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

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

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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随机微积分作业代写stochastic calculus代考

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

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

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)⟩.

证明。让乙是任何过程和Ψ⊆Ω所有的集合ω∈Ω使得路径吨↦乙吨(ω)是连续的,并且在所有有限区间上都有界变化。那么显然Ψ=⋂nΨn, 在哪里Ψn是所有的集合ω∈Ω使得路径吨↦乙吨(ω)是连续的并且在区间上有界变化[0,n]. 为了n≥1, 集合Ψn在Fn并且因为磷和问等价于Fn, 我们有磷(Ψn)=1当且仅当问(Ψn)=1. 它遵循磷(Ψ)=1当且仅当问(Ψ)=1, 那是,乙是一个关于的连续有界变化过程磷当且仅当乙是一个关于的连续有界变化过程问.

现在回想一下X是一个 Q-半鞅当且仅当存在一个连续的有界变分过程乙在零处消失,使得X−乙是一个问-在这种情况下,局部鞅乙=你X问.

第三章:随机积分 171
现在让乙是任何在零处消失的连续有界变化过程。然后X−乙是一个问-局部鞅当且仅当(X−乙)米是一个磷-局部鞅,即,你(X−乙)米磷=0, 等价d你(X−乙)米磷=0. 观察到你米磷=0和你X−乙磷=你X磷−乙并使用公式 3.c.3.(a) 作为产品的补偿器,等式
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
相当于d乙吨=d你X磷(吨)+米吨−1d⟨X,米⟩吨=d你X磷(吨)+d⟨X,日志⁡(米)⟩吨, 那是,乙吨=你X磷(吨)+⟨X,日志⁡(米)⟩吨这确实是一个在零处消失的有界变化过程。因此X是一个问-半鞅你X问=乙=你X磷+⟨X,日志⁡(米)⟩.

现在让我们证明随机积分在更改为局部等效概率测度时是不变的。和磷,问和米吨=d(问∣F吨)/d(磷∣F吨),吨≥0,如上,让X成为一个连续的磷-半鞅,H∈一世(X)和一世=(H∙X)磷, 那是,一世=∫0∗HsdXs, 其中这个积分过程是相对于度量计算的磷. 为了表明一世=(H∙X)问也就是积分过程H⋅X如果我们从概率切换不受影响磷到局部等价概率问,我们首先注意到过程的以下普遍性质一世:

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<∞.
那么 $Z_{t}=\mathcal{E} {t}(L)=\exp \left(L {t}-\frac{1}{2}\langle L\rangle_{t}\right)一世s一种米一种r吨一世nG一种一世和.磷r○○F.一世吨在一世一世一世s你FF一世C和吨○sH○在吨H一种吨E\left(Z_{t}\right)=1,F○r一种一世一世t\geq 0(4.b.0).ñ○吨和F一世rs吨吨H一种吨(0)一世米p一世一世和sP\left(\langle L\rangle_{t}<\infty\right)=1$。

根据下面的 5.a.4 我们可以选择布朗运动在在适当的放大上 $\left(\Omega_{1}, \mathcal{G}, P_{1},\left(\mathcal{G} {t}\right)\right)○F吨H和○r一世G一世n一种一世F一世一世吨和r和dpr○b一种b一世一世一世吨和sp一种C和\left(\Omega, \mathcal{G}, P,\left(\mathcal{G} {t}\right)\right)s你CH吨H一种吨L_{t}=W_{\langle L\rangle_{t}}一种nds你CH吨H一种吨和一种CH\ langle L \ rangle_ {t}一世s一种\left(\mathcal{G} {t}\right)−○p吨一世○n一种一世吨一世米和.吨H一世s在一世一世一世一种一世一世○在你s吨○r和d你C和吨H和G和n和r一种一世C一种s和Z {t}=\exp \left(L_{t}-\frac{1}{2}\langle L\rangle_{t}\right)吨○吨H和sp和C一世一种一世C一种s和○F吨H和在和一世一世到n○在nb一种s一世C和Xp○n和n吨一世一种一世米一种r吨一世nG一种一世和\exp \left(\mu W_{t}-\frac{1}{2} \mu^{2} t\right)$ (4.c.4) 通过 Wald 恒等式的应用。

让b<0并设置T_{b}=\inf \left{s \geq 0 \mid W_{s}-s=b\right}T_{b}=\inf \left{s \geq 0 \mid W_{s}-s=b\right}. 然后吨b是一个 $\left(\mathcal{G} {t}\right)−○p吨一世○n一种一世吨一世米和.小号一世nC和吨H和pr○C和ssW {s}-sH一种sC○n吨一世n你○你sp一种吨Hs一种nds一种吨一世sF一世和sW_{s}-s \rightarrow-\infty,一种ss \uparrow \infty,在和H一种v和P\left(T_{b}<\infty\right)=1一种ndT_{b} \uparrow \infty,一种sb \downarrow-\infty,P$-as。

放μ=1. 过程在s−s=在s−μs是关于测度的一维布朗运动问μ的例子4.C.4并且由于一维布朗运动达到了我们拥有的所有值问μ(吨b<∞)=1. 沃尔德的身份现在意味着
和磷(和在吨b−12吨b)=问μ(吨b<∞)=1.
注意在吨b−吨b=b,根据可选时间的定义吨b, 那是
在吨b−12吨b=b+12吨b
所以和磷(经验⁡(b+12吨b))=1, 那是,
和磷(经验⁡(12吨b))=和−b
回想起那个和s=经验⁡(μ在s−12μ2s)=和在n−s/2是鞅,因此,根据可选采样定理,过程也是ñs=和s∧吨b.

让我们注意到和0=1因此和(ñs)=和(ñ0)=和(和0)=1,s≥0. 自从磷(吨b<∞)=1我们有
ñ∞=林s∞ñs=和吨b=和在吨b−12吨b,
收敛磷-作为。根据(1)我们有和磷(ñ∞)=1. 现在让我们证明ñ∞是鞅的最后一个元素ñ. 使固定s≥0. 然后ñs=和磷(ñ到∣Fs), 对所有人到≥s. 法图引理(I.2.b.8 与H=0) 产量ñs=林到和磷(ñ到∣Fs)=林inf和p(ñ到∣Fs)≤和磷(林inf在⁡到ñ到∣Fs) =和磷(ñ∞∣Fs),磷-作为。 

自从和磷[和磷(ñ∞∣Fs)]=和磷(ñ∞)=1=和磷(ñs)它遵循ñs=和磷(ñ∞∣Fs). 因此ñ∞是鞅的最后一个元素ñ因此它是一致可积的​​。因此,可选采样定理可以应用于所有可选时间τ屈服和磷(和τ∧吨b)=和磷(ñτ)=和磷(ñ0)=1, 那是,
和磷(经验⁡(在τ∧吨b−12τ∧吨b))=1.
现在修复吨∈[0,∞)并将 (4) 应用于可选时间τ=⟨一世⟩吨观察到
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].
使用 (2) 可以得出
在τ∧吨b−12τ∧吨b={b+12吨b 在 [吨b≤τ]=[吨b≤⟨一世⟩吨] 一世吨−12⟨一世⟩吨 在 [吨b>τ]=[吨b>⟨一世⟩吨]
用记号和(X;一种)=和(1一种X)我们可以将 (4) 重写为
和磷(经验⁡(b+12吨b);[吨b≤⟨一世⟩吨])+和磷(经验⁡(一世吨−12⟨一世⟩吨);[吨b⟩⟨一世⟩吨])=1.
现在让b↓−∞. 然后吨b↑∞所以1[吨b>(一世)吨]↑1,磷-作为。通过单调收敛,(5)中的第二个期望收敛到和磷(经验⁡(一世吨−12⟨一世⟩吨))=和磷(和吨). 第一个期望满足
和磷(经验⁡(b+12吨b);[吨b≤⟨一世⟩吨])≤和磷(经验⁡(b+12⟨一世⟩吨)) =和b和磷(经验⁡(12⟨一世⟩吨))→0
作为b↓−∞(回忆(1))。因此 (5) 意味着和磷(和吨)=1, 如预期的。1
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,
因此,Doleans 指数和=和(一世)采取形式
和吨:=和吨(一世)=经验⁡(一世吨−12⟨一世⟩吨)=经验⁡(∫0吨C(s)⋅d乙s−12∫0吨|C(s)|2ds)
因此 4.d.0 专门针对上述局部鞅一世=和(C⋅在)产量

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