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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|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
$$
\tilde{M}{i}(t)=\sum{j=1}^{N} \int_{0}^{t}\left(\Phi(s) \psi_{N, d}(\Phi(s))\right){i, j}(s) d M{j}(s)
$$
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
$$
\tilde{M}{i}(t)=\sum{j=1}^{N} \int_{0}^{t}\left(\Phi(s) \psi_{N, d}(\Phi(s))\right){i, j}(s) d M{j}(s)
$$
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