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

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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

3.a.0. (a) If $X_{t}, Y_{t}$ are both submartingales, then so is the process $Z_{t}=X_{t} \vee Y_{t}$.
(a) If $X_{t}$ is a submartingale, then so is the process $X_{t}^{+}$.
Proof. (a) Let $X_{t}$ and $Y_{t}$ be submartingales and set $Z_{t}=\max \left{X_{t}, Y_{t}\right}$. Then $Z_{t}$ is $\mathcal{F}{t}$-measurable and $Z{t}^{+} \leq X_{t}^{+}+Y_{t}^{+}$, whence $E\left(Z_{t}^{+}\right) \leq E\left(X_{t}^{+}\right)+E\left(Y_{t}^{+}\right)<\infty$. If $s, t \in \mathcal{T}$ with $s \leq t$ then $Z_{t} \geq X_{t}$ and so $E_{\mathcal{F}{x}}\left(Z{t}\right) \geq E_{\mathcal{F}{n}}\left(X{t}\right) \geq X_{s}$. Similarly $E_{\mathcal{F}{n}}\left(Z{t}\right) \geq E_{\mathcal{F}{n}}\left(Y{t}\right) \geq Y_{s}$ and so $E_{\mathcal{F}{n}}\left(Z{t}\right) \geq X_{t} \vee Y_{t}=Z_{t}, P_{\text {-as. }}$ (b) follows from (a), since $X_{t}^{+}=X_{t} \vee 0$. I

## 微积分作业代写calclulus代考|Sampling at optional times.

3.b.1 Baby Optional Sampling Theorem. Let $X_{n}$ be a submartingale and $S, T$ bounded optional times with $S \leq T$. Then $X_{S}, X_{T} \in \mathcal{E}(P)$ and $E\left(X_{S} 1_{A}\right) \leq$ $E\left(X_{T} 1_{A}\right)$, for all sets $A \in \mathcal{F}{S}$, that is, $X{S} \leq E\left(X_{T} \mid \mathcal{F}{S}\right)$. In particular $E\left(X{S}\right) \leq$ $E\left(X_{T}\right)$

Proof. We have $E\left(X_{n}^{+}\right)<\infty$, for all $n \geq 1$, and so $E\left(X_{S}^{+}\right), E\left(X_{T}^{+}\right)<\infty$, especially $X_{S}, X_{T} \in \mathcal{E}(P)$. The submartingale condition for the stochastic sequence $X$ can be written as
$$E\left(1_{A}\left(X_{k+1}-X_{k}\right)\right) \geq 0, \quad \forall k \geq 1, A \in \mathcal{F}{k}$$ Assume now that $S \leq T \leq N, P$-as., where $N$ is some natural number. For each $\omega \in \Omega$ such that $T(\omega)<+\infty$ (and thus for $P$-ae. $\omega \in \Omega$ ) we have: $$X{T(\omega)}(\omega)-X_{S(\omega)}(\omega)=\sum_{k=S(\omega)}^{T(\omega)-1}\left(X_{k+1}(\omega)-X_{k}(\omega)\right)$$
The bounds in this sum depend on $\omega$. The boundedness $P([T \leq N])=1$ can be used to rewrite this with bounds independent of $\omega$ :
$$X_{T}-X_{S}=\sum_{k=1}^{N} 1_{[S \leq k<T]}\left(X_{k+1}-X_{k}\right)$$
Consequently, if $A \in \mathcal{F}{S}$ is any set, then $$1{A}\left(X_{T}-X_{S}\right)=\sum_{k=1}^{N} 1_{A \cap[S \leq k<T]}\left(X_{k+1}-X_{k}\right), P \text {-as. }$$
As $A \in \mathcal{F}{S}, A \cap[S \leq k] \in \mathcal{F}{k}$ and so the set $A \cap[S \leq k<T]=A \cap[S \leq k] \cap[T \leq k]^{c}$ is in $\mathcal{F}{k}$. By the submartingale condition, $E\left(1\right.$ An $\left.[S \leq k{k+1}-X_{k}\right)\right) \geq 0$, for all $k=1,2, \ldots, N$. Taking expectations in $(0)$ now yields $E\left(1_{A}\left(X_{T}-X_{S}\right) \geq 0\right.$, as desired.

3.a.0。(a) 如果X吨,和吨都是下鞅，那么过程也是和吨=X吨∨和吨.
(a) 如果X吨是下鞅，那么过程也是X吨+.

## 微积分作业代写calclulus代考|Sampling at optional times.

3.b.1 婴儿可选抽样定理。让Xn是一个亚鞅并且小号,吨有界可选时间小号≤吨. 然后X小号,X吨∈和(磷)和和(X小号1一种)≤ 和(X吨1一种), 对于所有集合 $A \in \mathcal{F} {S},吨H一种吨一世s,X {S} \leq E\left(X_{T} \mid \mathcal{F} {S}\right).一世np一种r吨一世C你一世一种rE\left(X {S}\right) \leqE\left(X_{T}\right)$

$$E\left(1_{A}\left(X_{k+1}-X_{k}\right)\right) \geq 0, \quad \forall k \geq 1, A \in \mathcal{F} {k}一种ss你米和n○在吨H一种吨小号≤吨≤ñ,磷−一种s.,在H和r和ñ一世ss○米和n一种吨你r一种一世n你米b和r.F○r和一种CHω∈Ωs你CH吨H一种吨吨(ω)<+∞(一种nd吨H你sF○r磷−一种和.ω∈Ω)在和H一种v和:X {T(\omega)}(\omega)-X_{S(\omega)}(\omega)=\sum_{k=S(\omega)}^{T(\omega)-1}\left( X_{k+1}(\omega)-X_{k}(\omega)\right) 吨H和b○你nds一世n吨H一世ss你米d和p和nd○nω.吨H和b○你nd和dn和ss磷([吨≤ñ])=1C一种nb和你s和d吨○r和在r一世吨和吨H一世s在一世吨Hb○你nds一世nd和p和nd和n吨○Fω: X_{T}-X_{S}=\sum_{k=1}^{N} 1_{[S \leq k<T]}\left(X_{k+1}-X_{k}\right)   因此，如果 A \in \mathcal{F} {S}一世s一种n和s和吨,吨H和n 1 {A}\left(X_{T}-X_{S}\right)=\sum_{k=1}^{N} 1_{A \cap[S \leq k<T]}\left(X_ {k+1}-X_{k}\right), P \text {-as. }$$