inequality we see that for any $t, s \in \mathbf{T}$ with $s<t$
$$
E\left[\varphi\left(X_{t}\right) \mid \mathcal{F}{s}\right] \geqq \varphi\left(E\left[X{t} \mid \mathcal{F}{s}\right]\right) \geqq \varphi\left(X{s}\right)
$$
Therefore we have our assertions.
随机分析代写
数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|PROBABILITY SPACE
让吨是一个有序集。我们只考虑这样的情况吨是的一个子集[−∞,∞]在这本书中。在几乎所有情况下，$\mathbf{T}$ is a subset of $[-\infty, \infty]$ in this book. In almost all cases, $\mathbf{T}$ is ${0,1,2, \ldots, m}, m \geqq 1$, or $\mathbf{Z}{\geq 0}={0,1,2, \ldots}$, or $[0, T], T>0$, or $[0, \infty)$, although we consider the case that $\mathbf{T}$ is $\mathbf{Z}{\leqq 0}={n \in$ $\mathbf{Z} ; n \leqq 0}$.
Let $(\Omega, \mathcal{F}, P)$ be a probability space.
Definition 2.1.1 We say that $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ is a filtration, if the following two conditions are satisfied.
(1) $\mathcal{F}{t}, t \in \mathbf{T}$ are sub- $\sigma$-algebras. (2) $\mathcal{F}{s} \subset \mathcal{F}{t}$ for any $s, t \in \mathbf{T}$ with $s{t}\right}_{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingale, if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right]=X{s}
$$
Definition 2.1.3 Let $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ submartingale (supermartingale), if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right] \geqq(\leqq) X{s}
$$
In the case that it is obvious what filtration $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ we handle, we call $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingales (sub-, supermartingales) martingales (sub-, supermartingales) 为简单起见。
以下是显而易见的。
数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|MARTINGALE
命题 2.1.11如果$X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are martingales, and if $a, b \in \mathbf{R}$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is a martingale.。
2如果$X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are submartingales (supermartingales), and if $a, b \geqq 0$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is also a submartingale (supermartingale).
Also, we have the following.
此外，我们还有以下内容。
命题 2.1.21(1) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a martingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.
(2) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a submartingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a non-decreasing convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.
证明 在这两种情况下，根据 Jensen 不等式，我们看到对于任何$t, s \in \mathbf{T}$ with $s<t$
$$
E\left[\varphi\left(X_{t}\right) \mid \mathcal{F}{s}\right] \geqq \varphi\left(E\left[X{t} \mid \mathcal{F}{s}\right]\right) \geqq \varphi\left(X{s}\right)
$$
因此我们有我们的断言。
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