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# 数学代写|随机分析代写Stochastic Analysis in Finance代考|MA547 Brownian Motion

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## 数学代写|随机分析代写Stochastic Analysis in Finance代考|Brownian Motion

We say that a probability measure $\mu$ on $(\mathbf{R}, \mathcal{B}(\mathbf{R})$ ) is a Gaussian distribution (of mean $m$ and variance $v, m \in \mathbf{R}, v \geqq 0$ ), if the characteristic function of $\mu$ is given by
$$\int_{\mathbf{R}} \exp (\sqrt{-1} \xi x) \mu(d x)=\exp \left(\sqrt{-1} m \xi-\frac{v}{2} \xi^2\right), \quad \xi \in \mathbf{R}$$
When $v>0$, the Gaussian distribution of mean $m$ and variance $v$ is a normal distribution and its mean is $m$ and its variance is $v$.
If $\mu$ is the Gaussian distribution of mean $m$ and variance 0 , then $\mu({m})=1$.
Definition 3.7.1 Let $\mathcal{X}$ be a family of random variables. We say that $\mathcal{X}$ is a Gaussian system, if the following is valid.

For any $n \geqq 1, a_1, \ldots, a_n \in \mathbf{R}$ and $X_1, \ldots, X_n \in \mathcal{X}$, the probability law of $a_1 X_1+\cdots+a_n X_n$ is a Gaussian distribution.

Proposition 3.7.1 Suppose that $X_n, n \in \mathbf{Z}{\geqq 0}$, are independent random variables and that each probability law of $X_n$ is a Gaussian distribution. Let $\tilde{\mathcal{H}}$ be the set of all linear combinations of $X_n, n \in \mathbf{Z}{\geqq 0}$, and let $\mathcal{H}$ be the set of all random variables which is given by $\mathcal{L}^2$-limit of sequences of random variables in $\tilde{\mathcal{H}}$. Then $\mathcal{H}$ is a Gaussian system.

Proof Suppose that the mean and the variance of $X_n, n=1,2, \ldots$, are $m_n$ and $v_n$ respectively. Then for any $a_n \in \mathbf{R}, n=1,2, \ldots$,
\begin{aligned} E\left[\exp \left(\sqrt{-1} \xi\left(\sum_{k=1}^n a_k X_k\right)\right)\right] &=\prod_{k=1}^n E\left[\exp \left(\sqrt{-1} \xi a_k X_k\right)\right] \ &=\exp \left(\sqrt{-1}\left(\sum_{k=1}^n a_k m_k\right) \xi-\frac{1}{2}\left(\sum_{k=1}^n a_k^2 v_k\right) \xi^2\right) . \end{aligned}
Therefore the probability law of $\sum_{k=1}^n a_k X_k$ is a Gaussian distribution.

## 数学代写|随机分析代写Stochastic Analysis in Finance代考|Optimal Stopping Time

Let $\left(\Omega, \mathcal{F}, P,\left{\mathcal{F}t\right}{t \in[0, \infty)}\right)$ be a standard filtered probability space.
Let $X:[0, \infty) \times \Omega \rightarrow \mathbf{R}$ be an adapted continuous process such that $E\left[\sup _{t \in[0, \infty)}\right.$ $\left.X(t)^2\right]<\infty$ For any stopping times $\sigma_i: \Omega \rightarrow[0, \infty], i=0,1$, such that $\sigma_0 \leqq \sigma_1$, we denote $\mathcal{S}{\sigma_0}^{\sigma_1}$ be the set of all stopping times $\tau: \Omega \rightarrow[0, \infty]$ for which $\sigma_0 \leqq \tau \leqq \sigma_1$. Let us take a $T \in(0, \infty)$ and fix it. Proposition 3.8.1 Suppose that $\sigma \in \mathcal{S}_0^T$. Then for any $\tau_1, \tau_2 \in \mathcal{S}\sigma^T$ there is a $\tau_3 \in \mathcal{S}\sigma^T$ such that $$E\left[X{\tau_1} \mid \mathcal{F}\sigma\right] \vee E\left[X{\tau_2} \mid \mathcal{F}\sigma\right] \leqq E\left[X{\tau_3} \mid \mathcal{F}\sigma\right] \text { a.s. }$$ Proof Let $\tau_1, \tau_2 \in \mathcal{S}\sigma^T$. Let $A=\left{E\left[X_{\tau_1} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right] \geqq E\left[X{\tau_2} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right]\right}$. Then we see that $A \in \mathcal{F}{\tau_1 \wedge \tau_2}$. Let $\tau_3: \Omega \rightarrow[0, T]$ be given by
$$\tau_3(\omega)=\left{\begin{array}{l} \tau_1(\omega), \text { if } \omega \in A, \ \tau_2(\omega), \text { otherwise } \end{array}\right.$$
Then we see that $\sigma \leqq \tau_1 \wedge \tau_2 \leqq \tau_3 \leqq T$ and that
$$\left{\tau_3<t\right}=\left(\left{\tau_1 \wedge \tau_2<t\right} \cap A \cap\left{\tau_1<t\right}\right) \cup\left(\left{\tau_1 \wedge \tau_2<t\right} \cap(\Omega \backslash A) \cap\left{\tau_2<t\right}\right) \in \mathcal{F}t .$$ So we see that $\tau_3 \in \mathcal{S}\sigma^T$. Also, we see that
\begin{aligned} & E\left[X_{\tau_3} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right]=E\left[1_A X{\tau_1}+\left(1-1_A\right) X_{\tau_2} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right] \ =& 1_A E\left[X{\tau_1} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right]+\left(1-1_A\right) E\left[X{\tau_2} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right]=E\left[X{\tau_1} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right] \vee E\left[X{\tau_2} \mid \mathcal{F}{\tau_1 \wedge \tau_2}\right] . \end{aligned} So we have our assertion. For any $\sigma \in \mathcal{S}_0^T$ let $\tilde{U}\sigma$ be defined by
$$\tilde{U}\sigma=\operatorname{ess.sup}\left{E\left[X\tau \mid \mathcal{F}\sigma\right] ; \tau \in \mathcal{S}\sigma^T\right} .$$
Since $\left|\tilde{U}\sigma\right| \leqq E\left[\sup {[0, T]} \mid X(t) | \mathcal{F}\sigma\right]$, we see by Proposition $2.6 .5$ that $\left{\tilde{U}\sigma ; \sigma \in \mathcal{S}_0^T\right}$ is uniformly integrable.

## 数学代写|随机分析代写Stochastic Analysis in Finance代考|Brownian Motion

$$\int_{\mathbf{R}} \exp (\sqrt{-1} \xi x) \mu(d x)=\exp \left(\sqrt{-1} m \xi-\frac{v}{2} \xi^2\right), \quad \xi \in \mathbf{R}$$

$$E\left[\exp \left(\sqrt{-1} \xi\left(\sum_{k=1}^n a_k X_k\right)\right)\right]=\prod_{k=1}^n E\left[\exp \left(\sqrt{-1} \xi a_k X_k\right)\right] \quad=\exp \left(\sqrt{-1}\left(\sum_{k=1}^n a_k m_k\right) \xi-\frac{1}{2}\left(\sum_{k=1}^n a_k^2 v_k\right) \xi^2\right)$$

## 数学代写|随机分析代写Stochastic Analysis in Finance代考|Optimal Stopping Time

$\$ \$$|tau_3( \omega) =\mid left { 给出 \tau_1(\omega), if \omega \in A, \tau_2(\omega), otherwise \正确的。 Thenweseethat \ \sigma \leqq \tau_1 \wedge \tau_2 \leqq \tau_3 \leqq T \$$ andthat
|in $\mid$ mathcal ${F} t$ 。
Soweseethat $\$ \tau_3 \in \mathcal{S} \sigma^T \$$. Also, weseethat$$
E\left[X_{\tau_3} \mid \mathcal{F}{\tau_1} \wedge \tau_2\right]=E\left[1_A X \tau_1+\left(1-1_A\right) X{\tau_2} \mid \mathcal{F}_{\tau_1} \wedge \tau_2\right]=\quad 1_A E\left[X \tau_1 \mid \mathcal{F} \tau_1 \wedge \tau_2\right]+\left(1-1_A\right) E\left[X \tau_2 \mid \mathcal{F} \tau_1 \wedge \tau_2\right]=E\left[X \tau_1 \mid \mathcal{F} \tau_1 \wedge \tau_2\right] \vee E\left[X \tau_2 \mid\right.
$$Sowehaveourassertion. Forany \ \sigma \in \mathcal{S}_0^T \$$ let $\$ \bar{U} \sigma \$$bedefinedby |tilde {U} \backslash sigma = \operatorname{ess.sup } \backslash left {E \backslash left [X \backslash tau \backslash mid \backslash mathcal {F} \backslash sigma \mid right ]; \mid tau \backslash in \mid \operatorname{mathcal}{\mathrm{S}} \backslash sigma^T \backslash right } 。 \ \$$

## MATLAB代写

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