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# 数学代写|随机分析作业代写stochastic analysis代考|Doob’s Inequality

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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,
$$E\left[M_{n}^{2}\right]=E\left[M_{0}^{2}\right]+\sum_{k=1}^{n} E\left[\left(M_{k}-M_{k-1}\right)^{2}\right], \quad n \geqq 0$$

## 数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|SUBMARTINGALE

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}

## 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}

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,
$$E\left[M_{n}^{2}\right]=E\left[M_{0}^{2}\right]+\sum_{k=1}^{n} E\left[\left(M_{k}-M_{k-1}\right)^{2}\right], \quad n \geqq 0$$