# 随机微积分作业代写stochastic calculus代考| ONE DIMENSIONAL BROWNIAN MOTION

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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|One dimensional Brownian motion starting at zero

2.a.0. The process $B=\left(B_{t}\right){t \geq 0}$ is a Brownian motion if and only if (a) $B{0}=0$ almost surely.
(b) For $0 \leq s<t$ the increment $B_{t}-B_{s}$ is normal with mean zero and variance $t-s: B_{t}-B_{s} \sim N(0, t-s) .$
(c) For all $0 \leq t_{1}<t_{2}<\ldots<t_{n}$ the variables $B_{t_{1}}, B_{t_{2}}-B_{t_{1}}, \ldots, B_{t_{n}}-B_{t_{n-1}}$ are independent.
(d) For every $\omega \in \Omega$ the path $t \in[0, \infty) \mapsto B_{t}(\omega) \in R$ is continuous.
Proof. $(\Rightarrow)$. Assume first that $B$ is a Brownian motion starting at zero. Because of $(\beta)$ and $(\delta)$ we have to show only that (b) and (c) hold.

Let $0 \leq s<t$. According to $(\gamma)$ we have $E\left(B_{s}\right)=E\left(B_{t}\right)=0, E\left(B_{s}^{2}\right)=$ $E\left(B_{s} B_{s}\right)=s \wedge s=s$, similarly $E\left(B_{t}^{2}\right)=t$ and finally $E\left(B_{s} B_{t}\right)=s$. Moreover, according to $(1),\left(B_{s}, B_{t}\right)$ is a two dimensional Gaussian variable and hence its linear image $B_{t}-B_{s}$ is a one dimensional normal variable with mean $E\left(B_{t}\right)-E\left(B_{s}\right)=0$ and variance $\operatorname{Var}\left(B_{t}-B_{s}\right)=E\left[\left(B_{t}-B_{s}\right)^{2}\right]=E\left[B_{t}^{2}-2 B_{t} B_{s}+B_{s}^{2}\right]=t-s$. Thus $B_{t}-B_{s} \sim N(0, t-s)$. This shows (b).
(c) Let $0 \leq t_{1}<\ldots<t_{n}$. By $(\alpha)$ the random vector $\left(B_{t_{1}}, B_{t_{2}}, \ldots, B_{t_{n}}\right)$ is Gaussian and hence so is its linear image $\left(B_{t_{1}}, B_{t_{2}}-B_{t_{1}}, \ldots, B_{t_{n}}-B_{t_{n-1}}\right)$. According to $1 . a .5$ the independence of the variables $B_{t_{1}}, B_{t_{2}}-B_{t_{1}}, \ldots, B_{t_{n}}-B_{t_{n-1}}$ follows if we can show that they are pairwise uncorrelated, that is, $E\left[B_{t_{1}}\left(B_{t_{j}}-B_{t_{j-1}}\right)\right]=0$ and $E\left[\left(B_{t_{j}}-B_{t_{j-1}}\right)\left(B_{t_{k}}-B_{t_{k-1}}\right)\right]=0$, for all $j \neq k$ (recall that all these variables have mean zero). Indeed, we may assume $k<j$ and thus $t_{k-1}<t_{k} \leq t_{j-1}<t_{j}$. It follows that $E\left[B_{t_{1}}\left(B_{t_{j}}-B_{t_{j-1}}\right)\right]=E\left(B_{t_{1}} B_{t_{j}}\right)-E\left(B_{t_{1}} B_{t_{j-1}}\right)=t_{1} \wedge t_{j}-t_{1} \wedge t_{j-1}=$ $t_{1}-t_{1}=0$ and similarly
$$E\left[\left(B_{t_{j}}-B_{t_{j-1}}\right)\left(B_{t_{k}}-B_{t_{k-1}}\right)\right]=t_{k}-t_{k}-t_{k-1}+t_{k-1}=0 .$$
$(\Leftarrow)$. Assume now that (a)-(d) are satisfied. We must verify $(\alpha)-(\delta)$. In view of (a) and (d) we need to show only $(\alpha)$ and $(\gamma)$. Note first that $B_{t}=B_{t}-B_{0}$ is a normal variable, according to (a) and (b). Let now $0 \leq t_{1}<\ldots<t_{n}$. According to (b), (c)

and the preceding remark, $B_{t_{1}}, B_{t_{2}}-B_{t_{1}}, \ldots, B_{t_{n}}-B_{t_{n-1}}$ are independent normal variables. It follows that $\left(B_{t_{1}}, B_{t_{2}}-B_{t_{1}}, \ldots, B_{t_{n}}-B_{t_{n-1}}\right)$ is a Gaussian vector (1.a.4) and hence so is its linear image $\left(B_{t_{1}}, B_{t_{2}}, \ldots, B_{t_{n}}\right)$. Thus $B$ is a Gaussian process. This shows $(\alpha)$.

To show $(\gamma)$, note now that (a) and (b) with $s=0$ imply that $B_{t} \sim N(0, t)$ and thus $E\left(B_{t}\right)=0$ and $E\left(B_{t}^{2}\right)=t$. Let $0 \leq s \leq t$. Then $E\left(B_{s} B_{t}\right)=E\left[B_{s}\left(B_{t}-B_{s}\right)\right]+$ $E\left(B_{s}^{2}\right)=E\left(B_{s}\right) E\left(B_{t}-B_{s}\right)+s=s$, where we have used the independence of $B_{s}$ and $B_{t}-B_{s}$, according to (c). This shows $(\gamma)$. I

Remark. Condition (b) implies that the increments $B_{t}-B_{s}$ are stationary, that is, the distribution of this increment depends on $s, t$ only through $t-s$. From (a) and (b) it follows that $B_{t} \sim N(0, t)$, for all $t>0$. Regarding the distribution $N(0,0)$ as the point measure $\epsilon_{0}$ concentrated at zero, this relation holds for $t=0$ also.

Brownian motion on $\left(\Omega, \mathcal{F},\left(\mathcal{F}{t}\right), P\right)$. The reader will note that no filtration is involved in the definition of a Brownian motion. However condition (c), the “independence of increments”, can be shown to be equivalent with the independence of $B{t}-B_{s}$ from the $\sigma$-field $\mathcal{F}{s}^{0}=\sigma\left(B{r} ; r \leq s\right)$, for all $0 \leq s<t$ (2.e.1 below). This motivates the following terminology: the process $B$ will be called a Brownian motion on the filtered probability space $\left(\Omega, \mathcal{F},\left(\mathcal{F}{t}\right), P\right)$ if it is adapted, satisfies conditions (a), (b), (d) of $2 . \mathrm{a} .0$ and the increment $B{t}-B_{s}$ is independent of the $\sigma$ field $\mathcal{F}{s}$, for all $0 \leq s{t}^{0}\right), P\right)$, where $\mathcal{F}{t}^{0}=\sigma\left(B{r} ; r \leq t\right)$. Let us now show that a Brownian motion starting at zero exists. We need the following

## 微积分作业代写calclulus代考|Miscellaneous properties

2.g.0. (a) $B_{t}-B_{s}$ is independent of the $\sigma$-field $\mathcal{F}{s}$, for all $0 \leq s{t}^{2}-t$ is a martingale and so $\langle B\rangle_{t}=t, t \geq 0$, under all measures $P_{x}, x \in R^{d}$.
Proof. Recall that $P=Q=P_{0}$ is the Wiener measure. Since the distribution of $B$ under $P_{x}$ is the distribution of $B^{x}=x+B$ under $P$ (2.c. 0 ) it will suffice to verify (a)-(c) for the process $B^{x}$ relative to the probability measure $P$ and it is now easy to see that we may assume $x=0 \in R^{d}$.
(a) $B_{t}-B_{s}$ is independent of the $\sigma$-field $\mathcal{F}{s}^{0}$ (2.e.1). According to B.10.2 this implies that $B{t}-B_{s}$ is independent of $\mathcal{F}{s}=\sigma\left(\mathcal{F}{s}^{0} \cup \mathcal{N}\right)$.
(b) Let $0 \leq s<t$ and recall that $E\left(B_{t}\right)=E\left(B_{s}\right)=0$. Writing $B_{t}=B_{s}+\left(B_{t}-B_{s}\right)$ and using (a) and I.2.b.2 it follows that $E\left(B_{t} \mid \mathcal{F}{s}\right)=B{s}+E\left(B_{t}-B_{s}\right)=B_{s}, P$-as. Thus $B$ is a $P$-martingale.
(c) $A_{t}=\langle B\rangle_{t}$ is the unique continuous increasing process $A$ such that $B^{2}-A$ is a local martingale. It will thus suffice to show that the process $B_{t}^{2}-t$ is also a martingale. To see this, we have to show that $E\left[B_{t}^{2}-t \mid \mathcal{F}{s}\right]=B{s}^{2}-s$, equivalently $E\left[B_{t}^{2}-B_{s}^{2} \mid \mathcal{F}{s}\right]=t-s$, for all $0 \leq s{t}^{2}-B_{s}^{2} \mid \mathcal{F}{s}\right]=E\left[\left(B{t}-B_{s}\right)^{2} \mid \mathcal{F}{s}\right]=E\left[\left(B{t}-B_{s}\right)^{2}\right]=\operatorname{Var}\left(B_{t}-B_{s}\right)=t-s .$
Here the first equality follows from the martingale property of $B$ (I.9.b.0), the second equality from the independence of the increment $B_{t}-B_{s}$ from the $\sigma$-field $\mathcal{F}{s}$ and the third equality from the fact that $B{t}-B_{s}$ is a normal variable with mean zero and variance $t-s$.
Remark. Since $E\left(B_{t}^{2}\right)=t, B$ is not $L^{2}$-bounded. However if $B$ is stopped at any bounded optional time $T$, then it becomes $L^{2}$-bounded, that is, $B_{t}^{T}=B_{t \wedge T}$ is an $L^{2}$-bounded martingale. Likewise, one dimensional Brownian motion is not uniformly integrable; indeed the family $\left{\left|B_{t}\right|: t \geq 0\right}$ is not $L^{1}$-bounded.

## 微积分作业代写calclulus代考|One dimensional Brownian motion starting at zero

2.a.0。过程乙=(乙吨)吨≥0是布朗运动当且仅当 (a)乙0=0几乎可以肯定。
(b) 为0≤s<吨增量乙吨−乙s是正常的，均值为零和方差吨−s:乙吨−乙s∼ñ(0,吨−s).
(c) 对所有人0≤吨1<吨2<…<吨n变量乙吨1,乙吨2−乙吨1,…,乙吨n−乙吨n−1是独立的。
(d) 对于每个ω∈Ω路径吨∈[0,∞)↦乙吨(ω)∈R是连续的。

(c) 让0≤吨1<…<吨n. 经过(一种)随机向量(乙吨1,乙吨2,…,乙吨n)是高斯的，因此它的线性图像也是(乙吨1,乙吨2−乙吨1,…,乙吨n−乙吨n−1). 根据1.一种.5变量的独立性乙吨1,乙吨2−乙吨1,…,乙吨n−乙吨n−1如果我们可以证明它们是成对不相关的，那么和[乙吨1(乙吨j−乙吨j−1)]=0和和[(乙吨j−乙吨j−1)(乙吨到−乙吨到−1)]=0， 对所有人j≠到（回想一下，所有这些变量的均值为零）。事实上，我们可以假设到<j因此吨到−1<吨到≤吨j−1<吨j. 它遵循和[乙吨1(乙吨j−乙吨j−1)]=和(乙吨1乙吨j)−和(乙吨1乙吨j−1)=吨1∧吨j−吨1∧吨j−1= 吨1−吨1=0同样地

(⇐). 现在假设满足（a）-（d）。我们必须验证(一种)−(d). 鉴于 (a) 和 (d) 我们只需要显示(一种)和(C). 首先注意乙吨=乙吨−乙0是一个正态变量，根据 (a) 和 (b)。现在让0≤吨1<…<吨n. 根据（b），（c）

## 微积分作业代写calclulus代考|Miscellaneous properties

2.g.0。（一种）乙吨−乙s独立于σ-field $\mathcal{F} {s},F○r一种一世一世0 \leq s{t}^{2}-t一世s一种米一种r吨一世nG一种一世和一种nds○\langle B\rangle_{t}=t, t \geq 0,你nd和r一种一世一世米和一种s你r和sP_{x}, x \in R^{d}.磷r○○F.R和C一种一世一世吨H一种吨P=Q=P_{0}一世s吨H和在一世和n和r米和一种s你r和.小号一世nC和吨H和d一世s吨r一世b你吨一世○n○F乙你nd和rP_{x}一世s吨H和d一世s吨r一世b你吨一世○n○FB^{x}=x+B你nd和r磷(2.C.0)一世吨在一世一世一世s你FF一世C和吨○v和r一世F和(一种)−(C)F○r吨H和pr○C和ssB^{x}r和一世一种吨一世v和吨○吨H和pr○b一种b一世一世一世吨和米和一种s你r和磷一种nd一世吨一世sn○在和一种s和吨○s和和吨H一种吨在和米一种和一种ss你米和x=0 \in R^{d}.(一种)B_{t}-B_{s}一世s一世nd和p和nd和n吨○F吨H和\西格玛−F一世和一世d\mathcal{F}{s}^{0}(2.和.1).一种CC○rd一世nG吨○乙.10.2吨H一世s一世米p一世一世和s吨H一种吨B{t}-B_{s}一世s一世nd和p和nd和n吨○F\mathcal{F}{s}=\sigma\left(\mathcal{F}{s}^{0} \cup \mathcal{N}\right).(b)一世和吨0 \leq s<t一种ndr和C一种一世一世吨H一种吨E\left(B_{t}\right)=E\left(B_{s}\right)=0.在r一世吨一世nGB_{t}=B_{s}+\left(B_{t}-B_{s}\right)一种nd你s一世nG(一种)一种nd一世.2.b.2一世吨F○一世一世○在s吨H一种吨E\left(B_{t} \mid \mathcal{F}{s}\right)=B{s}+E\left(B_{t}-B_{s}\right)=B_{s}, P−一种s.吨H你s乙一世s一种磷−米一种r吨一世nG一种一世和.(C)A_{t}=\langle B\rangle_{t}一世s吨H和你n一世q你和C○n吨一世n你○你s一世nCr和一种s一世nGpr○C和ss一种s你CH吨H一种吨B^{2}-A一世s一种一世○C一种一世米一种r吨一世nG一种一世和.一世吨在一世一世一世吨H你ss你FF一世C和吨○sH○在吨H一种吨吨H和pr○C和ssB_{t}^{2}-t一世s一种一世s○一种米一种r吨一世nG一种一世和.吨○s和和吨H一世s,在和H一种v和吨○sH○在吨H一种吨E\left[B_{t}^{2}-t \mid \mathcal{F}{s}\right]=B{s}^{2}-s,和q你一世v一种一世和n吨一世和E\left[B_{t}^{2}-B_{s}^{2} \mid \mathcal{F}{s}\right]=ts,F○r一种一世一世0 \leq s{t}^{2}-B_{s}^{2} \mid \mathcal{F}{s}\right]=E\left[\left(B{t}-B_{s} \right)^{2} \mid \mathcal{F}{s}\right]=E\left[\left(B{t}-B_{s}\right)^{2}\right]=\operatorname {Var}\left(B_{t}-B_{s}\right)=ts 。H和r和吨H和F一世rs吨和q你一种一世一世吨和F○一世一世○在sFr○米吨H和米一种r吨一世nG一种一世和pr○p和r吨和○F乙(一世.9.b.0),吨H和s和C○nd和q你一种一世一世吨和Fr○米吨H和一世nd和p和nd和nC和○F吨H和一世nCr和米和n吨B_{t}-B_{s}Fr○米吨H和\西格玛−F一世和一世d\mathcal{F}{s}一种nd吨H和吨H一世rd和q你一种一世一世吨和Fr○米吨H和F一种C吨吨H一种吨B{t}-B_{s}一世s一种n○r米一种一世v一种r一世一种b一世和在一世吨H米和一种n和和r○一种ndv一种r一世一种nC和ts.R和米一种r到.小号一世nC和E\left(B_{t}^{2}\right)=t, B一世sn○吨L^{2}−b○你nd和d.H○在和v和r一世F乙一世ss吨○pp和d一种吨一种n和b○你nd和d○p吨一世○n一种一世吨一世米和吨,吨H和n一世吨b和C○米和sL^{2}−b○你nd和d,吨H一种吨一世s,B_{t}^{T}=B_{t \wedge T}一世s一种nL^{2}−b○你nd和d米一种r吨一世nG一种一世和.一世一世到和在一世s和,○n和d一世米和ns一世○n一种一世乙r○在n一世一种n米○吨一世○n一世sn○吨你n一世F○r米一世和一世n吨和Gr一种b一世和;一世nd和和d吨H和F一种米一世一世和\left{\left|B_{t}\right|: t \geq 0\right}一世sn○吨L^{1}$-有界。