随机微积分作业代写stochastic calculus代考| OPTIONAL SAMPLING OF CLOSED SUBMARTINGALE SEQUENCES

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微积分作业代写calclulus代考|Sampling of closed submartingale sequences

5.b.0. Let $Z \in \mathcal{E}(P)$ with $E\left(Z^{+}\right)<\infty$ and set $X_{n}=E\left(Z \mid \mathcal{F}{n}\right)$ for $1 \leq n \leq \infty$. Then for any two optional times $S, T: \Omega \rightarrow \mathbb{N}{\infty}$ we have
(a) $E\left(X_{S}^{+}\right)<\infty$ and $X_{S}=E\left(Z \mid \mathcal{F}{S}\right)$. (b) $S \leq T$ implies $X{S}=E\left(X_{T} \mid \mathcal{F}{S}\right)$. Remark. If $Z \in L^{1}(P)$, then $\left(X{n}\right)$ is a martingale with last element $Z$.
Proof. (a) We have $X_{n}^{+} \leq E_{\mathcal{F}{n}}\left(Z^{+}\right)$(2.b.12) and so $E\left(1{A} X_{n}^{+}\right) \leq E\left(1_{A} Z^{+}\right)$, for all sets $A \in \mathcal{F}{n}$. It follows that $$E\left(X{S}^{+}\right)=\sum_{1 \leq n \leq \infty} E\left(X_{n}^{+} ;[S=n]\right) \leq \sum_{1 \leq n \leq \infty} E\left(Z^{+} ;[S=n]\right)=E\left(Z^{+}\right)<\infty$$
Thus $X_{S} \in \mathcal{E}(P)$. As $X_{S}$ is $\mathcal{F}{S}$-measurable (3.b.0.(c)), it will now suffice to show that $E\left(1{A} X_{S}\right)=E\left(1_{A} Z\right)$, for all sets $A \in \mathcal{F}{S}$. If $A \in \mathcal{F}{S}$, then $A \cap[S=n] \in \mathcal{F}{n}$, for all $n \in \mathbb{N}{\infty}$, and it follows that
\begin{aligned} E\left(1_{A} X_{S}\right) &=\sum_{1 \leq n \leq \infty} E\left(X_{S} ; A \cap[S=n]\right)=\sum_{1 \leq n \leq \infty} E\left(X_{n} ; A \cap[S=n]\right) \ &=\sum_{1 \leq n \leq \infty} E(Z ; A \cap[S=n])=E\left(1_{A} Z\right) . \end{aligned}
(b) If $S \leq T$, then $\mathcal{F}{S} \subseteq \mathcal{F}{T}$ and so, using (a), $E\left(X_{T} \mid \mathcal{F}{S}\right)=E\left(E\left(Z \mid \mathcal{F}{T}\right) \mid \mathcal{F}{S}\right)=$ $E\left(Z \mid \mathcal{F}{S}\right)=X_{S} \cdot \mathbf{I}$

微积分作业代写calclulus代考|Uniform integrability, last elements, closure

5.a.0. (i) If the submartingale $X=\left(X_{t}\right)$ is closed, then the family $\left{X_{t}^{+} \mid t \in \mathcal{T}\right}$ is uniformly integrable.
(ii) If the martingale $X=\left(X_{t}\right)$ is closed, then $X$ itself is uniformly integrable.
Proof. (i) Let $X=\left(X_{t}\right)$ be a submartingale with a last element. Then so is the process $\left(X_{t}^{+}\right)$, according to 3.a.0. Thus $0 \leq X_{t}^{+} \leq E\left(X_{\infty}^{+} \mid \mathcal{F}{t}\right), t \in \mathcal{T}$, and the uniform integrability of the family $\left{X{t}^{+} \mid t \in \mathcal{T}\right}$ now follows from the uniform integrability of the family $\left{E_{\mathcal{F}{t}}\left(X{\infty}^{+}\right) \mid t \in \mathcal{T}\right}$ (2.b.13). (ii) follows directly from 2.b.13. I

Consider now a submartingale $X=\left(X_{t}, \mathcal{F}{t}\right){t \in T}$, where the index set $\mathcal{T}$ does not have a last element. Choosing $\infty \notin \mathcal{T}$ and decreeing that $t \leq \infty$, for all $t \in \mathcal{T}$, $\mathcal{T}$ can be enlarged to a partially ordered index set $\mathcal{T} \cup{\infty}$ with last element $\infty$. The filtration $\left(\mathcal{F}{t}\right)$ can also be extended by setting $$\mathcal{F}{\infty}=\sigma\left(\bigcup_{t \in T} \mathcal{F}{t}\right)$$ The question is now if the submartingale $X$ can be extended also, that is, if there exists a random variable $X{\infty}$ such that the process $\left(X_{t}, \mathcal{F}{t}\right){t \in T \cup{\infty}}$ is still a submartingale, that is, such that $X_{\infty}$ is $\mathcal{F}{\infty}$-measurable, $E\left(X{\infty}^{+}\right)<\infty$ and $X_{t} \leq E\left(X_{\infty} \mid \mathcal{F}{t}\right)$, for all $t \in \mathcal{T}$. A random variable $X{\infty}$ having these properties will be called a last element for the submartingale $X$. The submartingale $X$ will be called closeable if there exists a last element $X_{\infty}$ for $X$. In this case $\left(X_{t}, \mathcal{F}{t}\right){t \in T \cup{\infty}}$ is a closed submartingale extending $X$.

A last element for the supermartingale $X$ is not uniquely determined: If $X_{\infty}$ is a last element for $X$ and $Z \geq 0$ is $\mathcal{F}{\infty}$-measurable with $E(Z)$ finite, then $X{\infty}+Z$ is also a last element for $X$.

Closeable supermartingales and martingales and last elements for these are defined similarly. Note that $X_{\infty}$ is a last element for the martingale $X$ if and only if $X_{\infty} \in L^{1}(P)$ is $\mathcal{F}{\infty}$-measurable and $X{t}=E\left(X_{\infty} \mid \mathcal{F}{t}\right)$, for all $t \in \mathcal{T}$. Equivalently $X{\infty}$ is a last element for the martingale $X$ if and only if it is a last element for $X$ both as a submartingale and as a supermartingale.

Note for example that each nonnegative supermartingale $X=\left(X_{t}\right)$ has a last element, namely $X_{\infty}=0$. However, if $X$ happens to be a martingale, then $X_{\infty}=0$ will be a last element for the martingale $X$ only if $X_{t}=0$, for all $t \in \mathcal{T}$.

In the case of martingale sequences, that is, $\mathcal{T}=\mathbb{N}$ with the usual order, $\mathcal{F}{\infty}=\sigma\left(\bigcup{n \geq 1} \mathcal{F}_{n}\right)$ and the convergence theorems yield the following:

微积分作业代写calclulus代考|Sampling of closed submartingale sequences

5.b.0。让和∈和(磷)和和(和+)<∞并设置 $X_{n}=E\left(Z \mid \mathcal{F} {n}\right)F○r1\leq n\leq\infty.吨H和nF○r一种n和吨在○○p吨一世○n一种一世吨一世米和sS, T: \Omega \rightarrow \mathbb{N} {\infty}在和H一种v和(一种)E\left(X_{S}^{+}\right)<\infty一种ndX_{S}=E\left(Z \mid \mathcal{F} {S}\right).(b)S \ leq T一世米p一世一世和sX {S}=E\left(X_{T} \mid \mathcal{F} {S}\right).R和米一种r到.一世FZ \ in L ^ {1} (P),吨H和n\left(X {n}\right)一世s一种米一种r吨一世nG一种一世和在一世吨H一世一种s吨和一世和米和n吨和.磷r○○F.(一种)在和H一种v和X_{n}^{+} \leq E_{\mathcal{F} {n}}\left(Z^{+}\right)(2.b.12)一种nds○E\left(1 {A} X_{n}^{+}\right) \leq E\left(1_{A} Z^{+}\right),F○r一种一世一世s和吨s一个 \in \mathcal{F} {n}.一世吨F○一世一世○在s吨H一种吨$ E\left(X {S}^{+}\right)=\sum_{1 \leq n \leq \infty} E\left(X_{n}^{+} ;[S=n]\right) \leq \sum_{1 \leq n \leq \infty} E\left(Z^{+} ;[S=n]\right)=E\left(Z^{+}\right)<\infty
$$因此X小号∈和(磷). 作为X小号是 \mathcal{F} {S}−米和一种s你r一种b一世和(3.b.0.(C)),一世吨在一世一世一世n○在s你FF一世C和吨○sH○在吨H一种吨E\left(1 {A} X_{S}\right)=E\left(1_{A} Z\right),F○r一种一世一世s和吨sA \in \mathcal{F} {S}.一世FA \in \mathcal{F} {S},吨H和nA \cap[S=n] \in \mathcal{F} {n},F○r一种一世一世n \in \mathbb{N} {\infty},一种nd一世吨F○一世一世○在s吨H一种吨和(1一种X小号)=∑1≤n≤∞和(X小号;一种∩[小号=n])=∑1≤n≤∞和(Xn;一种∩[小号=n]) =∑1≤n≤∞和(和;一种∩[小号=n])=和(1一种和).(b)一世FS \ leq T,吨H和n\mathcal{F} {S} \subseteq \mathcal{F} {T}一种nds○,你s一世nG(一种),E\left(X_{T} \mid \mathcal{F} {S}\right)=E\left(E\left(Z \mid \mathcal{F} {T}\right) \mid \mathcal{F } {S}\右）=E\left(Z \mid \mathcal{F} {S}\right)=X_{S} \cdot \mathbf{I} 微积分作业代写calclulus代考|Uniform integrability, last elements, closure 5.a.0. (i) 如果下鞅X=(X吨)关闭，然后是家庭\left{X_{t}^{+} \mid t \in \mathcal{T}\right}\left{X_{t}^{+} \mid t \in \mathcal{T}\right}是一致可积的​​。 (ii) 如果鞅X=(X吨)是关闭的，那么X本身是一致可积的​​。 证明。（我让X=(X吨)是具有最后一个元素的子鞅。那么过程也是如此(X吨+)，根据 3.a.0。因此 0 \leq X_{t}^{+} \leq E\left(X_{\infty}^{+} \mid \mathcal{F} {t}\right), t \in \mathcal{T},一种nd吨H和你n一世F○r米一世n吨和Gr一种b一世一世一世吨和○F吨H和F一种米一世一世和\left{X {t}^{+} \mid t \in \mathcal{T}\right}n○在F○一世一世○在sFr○米吨H和你n一世F○r米一世n吨和Gr一种b一世一世一世吨和○F吨H和F一种米一世一世和\left{E_{\mathcal{F} {t}}\left(X {\infty}^{+}\right) \mid t \in \mathcal{T}\right} (2.b.13) . (ii) 直接来自 2.b.13。一世 现在考虑一个子鞅 X=\left(X_{t}, \mathcal{F} {t}\right) {t \in T},在H和r和吨H和一世nd和Xs和吨\数学{T}d○和sn○吨H一种v和一种一世一种s吨和一世和米和n吨.CH○○s一世nG\infty \notin \mathcal{T}一种ndd和Cr和和一世nG吨H一种吨t \leq \infty,F○r一种一世一世t \in \mathcal{T},\数学{T}C一种nb和和n一世一种rG和d吨○一种p一种r吨一世一种一世一世和○rd和r和d一世nd和Xs和吨\mathcal{T} \cup{\infty}在一世吨H一世一种s吨和一世和米和n吨\infty.吨H和F一世一世吨r一种吨一世○n\left(\mathcal{F} {t}\right)C一种n一种一世s○b和和X吨和nd和db和s和吨吨一世nG \mathcal{F} {\infty}=\sigma\left(\bigcup_{t \in T} \mathcal{F} {t}\right)$$ 现在的问题是如果子鞅X也可以扩展，也就是说，如果存在随机变量 $X {\infty}s你CH吨H一种吨吨H和pr○C和ss\left(X_{t}, \mathcal{F} {t}\right) {t \in T \cup{\infty}}一世ss吨一世一世一世一种s你b米一种r吨一世nG一种一世和,吨H一种吨一世s,s你CH吨H一种吨X_{\infty}一世s\mathcal{F} {\infty}−米和一种s你r一种b一世和,E\left(X {\infty}^{+}\right)<\infty一种ndX_{t} \leq E\left(X_{\infty} \mid \mathcal{F} {t}\right),F○r一种一世一世t \in \mathcal{T}.一种r一种nd○米v一种r一世一种b一世和X {\infty}H一种v一世nG吨H和s和pr○p和r吨一世和s在一世一世一世b和C一种一世一世和d一种一世一种s吨和一世和米和n吨F○r吨H和s你b米一种r吨一世nG一种一世和X.吨H和s你b米一种r吨一世nG一种一世和X在一世一世一世b和C一种一世一世和dC一世○s和一种b一世和一世F吨H和r和和X一世s吨s一种一世一种s吨和一世和米和n吨X_{\infty}F○rX.一世n吨H一世sC一种s和\left(X_{t}, \mathcal{F} {t}\right) {t \in T \cup{\infty}}一世s一种C一世○s和ds你b米一种r吨一世nG一种一世和和X吨和nd一世nGX$。

Closeable supermartingales 和 martingales 以及它们的最后一个元素的定义类似。注意X∞是鞅的最后一个元素X当且仅当X∞∈一世1(磷)是 $\mathcal{F} {\infty}−米和一种s你r一种b一世和一种ndX {t}=E\left(X_{\infty} \mid \mathcal{F} {t}\right),F○r一种一世一世t \in \mathcal{T}.和q你一世v一种一世和n吨一世和X {\infty}一世s一种一世一种s吨和一世和米和n吨F○r吨H和米一种r吨一世nG一种一世和X一世F一种nd○n一世和一世F一世吨一世s一种一世一种s吨和一世和米和n吨F○rX$ 既作为下鞅又作为上鞅。