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数学代写|随机分析作业代写stochastic analysis代考|Representation of Continuous Local Martingale

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|Representation of Continuous Local Martingale

数学代写|随机分析作业代写stochastic analysis代考|progressively measurable processes

Theorem 5.2.1 Let $M_{i} \in \mathcal{M}{l o c}^{c}, i=1, \ldots, N$, and $B:[0, \infty) \times \Omega \rightarrow \mathbf{R}^{d}$ be a ddimensional $\left{\mathcal{F}{t}\right}_{t \in[0, \infty)}$-Brownian motion. Assume that there are progressively measurable processes $\Phi_{i k}:[0, \infty) \times \Omega \rightarrow \mathbf{R}, i=1, \ldots, N, k=1, \ldots, d$, such that
$$
\begin{aligned}
&P\left(\sum_{i=1}^{N} \sum_{k=1}^{d} \int_{0}^{T} \Phi_{i k}(t)^{2} d t<\infty\right)=1, \quad T>0, \
&\left\langle M_{i}, M_{j}\right\rangle(t)=\sum_{k=1}^{d} \int_{0}^{t} \Phi_{i k}(s) \Phi_{j k}(s) d s, \quad i, j=1, \ldots, N,
\end{aligned}
$$
and
$$
\left\langle M_{i}, B^{k}\right\rangle=0, \quad i=1, \ldots, N, k=1, \ldots, d
$$
Then there is a d-dimensional $\left{\mathcal{F}{t}\right}{t \in[0, \infty)}$-Brownian motion $\tilde{B}:[0, \infty) \times \Omega \rightarrow \mathbf{R}^{d}$ such that
$$
M_{i}(t)=\sum_{k=1}^{d} \int_{0}^{t} \Phi_{i k}(s) d \tilde{B}^{k}(s), \quad t \in[0, \infty), i=1, \ldots, N
$$

数学代写|随机分析作业代写stochastic analysis代考|progressively measurable processes

Then we see that $\left\langle N^{k}, B^{\ell}\right\rangle=0, k, \ell=1, \ldots, d$.
Let $A_{k \ell}, k, \ell=1, \ldots, d$, be progressively measurable processes given by
$$
\left(A_{k \ell}(t, \omega)\right){k, \ell=1, \ldots, d}=I{d}-\psi_{N, d}(\Phi(t, \omega)) \Phi(t, \omega)
$$
Also, let $\tilde{B} \in \mathcal{M}{\text {loc }}^{c}, k=1, \ldots, d$, be given by $$ \tilde{B}^{k}(t)=N^{k}(t)+\sum{\ell=1}^{d} \int_{0}^{t} A_{k \ell}(s) d B^{\ell}(s), \quad t \in[0, \infty)
$$
Then we see that
$$
\begin{aligned}
\left\langle\tilde{B}^{k}, \tilde{B}^{\ell}\right\rangle(t) &=\left\langle N^{k}, N^{\ell}\right\rangle(t)+\sum_{r=1}^{d} \int_{0}^{t} A_{k r}(s) A_{\ell r}(s) d s \
&=\sum_{i, j}^{N} \sum_{r=1}^{d} \int_{0}^{t} \Psi_{k, i}(s) \Psi_{\ell, j}(s) \Phi_{i, r}(s) \Phi_{j, r}(s) d s+\sum_{r=1}^{d} \int_{0}^{t} A_{k r}(s) A_{\ell r}(s) d s .
\end{aligned}
$$
Noting that $\psi_{N, d}(\Phi(s)) \Phi(s)$ is an orthogonal projection matrix, we see that
$$
\sum_{i, j}^{N} \sum_{r=1}^{d} \Psi_{k, i}(s) \Psi_{\ell, j}(s) \Phi_{i, r}(s) \Phi_{j, r}(s)+\sum_{r=1}^{d} A_{k r}(s) A_{\ell r}(s)=\delta_{k \ell}, \quad k, \ell=1, \ldots, d
$$
Therefore we see that $\tilde{B}$ is a $d$-dimensional $\left{\mathcal{F}{t}\right}{t \in[0, \infty)}$-Brownian motion. Let
$$
\tilde{M}{i}(t)=\sum{k=1}^{d} \int_{0}^{t} \Phi_{i, k}(s) d \tilde{B}^{k}(s)
$$
Then we see that
$$
\tilde{M}{i}(t)=\sum{k=1}^{d} \sum_{j=1}^{N} \int_{0}^{t} \Phi_{i, k}(s) \Psi_{k, j}(s) d M_{j}(s)+\sum_{k, \ell=1}^{d} \int_{0}^{t} \Phi_{i, k}(s) A_{k \ell}(s) d B^{\ell}(s) .
$$
Since $\Phi(s) \psi_{N, d}(\Phi(s)) \Phi(s)=\Phi(s)$, we see that
$$
\tilde{M}{i}(t)=\sum{j=1}^{N} \int_{0}^{t}\left(\Phi(s) \psi_{N, d}(\Phi(s))\right){i, j}(s) d M{j}(s)
$$

数学代写|随机分析作业代写stochastic analysis代考|Representation of Continuous Local Martingale

随机分析代写

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|PROGRESSIVELY MEASURABLE PROCESSES

定理 5.2.1 Let $M_{i} \in \mathcal{M}{l o c}^{c}, i=1, \ldots, N$, and $B:[0, \infty) \times \Omega \rightarrow \mathbf{R}^{d}$ be a ddimensional $\left{\mathcal{F}{t}\right}_{t \in[0, \infty)}$-Brownian motion. Assume that there are progressively measurable processes $\Phi_{i k}:[0, \infty) \times \Omega \rightarrow \mathbf{R}, i=1, \ldots, N, k=1, \ldots, d$, such that
$$
\begin{aligned}
&P\left(\sum_{i=1}^{N} \sum_{k=1}^{d} \int_{0}^{T} \Phi_{i k}(t)^{2} d t<\infty\right)=1, \quad T>0, \
&\left\langle M_{i}, M_{j}\right\rangle(t)=\sum_{k=1}^{d} \int_{0}^{t} \Phi_{i k}(s) \Phi_{j k}(s) d s, \quad i, j=1, \ldots, N,
\end{aligned}
$$
and
$$
\left\langle M_{i}, B^{k}\right\rangle=0, \quad i=1, \ldots, N, k=1, \ldots, d
$$
Then there is a d-dimensional $\left{\mathcal{F}{t}\right}{t \in[0, \infty)}$-Brownian motion $\tilde{B}:[0, \infty) \times \Omega \rightarrow \mathbf{R}^{d}$ such that
$$
M_{i}(t)=\sum_{k=1}^{d} \int_{0}^{t} \Phi_{i k}(s) d \tilde{B}^{k}(s), \quad t \in[0, \infty), i=1, \ldots, N
$$

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|PROGRESSIVELY MEASURABLE PROCESSES

然后我们看到$\left\langle N^{k}, B^{\ell}\right\rangle=0, k, \ell=1, \ldots, d$.
Let $A_{k \ell}, k, \ell=1, \ldots, d$, be progressively measurable processes given by
$$
\left(A_{k \ell}(t, \omega)\right){k, \ell=1, \ldots, d}=I{d}-\psi_{N, d}(\Phi(t, \omega)) \Phi(t, \omega)
$$
Also, let $\tilde{B} \in \mathcal{M}{\text {loc }}^{c}, k=1, \ldots, d$, be given by $$ \tilde{B}^{k}(t)=N^{k}(t)+\sum{\ell=1}^{d} \int_{0}^{t} A_{k \ell}(s) d B^{\ell}(s), \quad t \in[0, \infty)
$$
Then we see that
$$
\begin{aligned}
\left\langle\tilde{B}^{k}, \tilde{B}^{\ell}\right\rangle(t) &=\left\langle N^{k}, N^{\ell}\right\rangle(t)+\sum_{r=1}^{d} \int_{0}^{t} A_{k r}(s) A_{\ell r}(s) d s \
&=\sum_{i, j}^{N} \sum_{r=1}^{d} \int_{0}^{t} \Psi_{k, i}(s) \Psi_{\ell, j}(s) \Phi_{i, r}(s) \Phi_{j, r}(s) d s+\sum_{r=1}^{d} \int_{0}^{t} A_{k r}(s) A_{\ell r}(s) d s .
\end{aligned}
$$
Noting that $\psi_{N, d}(\Phi(s)) \Phi(s)$ is an orthogonal projection matrix, we see that
$$
\sum_{i, j}^{N} \sum_{r=1}^{d} \Psi_{k, i}(s) \Psi_{\ell, j}(s) \Phi_{i, r}(s) \Phi_{j, r}(s)+\sum_{r=1}^{d} A_{k r}(s) A_{\ell r}(s)=\delta_{k \ell}, \quad k, \ell=1, \ldots, d
$$
Therefore we see that $\tilde{B}$ is a $d$-dimensional $\left{\mathcal{F}{t}\right}{t \in[0, \infty)}$-Brownian motion. Let
$$
\tilde{M}{i}(t)=\sum{k=1}^{d} \int_{0}^{t} \Phi_{i, k}(s) d \tilde{B}^{k}(s)
$$
Then we see that
$$
\tilde{M}{i}(t)=\sum{k=1}^{d} \sum_{j=1}^{N} \int_{0}^{t} \Phi_{i, k}(s) \Psi_{k, j}(s) d M_{j}(s)+\sum_{k, \ell=1}^{d} \int_{0}^{t} \Phi_{i, k}(s) A_{k \ell}(s) d B^{\ell}(s) .
$$
Since $\Phi(s) \psi_{N, d}(\Phi(s)) \Phi(s)=\Phi(s)$, we see that
$$
\tilde{M}{i}(t)=\sum{j=1}^{N} \int_{0}^{t}\left(\Phi(s) \psi_{N, d}(\Phi(s))\right){i, j}(s) d M{j}(s)
$$

数学代写|随机分析作业代写stochastic analysis代考

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