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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|submartingale
First, we consider of the case that $m$ is an integer, $\mathbf{T}={0,1, \ldots, m}$ and our filtration is $\left{\mathcal{F}{n}\right}{n=0}^{m}$.
Proposition 2.2.1 Let $X=\left{X_{n}\right}_{n=0}^{m}$ be a submartingale. Then
$$
\lambda P\left(\max {k=0, \ldots, m} X{k}>\lambda\right) \leqq E\left[X_{m}, \max {k=0, \ldots, m} X{k}>\lambda\right] \leqq E\left[X_{m} \vee 0\right], \quad \lambda>0 .
$$
Proof Let $A_{k}=\bigcap_{\ell=0}^{k-1}\left{X_{\ell} \leqq \lambda\right} \cap\left{X_{k}>\lambda\right}, k=0,1, \ldots, m$. Then it is easy to see that $A_{k} \in \mathcal{F}{k}, k=0, \ldots, m$, and $A{k}$ ‘s are mutually disjoint. Also, we see that
$$
\left{\max {k=0, \ldots, m} X{k}>\lambda\right}=\bigcup_{k=0}^{m} A_{k}
$$
Then we see that
$$
\begin{aligned}
& \lambda P\left(\max {k=0, \ldots, m} X{k}>\lambda\right) \
=& \lambda \sum_{k=0}^{m} P\left(A_{k}\right) \leqq \sum_{k=0}^{m} E\left[X_{k}, A_{k}\right] \
\leqq & \sum_{k=0}^{m} E\left[E\left[X_{m} \mid \mathcal{F}{k}\right], A{k}\right]=\sum_{k=0}^{m} E\left[X_{m}, A_{k}\right]=E\left[X_{m}, \max {k=0, \ldots, m} X{k}>\lambda\right] .
\end{aligned}
$$
This implies our assertion
数学代写|随机分析作业代写stochastic analysis代考|(Doob’s inequality）
Theorem 2.2.1 (Doob’s inequality) Let $M=\left{M_{n}\right}_{n=0}^{m}$ be a martingale. Then for any $p \in(1, \infty)$
$$
E\left[\max {k=0,1, \ldots, m}\left|M{k}\right|^{p}\right]^{1 / p} \leqq \frac{p}{p-1} E\left[\left|M_{m}\right|^{p}\right]^{1 / p} .
$$
Proof Our assertion is obvious if $E\left[\left|M_{m}\right|^{p}\right]=\infty$. So we assume that $E\left[\left|M_{m}\right|^{p}\right]<$ $\infty$. Let $Z=\max {k=0,1, \ldots, m}\left|M{k}\right|$. Then by Jensen’s inequality, we see that $$ E\left[Z^{p}\right] \leqq \sum_{k=0}^{m} E\left[\left|M_{k}\right|^{p}\right]=\sum_{k=0}^{m} E\left[\left|E\left[M_{m} \mid \mathcal{F}{k}\right]\right|^{p}\right] \leqq \sum{k=0}^{m} E\left[\left|M_{m}\right|^{p}\right]<\infty $$ Let $X_{n}=\left|M_{n}\right|, n=0,1, \ldots, m$. Then by Proposition $2.1 .2$, we see that $X=$ $\left{X_{n}\right}_{n=0}^{m}$ is a submartingale. So by Proposition 2.2.1 we see that $$ \lambda P(Z>\lambda) \leqq E\left[\left|M_{m}\right|, Z>\lambda\right], \quad \lambda>0 .
$$
Then we see that
$$
\begin{aligned}
E\left[Z^{p}\right] &=E\left[\int_{0}^{\infty} 1_{[Z>\lambda}} p \lambda^{p-1} d \lambda\right]=p \int_{0}^{\infty} \lambda^{p-1} P(Z>\lambda) d \lambda \
& \leqq p \int_{0}^{\infty} \lambda^{p-2} E\left[\left|M_{m}\right|, Z>\lambda\right] d \lambda=p E\left[\left|M_{m}\right|\left(\int_{0}^{\infty} 1_{(Z>\lambda)} \lambda^{p-2} d \lambda\right)\right] \
&=\frac{p}{p-1} E\left[\left|M_{m}\right| Z^{p-1}\right] \leqq \frac{p}{p-1} E\left[\left|M_{m}\right|^{p}\right]^{1 / p} E\left[Z^{p}\right]^{(p-1) / p}
\end{aligned}
$$
Here we use Hölder’s inequality (Proposition 1.5.1). Since $E\left[Z^{p}\right]<\infty$, we have our assertion.
Now let us consider the case that $\mathbf{T}=\mathbf{Z}{\geqq 0}$ and our filtration is $\left{\mathcal{F}{n}\right}_{n=0}^{\infty}$. We say that $M=\left{M_{n}\right}_{n=0}^{\infty}$ is a square integrable martingale, if $M$ is a martingale and if $E\left[M_{n}^{2}\right]<\infty, n=0,1, \ldots .$
Proposition 2.2.2 If $M=\left{M_{n}\right}_{n=0}^{\infty}$ is a square integrable martingale, then
$$
E\left[\left(M_{n}-M_{m}\right)^{2} \mid \mathcal{F}{m}\right]=E\left[M{n}^{2} \mid \mathcal{F}{m}\right]-M{m}^{2}, \quad n \geqq m \geqq 1
$$
In particular,
$$
E\left[M_{n}^{2}\right]=E\left[M_{0}^{2}\right]+\sum_{k=1}^{n} E\left[\left(M_{k}-M_{k-1}\right)^{2}\right], \quad n \geqq 0
$$
随机分析代写
数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|SUBMARTINGALE
首先，我们考虑以下情况米是一个整数, $\mathbf{T}={0,1, \ldots, m}$ and our filtration is $\left{\mathcal{F}{n}\right}{n=0}^{m}$.
Proposition 2.2.1 Let $X=\left{X_{n}\right}_{n=0}^{m}$ be a submartingale. Then
$$
\lambda P\left(\max {k=0, \ldots, m} X{k}>\lambda\right) \leqq E\left[X_{m}, \max {k=0, \ldots, m} X{k}>\lambda\right] \leqq E\left[X_{m} \vee 0\right], \quad \lambda>0 .
$$
Proof Let $A_{k}=\bigcap_{\ell=0}^{k-1}\left{X_{\ell} \leqq \lambda\right} \cap\left{X_{k}>\lambda\right}, k=0,1, \ldots, m$. Then it is easy to see that $A_{k} \in \mathcal{F}{k}, k=0, \ldots, m$, and $A{k}$ ‘s are mutually disjoint. Also, we see that
$$
\left{\max {k=0, \ldots, m} X{k}>\lambda\right}=\bigcup_{k=0}^{m} A_{k}
$$
Then we see that
$$
\begin{aligned}
& \lambda P\left(\max {k=0, \ldots, m} X{k}>\lambda\right) \
=& \lambda \sum_{k=0}^{m} P\left(A_{k}\right) \leqq \sum_{k=0}^{m} E\left[X_{k}, A_{k}\right] \
\leqq & \sum_{k=0}^{m} E\left[E\left[X_{m} \mid \mathcal{F}{k}\right], A{k}\right]=\sum_{k=0}^{m} E\left[X_{m}, A_{k}\right]=E\left[X_{m}, \max {k=0, \ldots, m} X{k}>\lambda\right] .
\end{aligned}
$$
这意味着我们的断言
数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|(DOOB’S INEQUALITY）
Theorem 2.2.1 (Doob’s inequality) Let $M=\left{M_{n}\right}_{n=0}^{m}$ be a martingale. Then for any $p \in(1, \infty)$
$$
E\left[\max {k=0,1, \ldots, m}\left|M{k}\right|^{p}\right]^{1 / p} \leqq \frac{p}{p-1} E\left[\left|M_{m}\right|^{p}\right]^{1 / p} .
$$
Proof Our assertion is obvious if $E\left[\left|M_{m}\right|^{p}\right]=\infty$. So we assume that $E\left[\left|M_{m}\right|^{p}\right]<$ $\infty$. Let $Z=\max {k=0,1, \ldots, m}\left|M{k}\right|$. Then by Jensen’s inequality, we see that $$ E\left[Z^{p}\right] \leqq \sum_{k=0}^{m} E\left[\left|M_{k}\right|^{p}\right]=\sum_{k=0}^{m} E\left[\left|E\left[M_{m} \mid \mathcal{F}{k}\right]\right|^{p}\right] \leqq \sum{k=0}^{m} E\left[\left|M_{m}\right|^{p}\right]<\infty $$ Let $X_{n}=\left|M_{n}\right|, n=0,1, \ldots, m$. Then by Proposition $2.1 .2$, we see that $X=$ $\left{X_{n}\right}_{n=0}^{m}$ is a submartingale. So by Proposition 2.2.1 we see that $$ \lambda P(Z>\lambda) \leqq E\left[\left|M_{m}\right|, Z>\lambda\right], \quad \lambda>0 .
$$
Then we see that
$$
\begin{aligned}
E\left[Z^{p}\right] &=E\left[\int_{0}^{\infty} 1_{[Z>\lambda}} p \lambda^{p-1} d \lambda\right]=p \int_{0}^{\infty} \lambda^{p-1} P(Z>\lambda) d \lambda \
& \leqq p \int_{0}^{\infty} \lambda^{p-2} E\left[\left|M_{m}\right|, Z>\lambda\right] d \lambda=p E\left[\left|M_{m}\right|\left(\int_{0}^{\infty} 1_{(Z>\lambda)} \lambda^{p-2} d \lambda\right)\right] \
&=\frac{p}{p-1} E\left[\left|M_{m}\right| Z^{p-1}\right] \leqq \frac{p}{p-1} E\left[\left|M_{m}\right|^{p}\right]^{1 / p} E\left[Z^{p}\right]^{(p-1) / p}
\end{aligned}
$$
这里我们使用 Hölder 不等式磷r这p这s一世吨一世这n1.5.1. 自从和[从p]<∞，我们有我们的断言。
现在让我们考虑 $\mathbf{T}=\mathbf{Z}{\geqq 0}$ and our filtration is $\left{\mathcal{F}{n}\right}_{n=0}^{\infty}$. We say that $M=\left{M_{n}\right}_{n=0}^{\infty}$ is a square integrable martingale, if $M$ is a martingale and if $E\left[M_{n}^{2}\right]<\infty, n=0,1, \ldots .$
Proposition 2.2.2 If $M=\left{M_{n}\right}_{n=0}^{\infty}$ is a square integrable martingale, then
$$
E\left[\left(M_{n}-M_{m}\right)^{2} \mid \mathcal{F}{m}\right]=E\left[M{n}^{2} \mid \mathcal{F}{m}\right]-M{m}^{2}, \quad n \geqq m \geqq 1
$$
In particular,
$$
E\left[M_{n}^{2}\right]=E\left[M_{0}^{2}\right]+\sum_{k=1}^{n} E\left[\left(M_{k}-M_{k-1}\right)^{2}\right], \quad n \geqq 0
$$
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