# 随机微积分作业代写stochastic calculus代考| CONVERGENCE THEOREMS

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• 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|Upcrossing Lemma

$$E\left(U_{N}(\alpha, \beta)\right) \leq \frac{E\left(X_{N}^{+}\right)+|\alpha|}{\beta-\alpha}$$
Proof. For each $n \geq 1$ set $Y_{n}=\left(X_{n}-\alpha\right)^{+}$. Then $Y_{n} \geq 0$ is a submartingale and we have
$$X_{k} \leq \alpha \Longleftrightarrow Y_{k} \leq 0 \text { and } X_{k} \geq \beta \Longleftrightarrow Y_{k} \geq \beta-\alpha \text {. }$$
Thus the upcrossings of $(\alpha, \beta)$ by $\left(X_{n}(\omega)\right)$ happen at exactly the same times as the upcrossings of $(0, \beta-\alpha)$ by $\left(Y_{n}(\omega)\right)$. Setting $T_{0}=1$ and recalling that $T_{N}=N$ we can write
$$Y_{N} \geq Y_{N}-Y_{1}=\sum_{k=1}^{N}\left(Y_{T_{k}}-Y_{T_{k-1}}\right)=\sum_{k=1}^{N}\left(Y_{T_{k}}-Y_{S_{k}}\right)+\sum_{k=1}^{N}\left(Y_{S_{k}}-Y_{T_{k-1}}\right) .$$
If $S_{k}(\omega)<N$, then $Y_{S_{k}(\omega)}(\omega)=0$, and if $S_{k}(\omega)=N$, then $T_{k}(\omega)=N$ also. In any case we have $\left(Y_{T_{k}}-Y_{S_{k}}\right)(\omega) \geq 0$, for all $k=1,2, \ldots, N$. Moreover $U_{N}(\alpha, \beta)(\omega) \leq N$ and $\left(Y_{T_{k}}-Y_{S_{k}}\right)(\omega) \geq \beta-\alpha$, for all $k=1,2, \ldots, U_{N}(\alpha, \beta)(\omega)$, according to 4.a.1.(c). Thus $\sum_{k=1}^{N}\left(Y_{T_{k}}-Y_{S_{k}}\right) \geq(\beta-\alpha) U_{N}(\alpha, \beta)$ at each point of $\Omega$ and consequently
$$Y_{N} \geq(\beta-\alpha) U_{N}(\alpha, \beta)+\sum_{k=1}^{N}\left(Y_{S_{k}}-Y_{T_{k-1}}\right) .$$
Taking expectations we find that
$$E\left(Y_{N}\right) \geq(\beta-\alpha) E\left(U_{N}(\alpha, \beta)\right)+\sum_{k=1}^{N}\left(E\left(Y_{S_{k}}\right)-E\left(Y_{T_{k-1}}\right)\right) .$$
Applying the Optional Sampling Theorem 3.b.1 to the submartingale $\left(Y_{n}\right)$ and bounded optional times $T_{k-1} \leq S_{k}$ yields $E\left(Y_{S_{k}}\right) \geq E\left(Y_{T_{k-1}}\right)$, for all $k=1, \ldots, N$. We conclude that
$$E\left(Y_{N}\right) \geq(\beta-\alpha) E\left(U_{N}(\alpha, \beta)\right)$$
equivalently, that $E\left(U_{N}(\alpha, \beta)\right) \leq E\left(Y_{N}\right) /(\beta-\alpha)$. It remains to be shown merely that $E\left(Y_{N}\right) \leq E\left(X_{N}^{+}\right)+|\alpha|$. This follows immediately from $Y_{N}=\left(X_{N}-\alpha\right)^{+} \leq$ $X_{N}^{+}+|\alpha|$ upon taking expectations. $\boldsymbol{I}$

## 微积分作业代写calclulus代考|Upcrossings

4.a.0. Let $N \geq 1, B$ a Borel set and $S$ an optional time. Set
$$T(\omega)=N \wedge \inf \left{1 \leq kS(\omega) \text { and } X_{k}(\omega) \in B\right}$$
Then $T$ is an optional time.
Remark. Recall the convention $\inf (\emptyset)=+\infty$. It follows that $T(\omega)=N$, if there does not exist $k$ such that $S(\omega)<k<N$ and $X_{k}(\omega) \in B$, and $T(\omega)$ is the smallest such $k$, especially $T(\omega)<N$, if such $k$ exists.

Proof. Since the sequence $X$ is $\left(\mathcal{F}{n}\right)$-adapted, we have $\left[X{k} \in B\right] \in \mathcal{F}{k}$, for all $k \geq 1$. If $n \geq N$, we have $[T \leq n]=\Omega \in \mathcal{F}{n}$. Let now $nj$ such that $X_{n}(\omega) \geq \beta$ and the difference $k-j$ is maximal subject to these constraints. In other words $j{n}\right)$-optional. For $\omega \in \Omega$ set \begin{aligned} &S{1}(\omega)=N \wedge \inf \left{1 \leq kS_{1}(\omega) \text { and } X_{k}(\omega) \geq \beta\right} \end{aligned}
Assuming that $S_{n}(\omega), T_{n}(\omega)$ have already been defined, we set
\begin{aligned} &S_{n+1}(\omega)=N \wedge \inf \left{1 \leq kT_{n}(\omega) \text { and } X_{k}(\omega) \leq \alpha\right} \text { and } \ &T_{n+1}(\omega)=N \wedge \inf \left{1 \leq kS_{n+1}(\omega) \text { and } X_{k}(\omega) \geq \beta\right} \end{aligned}
Note that
$$S_{1}(\omega)<T_{1}(\omega)<S_{2}(\omega)<T_{2}(\omega)<\ldots<\left{\begin{array}{l} S_{j}(\omega)=N=T_{j}(\omega)=S_{j+1}(\omega)=\ldots \ T_{j}(\omega)=N=S_{j+1}(\omega)=T_{j+1}(\omega)=\ldots, \end{array}\right.$$ that is, the sequence $\left(S_{1}, T_{1}, S_{2}, T_{2}, \ldots\right)$ is strictly increasing until one of its terms hits the value $N$, when all the following terms stabilize at $N$. $\operatorname{Moreover} S_{j}(\omega), T_{j}(\omega)$ equal $N$ if and only if the condition for $k$ in the defining infimum can no longer be satisfied. Clearly $S_{N}(\omega)=T_{N}(\omega)=N$.

## 微积分作业代写calclulus代考|Upcrossing Lemma

X到≤一种⟺和到≤0 和 X到≥b⟺和到≥b−一种.

## 微积分作业代写calclulus代考|Upcrossings

4.a.0. 让ñ≥1,乙一个 Borel 集和小号一个可选的时间。放
T(\omega)=N \wedge \inf \left{1 \leq kS(\omega) \text { and } X_{k}(\omega) \in B\right}T(\omega)=N \wedge \inf \left{1 \leq kS(\omega) \text { and } X_{k}(\omega) \in B\right}

S_{1}(\omega)<T_{1}(\omega)<S_{2}(\omega)<T_{2}(\omega)<\ldots<\left{小号j(ω)=ñ=吨j(ω)=小号j+1(ω)=… 吨j(ω)=ñ=小号j+1(ω)=吨j+1(ω)=…,\正确的。
 即序列(小号1,吨1,小号2,吨2,…)严格增加，直到其中一项达到该值ñ, 当以下所有项稳定在ñ.而且⁡小号j(ω),吨j(ω)平等的ñ当且仅当到在定义下确界不能再满足。清楚地小号ñ(ω)=吨ñ(ω)=ñ.