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

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• 布朗运动 Brownian motion

## 微积分作业代写calclulus代考|a Sigma fields, information and conditional expectation

2.a.0. (a) If $X \in \mathcal{E}(P)$, then $1_{A} X \in \mathcal{E}(P)$, for all sets $A \in \mathcal{F}$.
(b) If $X \in \mathcal{E}(P)$ and $\alpha \in R$, then $\alpha X \in \mathcal{E}(P)$.
(c) If $X_{1}, X_{2} \in \mathcal{E}(P)$ and $E\left(X_{1}\right)+E\left(X_{2}\right)$ is defined, then $X_{1}+X_{2} \in \mathcal{E}(P)$.
Proof. We show only (c). We may assume that $E\left(X_{1}\right) \leq E\left(X_{2}\right)$. If $E\left(X_{1}\right)+E\left(X_{2}\right)$ is defined, then $E\left(X_{1}\right)>-\infty$ or $E\left(X_{2}\right)<\infty$. Let us assume that $E\left(X_{1}\right)>-\infty$ and so $E\left(X_{2}\right)>-\infty$, the other case being similar. Then $X_{1}, X_{2}>-\infty, P$-as. and hence $X_{1}+X_{2}$ is defined $P$-as. Moreover $E\left(X_{1}^{-}\right), E\left(X_{2}^{-}\right)<\infty$ and, since $\left(X_{1}+X_{2}\right)^{-} \leq X_{1}^{-}+X_{2}^{-}$, also $E\left(\left(X_{1}+X_{2}\right)^{-}\right)<\infty$. Thus $X_{1}+X_{2} \in \mathcal{E}(P)$.

## 微积分作业代写calclulus代考|Conditional expectation

2.b.0. A conditional expectation of $X$ given $\mathcal{G}$ exists and is $P$-as. uniquely determined. Henceforth it will be denoted $E(X \mid \mathcal{G})$ or $E_{\mathcal{G}}(X)$.

Proof. Uniqueness. Let $Z_{1}, Z_{2}$ be conditional expectations of $X$ given $\mathcal{G}$. Then $E\left(Z_{1} 1_{A}\right)=E\left(X 1_{A}\right)=E\left(Z_{2} 1_{A}\right)$, for all sets $A \in \mathcal{G}$. It will suffice to show that $P\left(Z_{1}0$. Then $E\left(Z_{1} 1_{A}\right) \leq \alpha P(A)<$ $\beta P(A) \leq E\left(Z_{2} 1_{A}\right)$, a contradiction.

Existence. (i) Assume first that $X \in L^{2}(P)$ and let $L^{2}(\mathcal{G}, P)$ be the space of all equivalence classes in $L^{2}(P)$ containing a $\mathcal{G}$-measurable representative. We claim

that the subspace $L^{2}(\mathcal{G}, P) \subseteq L^{2}(P)$ is closed. Indeed, let $Y_{n} \in L^{2}(\mathcal{G}, P), Y \in L^{2}(P)$ and assume that $Y_{n} \rightarrow Y$ in $L^{2}(P)$. Passing to a suitable subsequence of $Y_{n}$ if necessary, we may assume that $Y_{n} \rightarrow Y, P$-as. Set $\tilde{Y}=\lim \sup {n} Y{n}$. Then $\tilde{Y}$ is $\mathcal{G}$-measurable and $\tilde{Y}=Y, P$-as. This shows that $Y \in L^{2}(\mathcal{G}, P)$.

Let $Z$ be the orthogonal projection of $X$ onto $L^{2}(\mathcal{G}, P)$. Then $X=Z+U$, where $U \in L^{2}(\mathcal{G}, P)^{\perp}$, that is $E(U V)=0$, for all $V \in L^{2}(\mathcal{G}, P)$, especially $E\left(U 1_{A}\right)=0$, for all $A \in \mathcal{G}$. This implies that $E\left(X 1_{A}\right)=E\left(Z 1_{A}\right)$, for all $A \in \mathcal{G}$, and consequently $Z$ is a conditional expectation for $X$ given $\mathcal{G}$.
(ii) Assume now that $X \geq 0$ and let, for each $n \geq 1, Z_{n}$ be a conditional expectation of $X \wedge n \in L^{2}(P)$ given $\mathcal{G}$. Let $n \geq 1$. Then $E\left(Z_{n} 1_{A}\right)=E\left((X \wedge n) 1_{A}\right) \leq$ $E\left((X \wedge(n+1)) 1_{A}\right)=E\left(Z_{n+1} 1_{A}\right)$, for all sets $A \in \mathcal{G}$, and this combined with the $\mathcal{G}$-measurability of $Z_{n}, Z_{n+1}$ shows that $Z_{n} \leq Z_{n+1}, P$-as. (2.a.1.(a)). Set $Z=\limsup {n} Z{n}$. Then $Z \geq 0$ is $\mathcal{G}$-measurable and $Z_{n} \uparrow Z, P$-as. Let $A \in \mathcal{G}$. For each $n \geq 1$ we have $E\left(Z_{n} 1_{A}\right)=E\left((X \wedge n) 1_{A}\right)$ and letting $n \uparrow \infty$ it follows that $E\left(Z 1_{A}\right)=E\left(X 1_{A}\right)$, by monotone convergence. Thus $Z$ is a conditional expectation of $X$ given $\mathcal{G}$.
(iii) Finally, if $E(X)$ exists, let $Z_{1}, Z_{2}$ be conditional expectations of $X^{+}, X^{-}$given $\mathcal{G}$ respectively. Then $Z_{1}, Z_{2} \geq 0, E\left(Z_{1} 1_{A}\right)=E\left(X^{+} 1_{A}\right)$ and $E\left(Z_{2} 1_{A}\right)=E\left(X^{-} 1_{A}\right)$, for all sets $A \in \mathcal{G}$. Letting $A=\Omega$ we see that $E\left(Z_{1}\right)<\infty$ or $E\left(Z_{2}\right)<\infty$ and consequently the event $D=\left[Z_{1}<\infty\right] \cup\left[Z_{2}<\infty\right]$ has probability one. Clearly $D \in \mathcal{G}$. Thus the random variable $Z=1_{D}\left(Z_{1}-Z_{2}\right)$ is defined everywhere and $\mathcal{G}$-measurable. We have $Z^{+} \leq Z_{1}$ and $Z^{-} \leq Z_{2}$ and consequently $E\left(Z^{+}\right)<\infty$ or $E\left(Z^{-}\right)<\infty$, that is, $E(Z)$ exists. For each set $A \in \mathcal{G}$ we have $E\left(Z 1_{A}\right)=$ $E\left(Z_{1} 1_{A \cap D}\right)-E\left(Z_{2} 1_{A \cap D}\right)=E\left(X^{+} 1_{A \cap D}\right)-E\left(X^{-} 1_{A \cap D}\right)=E\left(X 1_{A \cap D}\right)=E\left(X 1_{A}\right) .$
Thus $Z$ is a conditional expectation of $X$ given $\mathcal{G}$.
Remark. By the very definition of the conditional expectation $E_{\mathcal{G}}(X)$ we have $E(X)=E\left(E_{\mathcal{G}}(X)\right)$, a fact often referred to as the double expectation theorem. Conditioning on the sub- $\sigma$-field $\mathcal{G}$ before evaluating the expectation $E(X)$ is a technique frequently applied in probability theory. Let us now consider some examples of conditional expectations. Throughout it is assumed that $X \in \mathcal{E}(P)$.

## 微积分作业代写calclulus代考|a Sigma fields, information and conditional expectation

2.a.0。(a) 如果X∈和(磷)， 然后1一种X∈和(磷), 对于所有集合一种∈F.
(b) 如果X∈和(磷)和一种∈R， 然后一种X∈和(磷).
(c) 如果X1,X2∈和(磷)和和(X1)+和(X2)被定义，那么X1+X2∈和(磷).

## 微积分作业代写calclulus代考|Conditional expectation

2.b.0。有条件的期望X给定G存在并且是磷-作为。唯一确定的。今后将表示和(X∣G)要么和G(X).

(ii) 现在假设X≥0并让，对于每个n≥1,和n成为有条件的期望X∧n∈一世2(磷)给定G. 让n≥1. 然后和(和n1一种)=和((X∧n)1一种)≤ 和((X∧(n+1))1一种)=和(和n+11一种), 对于所有集合一种∈G，这与G- 可测量性和n,和n+1表明和n≤和n+1,磷-作为。(2.a.1.(a))。设置 $Z=\limsup {n} Z {n}.吨H和nZ\geq 0一世s\数学{G}−米和一种s你r一种b一世和一种ndZ_ {n} \ uparrow Z, P−一种s.一世和吨一个 \in \mathcal{G}.F○r和一种CHn \ geq 1在和H一种v和E\left(Z_{n} 1_{A}\right)=E\left((X \wedge n) 1_{A}\right)一种nd一世和吨吨一世nGn \uparrow \infty一世吨F○一世一世○在s吨H一种吨E\left(Z 1_{A}\right)=E\left(X 1_{A}\right),b和米○n○吨○n和C○nv和rG和nC和.吨H你s和一世s一种C○nd一世吨一世○n一种一世和Xp和C吨一种吨一世○n○FXG一世v和n\数学{G}.(一世一世一世)F一世n一种一世一世和,一世F前任的）和X一世s吨s,一世和吨Z_{1}，Z_{2}b和C○nd一世吨一世○n一种一世和Xp和C吨一种吨一世○ns○FX^{+}, X^{-}G一世v和n\数学{G}r和sp和C吨一世v和一世和.吨H和nZ_{1}, Z_{2} \geq 0, E\left(Z_{1} 1_{A}\right)=E\left(X^{+} 1_{A}\right)一种ndE\left(Z_{2} 1_{A}\right)=E\left(X^{-} 1_{A}\right),F○r一种一世一世s和吨s一个 \in \mathcal{G}.一世和吨吨一世nGA = \ 欧米茄在和s和和吨H一种吨E\left(Z_{1}\right)<\infty○rE\left(Z_{2}\right)<\infty一种ndC○ns和q你和n吨一世和吨H和和v和n吨D=\left[Z_{1}<\infty\right] \cup\left[Z_{2}<\infty\right]H一种spr○b一种b一世一世一世吨和○n和.C一世和一种r一世和D \in \mathcal{G}.吨H你s吨H和r一种nd○米v一种r一世一种b一世和Z=1_{D}\left(Z_{1}-Z_{2}\right)一世sd和F一世n和d和v和r和在H和r和一种nd\数学{G}−米和一种s你r一种b一世和.在和H一种v和Z ^ {+} \ leq Z_ {1}一种ndZ ^ {-} \ leq Z_ {2}一种ndC○ns和q你和n吨一世和E\left(Z^{+}\right)<\infty○rE\left(Z^{-}\right)<\infty,吨H一种吨一世s,E (Z)和X一世s吨s.F○r和一种CHs和吨一个 \in \mathcal{G}在和H一种v和E\left(Z 1_{A}\right)=E\left(Z_{1} 1_{A \cap D}\right)-E\left(Z_{2} 1_{A \cap D}\right)=E\left(X^{+} 1_{A \cap D}\right)-E\left(X^{-} 1_{A \cap D}\right)=E\left(X 1_{A \cap D}\right)=E\left(X 1_ {A}\右）。吨H你s和一世s一种C○nd一世吨一世○n一种一世和Xp和C吨一种吨一世○n○FXG一世v和n\数学{G}.R和米一种r到.乙和吨H和v和r和d和F一世n一世吨一世○n○F吨H和C○nd一世吨一世○n一种一世和Xp和C吨一种吨一世○nE_{\mathcal{G}}(X)在和H一种v和E(X)=E\left(E_{\mathcal{G}}(X)\right),一种F一种C吨○F吨和nr和F和rr和d吨○一种s吨H和d○你b一世和和Xp和C吨一种吨一世○n吨H和○r和米.C○nd一世吨一世○n一世nG○n吨H和s你b−\西格玛−F一世和一世d\数学{G}b和F○r和和v一种一世你一种吨一世nG吨H和和Xp和C吨一种吨一世○n前任的）一世s一种吨和CHn一世q你和Fr和q你和n吨一世和一种pp一世一世和d一世npr○b一种b一世一世一世吨和吨H和○r和.一世和吨你sn○在C○ns一世d和rs○米和和X一种米p一世和s○FC○nd一世吨一世○n一种一世和Xp和C吨一种吨一世○ns.吨Hr○你GH○你吨一世吨一世s一种ss你米和d吨H一种吨X \in \mathcal{E}(P)$。