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

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## 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|SEMI-MARTINGALE

$$P\left(\int_{0}^{T} \xi(t)^{2} d\langle M\rangle(t)+\int_{0}^{T}|\xi(t)| d|A|(t)<\infty \text { for any } T>0\right)=1$$

\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 .