# 随机微积分作业代写stochastic calculus代考| ITO’S FORMULA

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• 随机偏微分方程
• 随机控制
• Ito积分
• black-Scholes-Merton option pricing formula
• Fokker–Planck equation
• 布朗运动 Brownian motion

## 微积分作业代写calclulus代考|Leviís characterization of Brownian motion

3.e.0. Let $B_{t}=\left(B_{t}^{1}, B_{t}^{2} \ldots, B_{t}^{d}\right)$ be a Brownian motion on $\left(\Omega, \mathcal{F},\left(\mathcal{F}{t}\right), P\right)$. Then (a) $B{t}^{i} B_{t}^{j}$ is a martingale, for all $i \neq j$. (b) $\left\langle B^{i}, B^{j}\right\rangle=0$, for all $i \neq j$.
Proof. Each coordinate process $B_{t}^{j}$ is a one dimensional Brownian motion and hence a continuous, square integrable martingale (II.2.g.0). Especially $B_{t}^{j} \in L^{2}(P)$ and hence $B_{t}^{i} B_{t}^{j} \in L^{1}(P)$ for all $i, j$ and $t \geq 0$. The covariation $\left\langle B^{i}, B^{j}\right\rangle$ is the unique continuous, bounded variation process $A$ such that $A_{0}=0$ and $B^{i} B^{j}-A$ is a local martingale. Thus (b) follows from (a).

Let us now show (a). Assume that $i \neq j$. In order to see that $B_{t}^{i} B_{t}^{j}$ is a martingale we must show that $E_{P}\left[B_{t}^{i} B_{t}^{j}-B_{s}^{i} B_{s}^{j} \mid \mathcal{F}{s}\right]=0$. Write $$B{t}^{i} B_{t}^{j}-B_{s}^{i} B_{s}^{j}=\left(B_{t}^{i}-B_{s}^{i}\right)\left(B_{t}^{j}-B_{s}^{j}\right)+B_{s}^{i}\left(B_{t}^{j}-B_{s}^{j}\right)+B_{s}^{j}\left(B_{t}^{i}-B_{s}^{i}\right)$$
The first summand is a function of the increment $B_{t}-B_{s}$ and hence independent of the $\sigma$-field $\mathcal{F}{s}$. Since the two factors are themselves independent we obtain \begin{aligned} E{P}\left[\left(B_{t}^{i}-B_{s}^{i}\right)\left(B_{t}^{j}-B_{s}^{j}\right) \mid \mathcal{F}{s}\right] &=E{P}\left[\left(B_{t}^{i}-B_{s}^{i}\right)\left(B_{t}^{j}-B_{s}^{j}\right)\right] \ &=E_{P}\left[B_{t}^{i}-B_{s}^{i}\right] E_{P}\left[B_{t}^{j}-B_{s}^{j}\right]=0 \end{aligned}
Moreover the $\mathcal{F}{s}$-measurability of $B{s}^{i}, B_{s}^{j}$ implies that
\begin{aligned} &E_{P}\left[B_{s}^{i}\left(B_{t}^{j}-B_{s}^{j}\right) \mid \mathcal{F}{s}\right]=B{s}^{i} E_{P}\left[B_{t}^{j}-B_{s}^{j} \mid \mathcal{F}{s}\right]=0 \quad \text { and } \ &E{P}\left[B_{s}^{j}\left(B_{t}^{i}-B_{s}^{i}\right) \mid \mathcal{F}{s}\right]=B{s}^{j} E_{P}\left[B_{t}^{i}-B_{s}^{i} \mid \mathcal{F}{s}\right]=0 \end{aligned} Conditioning on the $\sigma$-field $\mathcal{F}{s}$ in $(0)$ now yields $E_{P}\left[B_{t}^{i} B_{t}^{j}-B_{s}^{i} B_{s}^{j} \mid \mathcal{F}_{s}\right]=0$.

## 微积分作业代写calclulus代考|Harmonic functions of Brownian motion

3.f The multiplicative compensator $U_{X}$.
3.f.0. Let $X \in \mathcal{S}{+}$. Then there is a unique continuous bounded variation process $A$ such that $A{0}=1, A>0$ and $X_{t} / A_{t}$ is a local martingale. The process $A$ is called the multiplicative compensator of the semimartingale $X$ and denoted $A=U_{X}$. The relationship to the (additive) compensator $u_{X}$ of $X$ is as follows:
$$U_{X}(t)=\exp \left(\int_{0}^{t} \frac{1}{X_{s}} d u_{X}(s)\right) \quad \text { and } \quad u_{X}(t)=\int_{0}^{t} X_{s} d \log \left(U_{X}(s)\right)$$
Proof. Uniqueness. Here we will also see how to find such a process $A$. Assume that $A$ is a process with the above properties and set $Z=1 / A$. Since the continuous, positive bounded variation process $A$ is $P$-as. pathwise bounded away from zero on finite intervals, it follows that $Z=1 / A$ is itself a continuous bounded variation process with $Z_{0}=A_{0}=1$. Thus $\langle Z, X\rangle=0$ and $u_{Z}=Z$. As $Z X$ is a local martingale, $u_{Z X}=0$ and formula 3.c.3.(a) for the compensator $u_{Z X}$ yields
$$0=d u_{Z X}(t)=X_{t} d Z_{t}+Z_{t} d u_{X}(t), \quad \text { that is, } \quad Z_{s}^{-1} d Z_{s}=-X_{s}^{-1} d u_{X}(s)$$
Chapter III: Stochastic Integration 169
Since $Z$ is of bounded variation this can be rewritten as $d \log \left(Z_{s}\right)=-X_{s}^{-1} d u_{X}(s)$. Observing that $\log \left(Z_{0}\right)=0$ integration yields
$$\log \left(Z_{t}\right)=-\int_{0}^{t} \frac{1}{X_{s}} d u_{X}(s) \quad \text { and so } \log \left(A_{t}\right)=-\log \left(Z_{t}\right)=\int_{0}^{t} \frac{1}{X_{s}} d u_{X}(s) .$$
This shows that a process $A$ with the above properties must be given by the first formula in $(0)$. In particular $A$ is uniquely determined.
Existence. Set $A_{t}=\exp \left(\int_{0}^{t} X_{s}^{-1} d u_{X}(s)\right)$. We verify that $A$ has the desired properties. We merely have to reverse the considerations of (a) above. Clearly $A$ is a strictly positive, continuous bounded variation process with $A_{0}=1$. Set $Z=1 / A$. To show that $Z X$ is a local martingale note that it is a continuous semimartingale and $Z_{t}=\exp \left(-\int_{0}^{t} X_{s}^{-1} d u_{X}(s)\right)$ and thus $d \log \left(Z_{s}\right)=-X_{s}^{-1} d u_{X}(s)$. Since $Z$ is a bounded variation process, this can be rewritten as
$$Z_{s}^{-1} d Z_{s}=-X_{s}^{-1} d u_{X}(s) \text { and so } \quad X_{s} d Z_{s}+Z_{s} d u_{X}(s)=0$$
that is, $d u_{Z X}(s)=0$ and so $u_{Z X}=0$. Thus $Z X$ is a local martingale. It remains to verify the second equation in $(0)$, that is, $d u_{X}(t)=X_{t} d \log \left(U_{X}(t)\right)$. This follows at once from the first equation in $(0)$ upon taking the logarithm, differentiating and multiplying with $X_{t}$.

## 微积分作业代写calclulus代考|Leviís characterization of Brownian motion

3.e.0。让乙吨=(乙吨1,乙吨2…,乙吨d)成为一个布朗运动(Ω,F,(F吨),磷). 那么（一）乙吨一世乙吨j是鞅，对于所有人一世≠j. (二)⟨乙一世,乙j⟩=0， 对所有人一世≠j.

## 微积分作业代写calclulus代考|Harmonic functions of Brownian motion

3.f乘法补偿器üX.
3.f.0。让X∈小号+. 则有一个独特的连续有界变分过程一种这样一种0=1,一种>0和X吨/一种吨是局部鞅。过程一种称为半鞅的乘法补偿器X并表示一种=üX. 与（加法）补偿器的关系你X的X如下：
üX(吨)=经验⁡(∫0吨1Xsd你X(s)) 和 你X(吨)=∫0吨Xsd日志⁡(üX(s))

0=d你和X(吨)=X吨d和吨+和吨d你X(吨), 那是， 和s−1d和s=−Xs−1d你X(s)

169和是有界变化的，这可以重写为d日志⁡(和s)=−Xs−1d你X(s). 观察到日志⁡(和0)=0积分收益率