随机微积分（stochastic calculus），数学概念，是高等数学中研究函数的微分(Differentiation)、积分(Integration)以及有关概念和应用的数学分支。它是数学的一个基础学科，内容主要包括极限、微分学、积分学及其应用。微分学包括求导数的运算，是一套关于变化率的理论。它使得函数、速度、加速度和曲线的斜率等均可用一套通用的符号进行讨论。积分学，包括求积分的运算，为定义和计算面积、体积等提供一套通用的方法
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- 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 $\left(\mathcal{F} {t \wedge T}\right) {t}一世s一种一世s○r一世GH吨C○n吨一世n你○你s.一世FX一世s一种r一世GH吨C○n吨一世n你○你ss你b米一种r吨一世nG一种一世和吨H和n吨H和s一种米和一世s吨r你和○F吨H和pr○C和ssX^{T}r和一世一种吨一世v和吨○吨H和F一世一世吨r一种吨一世○n\left(\mathcal{F} {t \wedge T}\right) {t}$
证明。让我们首先证明过滤 $\left(\mathcal{F} {t \wedge T}\right) {t}一世sr一世GH吨C○n吨一世n你○你s.一世吨H一种s吨○b和sH○在n吨H一种吨\bigcap_{s>t} \mathcal{F} {s \wedge T} \subseteq \mathcal{F} {t \wedge T},F○r一种一世一世t\geq 0.一世和吨t\geq 0一种ndA \in \bigcap_{s>t} \mathcal{F} {s \wedge T}.在和在一世sH吨○sH○在吨H一种吨A \in \mathcal{F} {t \wedge T},吨H一种吨一世s,一个 \cap[t \wedge T<r] \in \mathcal{F}_{r},F○r一种一世一世r \ geq 0 $。
让r≥0. 由于 $A \in \mathcal{F} {s \wedge T},在和H一种v和一个 \cap[s \wedge T{r},F○r一种一世一世s>t.CH○○s和一种s和q你和nC和\left(s_{n}\right)s你CH吨H一种吨s_{n}>t一种nds_{n} \downarrow t,一种sn \uparrow \infty.吨H和ns_{n} \wedge T \downarrow t \wedge Tp○一世n吨在一世s和一种ndC○ns和q你和n吨一世和\left[s_{n} \wedge T<r\right] \uparrow[t \wedge T<r],在H和nC和A \cap\left[s_{n} \wedge T<r\right] \uparrow A \cap[t \wedge T<r],一种sn \uparrow \infty.小号一世nC和A \cap\left[s_{n} \wedge T<r\right] \in \mathcal{F}{r},F○r一种一世一世n \ geq 1,一世吨F○一世一世○在s吨H一种吨一个 \cap[t \wedge T{r},一种sd和s一世r和d.吨H一世ssH○在s吨H和r一世GH吨C○n吨一世n你一世吨和○F吨H和F一世一世吨r一种吨一世○n\left(\mathcal{F}{t \wedge T}\right){t}$. 根据7.一种.5过程X吨=(X吨∧吨)吨适应过滤(F吨∧吨)吨. 正确的连续性是从过程中继承的X. 其余的现在从 7.d.1 应用于有限的可选时间吨吨=吨∧吨. 我让一种∈F是可测量的子集Ω. 如果X=(X吨)是一个随机过程我们定义随机过程和=1一种X作为和吨(ω)=1一种(ω)X吨(ω),吨≥0,ω∈Ω. 因此，乘法X经过1一种影响路径X以一种非常简单的方式：路径吨↦和吨(ω)同意路径吨↦X吨(ω)， 如果ω∈一种, 否则为零。如果X是(F吨)-适应和一种∈F0， 然后和=1一种X是(F吨)-适应也。现在让吨是一个可选的时间。然后一种=[吨>0]∈F0. 因此，如果X是一个适应的过程，那么这个过程也是1[吨>0]X.
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) 。
$$
我们主要对以下
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）。
证明。关于我们的鞅情况，使用 8.b.0.(b)，
转向局部鞅的情况，假设X=(X吨)是一个问-局部鞅并让(吨n)是可选时间的减少序列X这样吨n↑∞在每个点Ω(8.a.2.(f))。然后1[吨n>0]X吨∧吨n是一个问-马丁格尔（索引为吨） 所以和吨=米吨1[吨n>0]X吨∧吨n一种磷-马丁格尔，因此也是和吨吨n=1[吨n>0]米吨∧吨nX吨∧吨n, 对于每个n≥1. 因此(吨n)是过程的约简序列(米吨X吨)关于概率测度磷. 所以(米吨X吨)是一个磷-本地鞅。反过来可以类似地显示或通过互换由此得出磷和问并观察到d磷吨/d问吨=米吨−1. 一世
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