# 随机微积分作业代写stochastic calculus代考| CHANGE OF MEASURE

my-assignmentexpert™ 随机微积分stochastic calculus作业代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。my-assignmentexpert™， 最高质量的随机微积分stochastic calculus作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于随机微积分stochastic calculus作业种类很多，难度波动比较大，同时其中的大部分作业在字数上都没有具体要求，因此随机微积分stochastic calculus作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

my-assignmentexpert™ 为您的留学生涯保驾护航 在经济学作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的微积分calculus代写服务。我们的专家在随机微积分stochastic calculus 代写方面经验极为丰富，各种随机微积分stochastic calculus相关的作业也就用不着 说。

• 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|Locally equivalent change of probability

4.a.1 Girsanov’s Formula. Assume that $Q$ is locally equivalent to $P$ and let $X$ be a continuous $P$-semimartingale. Then $X$ is also a continuous $Q$-semimartingale and its compensator $u_{X}^{Q}$ with respect to $Q$ is given by $u_{X}^{Q}=u_{X}^{P}+\langle X, \log (M)\rangle$.

Proof. Let $B$ be any process and $\Psi \subseteq \Omega$ the set of all $\omega \in \Omega$ such that the path $t \mapsto B_{t}(\omega)$ is continuous and of bounded variation on all finite intervals. Clearly then $\Psi=\bigcap_{n} \Psi_{n}$, where $\Psi_{n}$ is the set of all $\omega \in \Omega$ such that the path $t \mapsto B_{t}(\omega)$ is continuous and of bounded variation on the interval $[0, n]$. For $n \geq 1$, the set $\Psi_{n}$ is in $\mathcal{F}{n}$ and since $P$ and $Q$ are equivalent on $\mathcal{F}{n}$, we have $P\left(\Psi_{n}\right)=1$ if and only if $Q\left(\Psi_{n}\right)=1$. It follows that $P(\Psi)=1$ if and only if $Q(\Psi)=1$, that is, $B$ is a continuous bounded variation process with respect to $P$ if and only if $B$ is a continuous bounded variation process with respect to $Q$.

Recall now that $X$ is a Q-semimartingale if and only if there exists a continuous bounded variation process $B$ vanishing at zero such that $X-B$ is a $Q$-local martingale in which case $B=u_{X}^{Q}$.

Chapter III: Stochastic Integration 171
Now let $B$ be any continuous bounded variation process vanishing at zero. Then $X-B$ is a $Q$-local martingale if and only if $(X-B) M$ is a $P$-local martingale, that is, $u_{(X-B) M}^{P}=0$, equivalently $d u_{(X-B) M}^{P}=0$. Observing that $u_{M}^{P}=0$ and $u_{X-B}^{P}=u_{X}^{P}-B$ and using formula 3.c.3.(a) for the compensator of a product, the equality
\begin{aligned} 0=d u_{(X-B) M}^{P} &=M_{t} d u_{X-B}^{P}(t)+d\langle X-B, M\rangle_{t} \ &=M_{t} d u_{X}^{P}(t)-M_{t} d B_{t}+d\langle X, M\rangle_{t} \end{aligned}
is equivalent with $d B_{t}=d u_{X}^{P}(t)+M_{t}^{-1} d\langle X, M\rangle_{t}=d u_{X}^{P}(t)+d\langle X, \log (M)\rangle_{t}$, that is, $B_{t}=u_{X}^{P}(t)+\langle X, \log (M)\rangle_{t}$ which is indeed a bounded variation process vanishing at zero. Thus $X$ is a $Q$-semimartingale with $u_{X}^{Q}=B=u_{X}^{P}+\langle X, \log (M)\rangle$.

Let us now show that stochastic integrals are invariant under change to a locally equivalent probability measure. With $P, Q$ and $M_{t}=d\left(Q \mid \mathcal{F}{t}\right) / d\left(P \mid \mathcal{F}{t}\right), t \geq 0$, as above, let $X$ be a continuous $P$-semimartingale, $H \in L(X)$ and $I=(H \bullet X)^{P}$, that is, $I=\int_{0}^{*} H_{s} d X_{s}$, where this integral process is computed with respect to the measure $P$. To show that $I=(H \bullet X)^{Q}$ also, that is, the integral process $H \cdot X$ is unaffected if we switch from the probability $P$ to the locally equivalent probability $Q$, we first note the following universal property of the process $I$ :

## 微积分作业代写calclulus代考|Abstract Novikov Condition

4.d.0 Abstract Novikov Condition. Let $L$ be a continuous local martingale with $L_{0}=0$ which is adapted to the augmented filtration generated by some Brownian motion and assume that $\quad E_{P}\left[\exp \left(\frac{1}{2}\langle L\rangle_{t}\right)\right]<\infty, \quad \forall 0 \leq t<\infty$.
Then $Z_{t}=\mathcal{E}{t}(L)=\exp \left(L{t}-\frac{1}{2}\langle L\rangle_{t}\right)$ is a martingale.
Proof. It will suffice to show that $E\left(Z_{t}\right)=1$, for all $t \geq 0$ (4.b.0). Note first that (0) implies $P\left(\langle L\rangle_{t}<\infty\right)=1$.

According to 5.a.4 below we can choose a Brownian motion $W$ on a suitable enlargement $\left(\Omega_{1}, \mathcal{G}, P_{1},\left(\mathcal{G}{t}\right)\right)$ of the original filtered probability space $\left(\Omega, \mathcal{G}, P,\left(\mathcal{G}{t}\right)\right)$ such that $L_{t}=W_{\langle L\rangle_{t}}$ and such that each $\langle L\rangle_{t}$ is a $\left(\mathcal{G}{t}\right)$-optional time. This will allow us to reduce the general case $Z{t}=\exp \left(L_{t}-\frac{1}{2}\langle L\rangle_{t}\right)$ to the special case of the well known basic exponential martingale $\exp \left(\mu W_{t}-\frac{1}{2} \mu^{2} t\right)$ (4.c.4) via an application of Wald’s identity.

Let $b<0$ and set $T_{b}=\inf \left{s \geq 0 \mid W_{s}-s=b\right}$. Then $T_{b}$ is a $\left(\mathcal{G}{t}\right)$-optional time. Since the process $W{s}-s$ has continuous paths and satisfies $W_{s}-s \rightarrow-\infty$, as $s \uparrow \infty$, we have $P\left(T_{b}<\infty\right)=1$ and $T_{b} \uparrow \infty$, as $b \downarrow-\infty, P$-as.

Set $\mu=1$. The process $W_{s}-s=W_{s}-\mu s$ is a one dimensional Brownian motion with respect to the measure $Q^{\mu}$ of example $4 . c .4$ and since one dimensional Brownian motion hits all values we have $Q^{\mu}\left(T_{b}<\infty\right)=1$. Wald’s identity now implies that
$$E_{P}\left(e^{W_{T_{b}}-\frac{1}{2} T_{b}}\right)=Q^{\mu}\left(T_{b}<\infty\right)=1 .$$
Note that $W_{T_{b}}-T_{b}=b$, by definition of the optional time $T_{b}$, that is
$$W_{T_{b}}-\frac{1}{2} T_{b}=b+\frac{1}{2} T_{b}$$
and so $E_{P}\left(\exp \left(b+\frac{1}{2} T_{b}\right)\right)=1$, that is,
$$E_{P}\left(\exp \left(\frac{1}{2} T_{b}\right)\right)=e^{-b}$$
Recall that $Y_{s}=\exp \left(\mu W_{s}-\frac{1}{2} \mu^{2} s\right)=e^{W_{n}-s / 2}$ is a martingale and hence, by the Optional Sampling Theorem, so is the process $N_{s}=Y_{s \wedge} T_{b}$.

Let us note that $Y_{0}=1$ and hence $E\left(N_{s}\right)=E\left(N_{0}\right)=E\left(Y_{0}\right)=1, s \geq 0$. Since $P\left(T_{b}<\infty\right)=1$ we have
$$N_{\infty}=\lim {s{ }{\infty}} N_{s}=Y_{T_{b}}=e^{W_{T_{b}}-\frac{1}{2} T_{b}},$$
with convergence $P$-as. According to (1) we have $E_{P}\left(N_{\infty}\right)=1$. Let us now show that $N_{\infty}$ is a last element for the martingale $N$. Fix $s \geq 0$. Then $N_{s}=E_{P}\left(N_{k} \mid \mathcal{F}{s}\right)$, for all $k \geq s$. Fatou’s Lemma (I.2.b.8 with $h=0$ ) yields \begin{aligned} N{s} &=\lim {k} E{P}\left(N_{k} \mid \mathcal{F}{s}\right)=\liminf E{p}\left(N_{k} \mid \mathcal{F}{s}\right) \leq E{P}\left(\liminf \operatorname{in}{k} N{k} \mid \mathcal{F}{s}\right) \ &=E{P}\left(N_{\infty} \mid \mathcal{F}_{s}\right), \quad P \text {-as. } \end{aligned}

Since $E_{P}\left[E_{P}\left(N_{\infty} \mid \mathcal{F}{s}\right)\right]=E{P}\left(N_{\infty}\right)=1=E_{P}\left(N_{s}\right)$ it follows that $N_{s}=E_{P}\left(N_{\infty} \mid \mathcal{F}{s}\right)$. Thus $N{\infty}$ is a last element for the martingale $N$ which is thus uniformly integrable. Consequently the Optional Sampling Theorem can be applied with all optional times $\tau$ to yield $E_{P}\left(Y_{\tau \wedge T_{b}}\right)=E_{P}\left(N_{\tau}\right)=E_{P}\left(N_{0}\right)=1$, that is,
$$E_{P}\left(\exp \left(W_{\tau \wedge T_{b}}-\frac{1}{2} \tau \wedge T_{b}\right)\right)=1 .$$
Now fix $t \in[0, \infty)$ and apply (4) to the optional time $\tau=\langle L\rangle_{t}$ observing that
$$\tau \wedge T_{b}= \begin{cases}T_{b} & \text { on }\left[T_{b} \leq \tau\right]=\left[T_{b} \leq\langle L\rangle_{t}\right] \ \langle L\rangle_{t} & \text { on }\left[T_{b}>\tau\right]=\left[T_{b}>\langle L\rangle_{t}\right] .\end{cases}$$
Using (2) it follows that
$$W_{\tau \wedge T_{b}}-\frac{1}{2} \tau \wedge T_{b}= \begin{cases}b+\frac{1}{2} T_{b} & \text { on }\left[T_{b} \leq \tau\right]=\left[T_{b} \leq\langle L\rangle_{t}\right] \ L_{t}-\frac{1}{2}\langle L\rangle_{t} & \text { on }\left[T_{b}>\tau\right]=\left[T_{b}>\langle L\rangle_{t}\right]\end{cases}$$
With the notation $E(X ; A)=E\left(1_{A} X\right)$ we can rewrite (4) as
$$\left.E_{P}\left(\exp \left(b+\frac{1}{2} T_{b}\right) ;\left[T_{b} \leq\langle L\rangle_{t}\right]\right)+E_{P}\left(\exp \left(L_{t}-\frac{1}{2}\langle L\rangle_{t}\right) ;\left[T_{b}\right\rangle\langle L\rangle_{t}\right]\right)=1 .$$
Now let $b \downarrow-\infty$. Then $T_{b} \uparrow \infty$ and so $1_{\left[T_{b}>(L){t}\right]} \uparrow 1, P$-as. By Monotone Convergence the second expectation in (5) converges to $E{P}\left(\exp \left(L_{t}-\frac{1}{2}\langle L\rangle_{t}\right)\right)=E_{P}\left(Z_{t}\right)$. The first expectation satisfies
\begin{aligned} E_{P}\left(\exp \left(b+\frac{1}{2} T_{b}\right) ;\left[T_{b} \leq\langle L\rangle_{t}\right]\right) & \leq E_{P}\left(\exp \left(b+\frac{1}{2}\langle L\rangle_{t}\right)\right) \ &=e^{b} E_{P}\left(\exp \left(\frac{1}{2}\langle L\rangle_{t}\right)\right) \rightarrow 0 \end{aligned}
as $b \downarrow-\infty$ (recall (1)). Thus (5) implies that $E_{P}\left(Z_{t}\right)=1$, as desired. $\mathbf{1}$
Application to Girsanov’s Theorem. It remains to apply this condition to decide when the drift rate process $\gamma(s)$ satisfies the assumption of Girsanov’s Theorem 4.c.2 . Here we are dealing with the special case of the continuous local martingale $L=\gamma \bullet W$, where $W$ is some $d$-dimensional Brownian motion. For this local martingale we have
$$\langle L\rangle_{t}=\int_{0}^{t}|\gamma(s)|^{2} d s,$$
and consequently the Doleans exponential $Z=\mathcal{E}(L)$ assumes the form
$$Z_{t}:=\mathcal{E}{t}(L)=\exp \left(L{t}-\frac{1}{2}\langle L\rangle_{t}\right)=\exp \left(\int_{0}^{t} \gamma(s) \cdot d B_{s}-\frac{1}{2} \int_{0}^{t}|\gamma(s)|^{2} d s\right)$$
Thus 4.d.0 specialized to the above local martingale $L=\mathcal{E}(\gamma \cdot W)$ yields

## 微积分作业代写calclulus代考|Locally equivalent change of probability

4.a.1 吉尔萨诺夫公式。假使，假设问本地等价于磷然后让X成为一个连续的磷-半鞅。然后X也是一个连续的问-半鞅及其补偿器你X问关于问是（谁）给的你X问=你X磷+⟨X,日志⁡(米)⟩.

0=d你(X−乙)米磷=米吨d你X−乙磷(吨)+d⟨X−乙,米⟩吨 =米吨d你X磷(吨)−米吨d乙吨+d⟨X,米⟩吨

## 微积分作业代写calclulus代考|Abstract Novikov Condition

4.d.0 抽象诺维科夫条件。让一世是一个连续的局部鞅一世0=0它适用于由某些布朗运动产生的增强过滤，并假设和磷[经验⁡(12⟨一世⟩吨)]<∞,∀0≤吨<∞.

ñ∞=林s∞ñs=和吨b=和在吨b−12吨b,

τ∧吨b={吨b 在 [吨b≤τ]=[吨b≤⟨一世⟩吨] ⟨一世⟩吨 在 [吨b>τ]=[吨b>⟨一世⟩吨].

Girsanov 定理的应用。仍然应用此条件来决定何时漂移率过程C(s)满足 Girsanov 定理 4.c.2 的假设。这里我们处理的是连续局部鞅的特殊情况一世=C∙在， 在哪里在是一些d维布朗运动。对于这个本地鞅，我们有
⟨一世⟩吨=∫0吨|C(s)|2ds,