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数学代写|随机分析作业代写stochastic analysis代考|Continuous Semi-martingales

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|Continuous Semi-martingales

数学代写|随机分析作业代写stochastic analysis代考|semi-martingale

Definition 4.2.1 We say that a stochastic process $X:[0, \infty) \times \Omega \rightarrow \mathbf{R}$ is a continuous semi-martingale, if there are $M \in \mathcal{M}{l o c}^{c}$ and $A \in \mathcal{A}{c}$ such that $X(t, \omega)=$ $X(0, \omega)+M(t, \omega)+A(t, \omega), t \in[0, \infty), \omega \in \Omega$

For any continuous semi-martingale $X$ the decomposition $X=X_{0}+M+A$, $M \in \mathcal{M}{l o c}^{c}, A \in \mathcal{A}{c}$ is unique because of Proposition 3.6.4. We say that a progressively measurable process $\xi$ is $X$-integrable, if
$$
P\left(\int_{0}^{T} \xi(t)^{2} d\langle M\rangle(t)+\int_{0}^{T}|\xi(t)| d|A|(t)<\infty \text { for any } T>0\right)=1
$$

Here $|A|$ is the total variation process given in Proposition 3.6.9. If $\xi$ is $X$-integrable, we can define $\xi \bullet M$ and $\int_{0}^{t} \xi(s) d A(s), t \in[0, \infty)$. Moreover, we see that $\xi \bullet M \in$ $\mathcal{M}{l o c}^{c}$ and $\int{0}^{t} \xi(s) d A(s) \in \mathcal{A}{c}$. So we define a continuous semi-martingale $(\xi \bullet X)$ by $$ (\xi \bullet X)(t)=(\xi \bullet M)(t)+\int{0}^{t} \xi(s) d A(s), \quad t \in[0, \infty)
$$
We often denote $(\xi \bullet X)(t)$ by $\int_{0}^{t} \xi(s) d X(s)$.
Note that any adapted continuous process is $X$-integrable for any continuous semi-martingale $X$.

Let $X$ and $Y$ be continuous semi-martingales such that $X=X_{0}+M+A, Y=$ $Y_{0}+N+\tilde{A}, M, N \in \mathcal{M}{l o c}^{c}$, and $A, \tilde{A} \in \mathcal{A}{c}$. Then we define $\langle X, Y\rangle \in \mathcal{A}_{c}$ by
$$
\langle X, Y\rangle=\langle M, N\rangle .
$$

数学代写|随机分析作业代写stochastic analysis代考|X -integrable progres

Proposition 4.2.1 Let $X$ be a continuous semi-martingale such that $X=X_{0}+M+$ $A, M \in \mathcal{M}{\text {loc }}^{c}, A \in \mathcal{A}{c}$. Suppose that $\xi, \xi_{n}, n=1,2, \ldots$, are $X$-integrable progressively measurable processes and that
$$
P\left(\int_{0}^{T}\left|\xi(t)-\xi_{n}(t)\right|^{2} d\langle M\rangle(t)+\int_{0}^{T}\left|\xi(t)-\xi_{n}(t)\right| d|A|(t)>\varepsilon\right) \rightarrow 0, \quad n \rightarrow \infty
$$
for any $T>0$ and $\varepsilon>0$. Then
$$
P\left(\sup {t \in[0, T]}\left|(\xi \bullet X)(t)-\left(\xi{n} \bullet X\right)(t)\right| \geqq \varepsilon\right) \rightarrow 0, \quad n \rightarrow \infty
$$
for any $T>0$ and $\varepsilon>0$.
Proof Let
$$
\eta_{n}(t)=\left(\left(\int_{0}^{t}\left|\xi(s)-\xi_{n}(s)\right|^{2} d\langle M\rangle(s)\right)^{1 / 2}+\int_{0}^{t}\left|\xi(s)-\xi_{n}(s)\right| d|A|(s)\right), \quad t \geqq 0 .
$$
Then we see that $P\left(\eta_{n}(T)>\delta\right) \rightarrow 0, n \rightarrow \infty$ for any $\delta>0$ and $T>0$. Let
$$
\tau_{n, m}=\inf \left{t>0 ; \eta_{n}(t)>\frac{1}{m}\right}, \quad n, m=1,2, \ldots
$$
Then we see that for any $T>0$

$$
\begin{aligned}
&E\left[\sup {t \in[0, T]}\left|(\xi \bullet X)^{\tau{n, m}}(t)-\left(\xi_{n} \bullet X\right)^{\tau_{n, m}}(t)\right|\right] \
&\leqq E\left[\sup {t \in[0, T]}\left|\left(\left(\xi-\xi{n}\right) \bullet M^{\tau_{n, m}}\right)(t)\right|^{2}\right]^{1 / 2}+E\left[\sup {t \in[0, T]}\left|\int{0}^{t}\left(\xi(s)-\xi_{n}(s)\right) d A^{\tau_{n, m}}(s)\right|\right] \
&\leqq 2 E\left[\int_{0}^{T \wedge \tau_{n, m}}\left|\xi(t)-\xi_{n}(t)\right|^{2} d\langle M\rangle(t)\right]^{1 / 2}+E\left[\int_{0}^{T \wedge \tau_{n, m}}\left|\xi(t)-\xi_{n}(t)\right| d|A|(t)\right] \
&\leqq \frac{3}{m}
\end{aligned}
$$
Therefore we see that for any $T>0$ and $\varepsilon>0$
$$
\begin{aligned}
&P\left(\sup {t \in[0, T]}\left|(\xi \bullet X)(t)-\left(\xi{n} \bullet X\right)(t)\right|>\varepsilon\right) \
&\leqq P\left(\tau_{n, m}{t \in[0, T]}\left|(\xi \bullet X)^{\tau{n, m}}(t)-\left(\xi_{n} \bullet X\right)^{\tau_{n, m}}(t)\right|\right] \ &\leqq P\left(\eta_{n}(T)>1 / m\right)+\frac{3}{m \varepsilon}
\end{aligned}
$$
So we see that
$$
\varlimsup_{n \rightarrow \infty} P\left(\sup {t \in[0, T]}\left|(\xi \bullet X)(t)-\left(\xi{n} \bullet X\right)(t)\right|>\varepsilon\right) \leqq \frac{3}{m \varepsilon}, \quad m \geqq 1 .
$$
This implies our assertion.

数学代写|随机分析作业代写stochastic analysis代考|Continuous Semi-martingales

随机分析代写

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|SEMI-MARTINGALE

定义 4.2.1 我们说一个随机过程$X:[0, \infty) \times \Omega \rightarrow \mathbf{R}$ is a continuous semi-martingale, if there are $M \in \mathcal{M}{l o c}^{c}$ and $A \in \mathcal{A}{c}$ such that $X(t, \omega)=$ $X(0, \omega)+M(t, \omega)+A(t, \omega), t \in[0, \infty), \omega \in \Omega$

对于任何连续半鞅X分解 $X=X_{0}+M+A$, $M \in \mathcal{M}{l o c}^{c}, A \in \mathcal{A}{c}$ is unique because of Proposition 3.6.4. We say that a progressively measurable process $\xi$ is $X$-integrable, if
$$
P\left(\int_{0}^{T} \xi(t)^{2} d\langle M\rangle(t)+\int_{0}^{T}|\xi(t)| d|A|(t)<\infty \text { for any } T>0\right)=1
$$

这里|一种|是命题 3.6.9 中给出的总变差过程。如果X是X-可积,我们可以定义 $\xi \bullet M$ and $\int_{0}^{t} \xi(s) d A(s), t \in[0, \infty)$. Moreover, we see that $\xi \bullet M \in$ $\mathcal{M}{l o c}^{c}$ and $\int{0}^{t} \xi(s) d A(s) \in \mathcal{A}{c}$. So we define a continuous semi-martingale $(\xi \bullet X)$ by $$ (\xi \bullet X)(t)=(\xi \bullet M)(t)+\int{0}^{t} \xi(s) d A(s), \quad t \in[0, \infty)
我们经常表示(X∙X)(吨)经过∫0吨X(s)dX(s).
请注意,任何适应的连续过程都是X-对于任何连续半鞅可积X.

$X=X_{0}+M+A, Y=$ $Y_{0}+N+\tilde{A}, M, N \in \mathcal{M}{l o c}^{c}$, and $A, \tilde{A} \in \mathcal{A}{c}$. Then we define $\langle X, Y\rangle \in \mathcal{A}_{c}$ by
$$
\langle X, Y\rangle=\langle M, N\rangle .
$$

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|X -INTEGRABLE PROGRES

命题 4.2.1 让X是一个连续的半鞅,使得$X=X_{0}+M+$ $A, M \in \mathcal{M}{\text {loc }}^{c}, A \in \mathcal{A}{c}$. Suppose that $\xi, \xi_{n}, n=1,2, \ldots$, are $X$-integrable progressively measurable processes and that
$$
P\left(\int_{0}^{T}\left|\xi(t)-\xi_{n}(t)\right|^{2} d\langle M\rangle(t)+\int_{0}^{T}\left|\xi(t)-\xi_{n}(t)\right| d|A|(t)>\varepsilon\right) \rightarrow 0, \quad n \rightarrow \infty
$$
for any $T>0$ and $\varepsilon>0$. Then
$$
P\left(\sup {t \in[0, T]}\left|(\xi \bullet X)(t)-\left(\xi{n} \bullet X\right)(t)\right| \geqq \varepsilon\right) \rightarrow 0, \quad n \rightarrow \infty
$$
for any $T>0$ and $\varepsilon>0$.
Proof Let
$$
\eta_{n}(t)=\left(\left(\int_{0}^{t}\left|\xi(s)-\xi_{n}(s)\right|^{2} d\langle M\rangle(s)\right)^{1 / 2}+\int_{0}^{t}\left|\xi(s)-\xi_{n}(s)\right| d|A|(s)\right), \quad t \geqq 0 .
$$
Then we see that $P\left(\eta_{n}(T)>\delta\right) \rightarrow 0, n \rightarrow \infty$ for any $\delta>0$ and $T>0$. Let
$$
\tau_{n, m}=\inf \left{t>0 ; \eta_{n}(t)>\frac{1}{m}\right}, \quad n, m=1,2, \ldots
$$
然后我们看到对于任何吨>0

$$
\begin{aligned}
&E\left[\sup {t \in[0, T]}\left|(\xi \bullet X)^{\tau{n, m}}(t)-\left(\xi_{n} \bullet X\right)^{\tau_{n, m}}(t)\right|\right] \
&\leqq E\left[\sup {t \in[0, T]}\left|\left(\left(\xi-\xi{n}\right) \bullet M^{\tau_{n, m}}\right)(t)\right|^{2}\right]^{1 / 2}+E\left[\sup {t \in[0, T]}\left|\int{0}^{t}\left(\xi(s)-\xi_{n}(s)\right) d A^{\tau_{n, m}}(s)\right|\right] \
&\leqq 2 E\left[\int_{0}^{T \wedge \tau_{n, m}}\left|\xi(t)-\xi_{n}(t)\right|^{2} d\langle M\rangle(t)\right]^{1 / 2}+E\left[\int_{0}^{T \wedge \tau_{n, m}}\left|\xi(t)-\xi_{n}(t)\right| d|A|(t)\right] \
&\leqq \frac{3}{m}
\end{aligned}
$$
Therefore we see that for any $T>0$ and $\varepsilon>0$
$$
\begin{aligned}
&P\left(\sup {t \in[0, T]}\left|(\xi \bullet X)(t)-\left(\xi{n} \bullet X\right)(t)\right|>\varepsilon\right) \
&\leqq P\left(\tau_{n, m}{t \in[0, T]}\left|(\xi \bullet X)^{\tau{n, m}}(t)-\left(\xi_{n} \bullet X\right)^{\tau_{n, m}}(t)\right|\right] \ &\leqq P\left(\eta_{n}(T)>1 / m\right)+\frac{3}{m \varepsilon}
\end{aligned}
$$
So we see that
$$
\varlimsup_{n \rightarrow \infty} P\left(\sup {t \in[0, T]}\left|(\xi \bullet X)(t)-\left(\xi{n} \bullet X\right)(t)\right|>\varepsilon\right) \leqq \frac{3}{m \varepsilon}, \quad m \geqq 1 .
$$
这暗示了我们的断言。

数学代写|随机分析作业代写stochastic analysis代考

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