# 随机微积分作业代写stochastic calculus代考| LOCAL MARTINGALES

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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|Localization

8.a.0. The filtration $\left(\mathcal{F}{t \wedge T}\right){t}$ is also right continuous. If $X$ is a right continuous submartingale then the same is true of the process $X^{T}$ relative to the filtration $\left(\mathcal{F}{t \wedge T}\right){t}$

Proof. Let us first show that the filtration $\left(\mathcal{F}{t \wedge T}\right){t}$ is right continuous. It has to be shown that $\bigcap_{s>t} \mathcal{F}{s \wedge T} \subseteq \mathcal{F}{t \wedge T}$, for all $t \geq 0$. Let $t \geq 0$ and $A \in \bigcap_{s>t} \mathcal{F}{s \wedge T}$. We wish to show that $A \in \mathcal{F}{t \wedge T}$, that is, $A \cap[t \wedge T<r] \in \mathcal{F}_{r}$, for all $r \geq 0$.

Let $r \geq 0$. Since $A \in \mathcal{F}{s \wedge T}$, we have $A \cap[s \wedge T{r}$, for all $s>t$. Choose a sequence $\left(s_{n}\right)$ such that $s_{n}>t$ and $s_{n} \downarrow t$, as $n \uparrow \infty$. Then $s_{n} \wedge T \downarrow t \wedge T$ pointwise and consequently $\left[s_{n} \wedge T<r\right] \uparrow[t \wedge T<r]$, whence $A \cap\left[s_{n} \wedge T<r\right] \uparrow A \cap[t \wedge T<r]$, as $n \uparrow \infty$. Since $A \cap\left[s_{n} \wedge T<r\right] \in \mathcal{F}{r}$, for all $n \geq 1$, it follows that $A \cap[t \wedge T{r}$, as desired. This shows the right continuity of the filtration $\left(\mathcal{F}{t \wedge T}\right){t}$. According to $7 . a .5$ the process $X^{T}=\left(X_{t \wedge T}\right){t}$ is adapted to the filtration $\left(\mathcal{F}{t \wedge T}\right){t}$. The right continuity is inherited from the process $X$. The rest now follows from 7.d.1 applied to the bounded optional times $T{t}=t \wedge T$. I Let $A \in \mathcal{F}$ be a measurable subset of $\Omega$. If $X=\left(X_{t}\right)$ is a stochastic process we define the stochastic process $Y=1_{A} X$ as $Y_{t}(\omega)=1_{A}(\omega) X_{t}(\omega), t \geq 0, \omega \in \Omega$. Thus, multiplication of $X$ by $1_{A}$ affects the paths of $X$ in a very simple way: the path $t \mapsto Y_{t}(\omega)$ agrees with the path $t \mapsto X_{t}(\omega)$, if $\omega \in A$, and is identically zero otherwise. If $X$ is $\left(\mathcal{F}{t}\right)$-adapted and $A \in \mathcal{F}{0}$, then $Y=1_{A} X$ is $\left(\mathcal{F}{t}\right)$-adapted also. Now let $T$ be an optional time. Then $A=[T>0] \in \mathcal{F}{0}$. Thus, if $X$ is an adapted process, then so is the process $1_{[T>0]} X$.

## 微积分作业代写calclulus代考|Bayes Theorem

8. Assume that $P$ and $Q$ are locally equivalent. Then
(a) $M_{t}$ is a strictly positive $P$-martingale.
(b) For $0 \leq t{T}\right)$ we have $$E{Q}\left(f \mid \mathcal{F}{t}\right)=\frac{E{P}\left(M_{T} f \mid \mathcal{F}{t}\right)}{M{t}}=\frac{E_{P}\left(M_{T} f \mid \mathcal{F}{t}\right)}{E{P}\left(M_{T} \mid \mathcal{F}{t}\right)} .$$ Proof. (a) Fix $0 \leq t{t}$ and $Q_{t}$ on $\mathcal{F}{t}$ implies that $M{t}=d Q_{t} / d P_{t}>0, P$-as. and $M_{t} \in L^{1}\left(P, \mathcal{F}{t}\right)$, especially $M{t}$ is $\mathcal{F}{t}$-measurable, by definition of the Radon-Nikodym derivative. Moreover for $A \in \mathcal{F}{t} \subseteq \mathcal{F}{T}$ we have $E{P}\left(1_{A} M_{t}\right)=Q(A)=E_{P}\left(1_{A} M_{T}\right)$ (see (0)). This shows that $M_{t}=E_{P}\left(M_{T} \mid \mathcal{F}{t}\right)$. (b) Let $0 \leq t{T}\right)$ and $h=E_{Q}\left(f \mid \mathcal{F}{t}\right)$. We want to show that $M{t} h=E_{P}\left(M_{T} f \mid \mathcal{F}{t}\right)$. Let $A \in \mathcal{F}{t}$ and note that $h$ is $\mathcal{F}{t}$-measurable and satisfies $E{Q}\left(1_{A} h\right)=E_{Q}\left(1_{A} f\right)$. Using the $\mathcal{F}{t}$-measurability of $1{A} h$, the $\mathcal{F}{T}$-measurability of $1{A} f$ and $(0)$, we can now write this as $E_{P}\left(1_{A} M_{t} h\right)=E_{P}\left(1_{A} M_{T} f\right)$. Since $M_{t} h$ is $\mathcal{F}{t}$-measurable this implies that $M{t} h=E_{P}\left(M_{T} f \mid \mathcal{F}{t}\right)$, as desired. Remark. Recalling that $M{T}=d P_{T} / d Q_{T}$, (b) can also be written as
$$E_{Q}\left(f \mid \mathcal{F}{t}\right)=\frac{E{P}\left(\left(d Q_{T} / d P_{T}\right) f \mid \mathcal{F}{t}\right)}{E{P}\left(d Q_{T} / d P_{T} \mid \mathcal{F}{t}\right)}, \quad \forall 0 \leq t{T}\right) .$$
We are mainly interested in the following
8.b.1 Corollary. The adapted process $\left(X_{t}\right)$ is a $Q$-martingale (Q-local martingale) if and only if the process $\left(M_{t} X_{t}\right)$ is a $P$-martingale ( $P$-local martingale).
Proof. Regarding the martingale case we have, using 8.b.0.(b),
Turning to the case of local martingales, assume that $X=\left(X_{t}\right)$ is a $Q$-local martingale and let $\left(T_{n}\right)$ be a reducing sequence of optional times for $X$ such that $T_{n} \uparrow \infty$ at each point of $\Omega$ (8.a.2.(f)). Then $1_{\left[T_{n}>0\right]} X_{t \wedge T_{n}}$ is a $Q$-martingale (indexed by $t$ ) and so $Y_{t}=M_{t} 1_{\left[T_{\mathrm{n}}>0\right]} X_{t \wedge T_{\mathrm{n}}}$ a $P$-martingale and hence so is $Y_{t}^{T_{n}}=1_{\left[T_{\mathrm{n}}>0\right]} M_{t \wedge T_{\mathrm{n}}} X_{t \wedge T_{\mathrm{n}}}$, for each $n \geq 1$. Thus $\left(T_{n}\right)$ is a reducing sequence for the process $\left(M_{t} X_{t}\right)$ with respect to the probability measure $P$. Consequently $\left(M_{t} X_{t}\right)$ is a $P$-local martingale. The converse can be shown similarly or follows from this by interchanging $P$ and $Q$ and observing that $d P_{t} / d Q_{t}=M_{t}^{-1}$. I

## 微积分作业代写calclulus代考|Localization

8.a.0. 过滤 $\left(\mathcal{F} {t \wedge T}\right) {t}一世s一种一世s○r一世GH吨C○n吨一世n你○你s.一世FX一世s一种r一世GH吨C○n吨一世n你○你ss你b米一种r吨一世nG一种一世和吨H和n吨H和s一种米和一世s吨r你和○F吨H和pr○C和ssX^{T}r和一世一种吨一世v和吨○吨H和F一世一世吨r一种吨一世○n\left(\mathcal{F} {t \wedge T}\right) {t}$

## 微积分作业代写calclulus代考|Bayes Theorem

8. 假设磷和问是本地等价的。那么
（一）米吨是严格正数磷-鞅。
(b) 为0 \leq t{T}\右）0 \leq t{T}\右）我们有 $$E{Q}\left(f \mid \mathcal{F} {t}\right)=\frac{E {P}\left(M_{T} f \mid \mathcal{F} {t }\right)}{M {t}}=\frac{E_{P}\left(M_{T} f \mid \mathcal{F} {t}\right)}{E {P}\left(M_ {T} \mid \mathcal{F} {t}\right)} 。证明。(a) 在 \mathcal{F}{t} 上修正 0 \leq t{t} 和 Q_{t} 意味着 M{t}=d Q_{t} / d P_{t}> 0，P-as。和 M_{t} \in L^{1}\left(P, \mathcal{F}{t}\right)，特别是 M{t} 是 \mathcal{F}{t}-根据 Radon-Nikodym 导数的定义，可测量。此外，对于 A \in \mathcal{F}{t} \subseteq \mathcal{F}{T} 我们有 E{P}\left(1_{A} M_{t}\right)=Q(A )=E_{P}\left(1_{A} M_{T}\right)（见 (0)）。这表明 M_{t}=E_{P}\left(M_{T} \mid \mathcal{F}{t}\right)。(b) 设 0 \leq t{T}\right) 和 h=E_{Q}\left(f \mid \mathcal{F}{t}\right)。我们想证明 M{t} h=E_{P}\left(M_{T} f \mid \mathcal{F}{t}\right)。令 A \in \mathcal{F}{t} 并注意 h 是 \mathcal{F}{t}-可测量的并且满足 E{Q}\left(1_{A} h\right )=E_{Q}\left(1_{A} f\right)。使用 1{A} h 的 \mathcal{F}{t}-可测性，1{A} f 和 (0) 的 \mathcal{F}{T}-可测性，我们现在可以写成 E_{P}\left(1_{A} M_{t} h \right)=E_{P}\left(1_{A} M_{T} f\right)。由于 M_{t} h 是 \mathcal{F}{t}-可测量的，这意味着 M{t} h=E_{P}\left(M_{T} f \mid \mathcal{F} {t}\right)，根据需要。评论。回想M{T}=d P_{T} / d Q_{T}, (b) 也可以写成Proof. (a) Fix 0 \leq t{t} and Q_{t} on \mathcal{F}{t} implies that M{t}=d Q_{t} / d P_{t}>0, P-as. and M_{t} \in L^{1}\left(P, \mathcal{F}{t}\right), especially M{t} is \mathcal{F}{t}-measurable, by definition of the Radon-Nikodym derivative. Moreover for A \in \mathcal{F}{t} \subseteq \mathcal{F}{T} we have E{P}\left(1_{A} M_{t}\right)=Q(A)=E_{P}\left(1_{A} M_{T}\right) (see (0)). This shows that M_{t}=E_{P}\left(M_{T} \mid \mathcal{F}{t}\right). (b) Let 0 \leq t{T}\right) and h=E_{Q}\left(f \mid \mathcal{F}{t}\right). We want to show that M{t} h=E_{P}\left(M_{T} f \mid \mathcal{F}{t}\right). Let A \in \mathcal{F}{t} and note that h is \mathcal{F}{t}-measurable and satisfies E{Q}\left(1_{A} h\right)=E_{Q}\left(1_{A} f\right). Using the \mathcal{F}{t}-measurability of 1{A} h, the \mathcal{F}{T}-measurability of 1{A} f and (0), we can now write this as E_{P}\left(1_{A} M_{t} h\right)=E_{P}\left(1_{A} M_{T} f\right). Since M_{t} h is \mathcal{F}{t}-measurable this implies that M{t} h=E_{P}\left(M_{T} f \mid \mathcal{F}{t}\right), as desired. Remark. Recalling that M{T}=d P_{T} / d Q_{T}, (b) can also be written as E_{Q}\left(f \mid \mathcal{F}{t}\right)=\frac{E{P}\left(\left(d Q_{T} / d P_{T}\right) f \mid \mathcal{F}{t}\right)}{E{P}\left(d Q_{T} / d P_{T} \mid \mathcal{F}{t}\right)}, \quad \forall 0 \leq t{T}\right) 。$$

8.b.1 推论感兴趣。适应过程(X吨)是一个问-martingale (Q-local martingale) 当且仅当过程(米吨X吨)是一个磷- 鞅 (磷-局部鞅）。