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数学代写|随机分析作业代写stochastic analysis代考|Conditional Expectation

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数学代写|随机分析作业代写stochastic analysis代考|non-negative random variable

Let $(\Omega, \mathcal{F}, P)$ be a probability space. Then we have the following.
Theorem 1.3.1 Let $\mathcal{G}$ be a sub- $\sigma$-algebra, and $X$ be a non-negative random variable. Then there exists a non-negative random variable $Y$ satisfying the following two conditions.
(1) $Y$ is G-measurable.
(2) $E[Y, B]=E[X, B]$ for any $B \in \mathcal{G}$.
Moreover, if $Y^{\prime}$ is another non-negative random variable satisfying Conditions
(1) and (2), then $Y^{\prime}=Y$ a.s.
Proof Let $X_{n}=X \wedge n, n=0,1,2, \ldots$ Since $E\left[X_{n}^{2}\right]<\infty$, by Proposition $1.2 .3$ there are $Y_{n} \in \mathcal{L}{\mathcal{G}}^{2}, n=1,2, \ldots$, such that $E\left[Y{n}, B\right]=E\left[X_{n}, B\right]$ for any $B \in \mathcal{G}$. Let $Y_{0}=0$. Then $Y_{0} \in \mathcal{L}{G}^{2}$ and $E\left[Y{0}, B\right]=E\left[X_{0}, B\right]=0$ for any $B \in \mathcal{G}$. Let $A_{n, m}=\left{Y_{n}-Y_{m}<0\right}$ for $n>m \geqq 0$. Then we see that $A_{n, m} \in \mathcal{G}$. Since $X_{n} \geqq X_{m}$, we see that
$$E\left[Y_{n}-Y_{m}, A_{n, m}\right]=E\left[X_{n}-X_{m}, A_{n, m}\right] \geqq 0$$

This shows that $P\left(Y_{n}-Y_{m}<0\right)=0, n>m \geqq 0$, and so $P\left(Y_{n} \geqq Y_{m}\right)=1$. Let $\Omega_{0}=\bigcap_{n=0}^{\infty}\left{Y_{n+1} \geqq Y_{n}\right}$ and let $Y=\lim {n \rightarrow \infty} 1{\Omega_{0}} Y_{n}$. Since $X_{n}, n=1,2, \ldots$, and $1_{\Omega_{0}} Y_{n}, n=1,2, \ldots$, are non-decreasing sequences of non-negative random variables, we see that for $B \in \mathcal{G}$
$$E[Y, B]=\lim {n \rightarrow \infty} E\left[1{\Omega_{0}} Y_{n}, B\right]=\lim {n \rightarrow \infty} E\left[Y{n}, B\right]=\lim {n \rightarrow \infty} E\left[X{n}, B\right]=E[X, B]$$
This implies our first assertion.
Suppose that $Y^{\prime}$ is a non-negative random variable satisfying Conditions (1) and (2). Let $A_{n}=\left{Y^{\prime} \geqq Y+\frac{1}{n}, Y \leqq n\right}, n \geqq 1$. Then we see that $A_{n} \in \mathcal{G}$ and $E\left[Y, A_{n}\right]<\infty$. Therefore we see that $$E\left[Y, A_{n}\right]+\frac{1}{n} P\left(A_{n}\right) \leqq E\left[Y, A_{n}\right]+E\left[Y^{\prime}-Y, A_{n}\right]=E\left[Y^{\prime}, A_{n}\right]=E\left[X, A_{n}\right]=E\left[Y, A_{n}\right] .$$ This implies that $P\left(A_{n}\right)=0$. Note that $$\bigcup_{n=1}^{\infty} A_{n}=\left{Y^{\prime}>Y\right}$$
So we see that $P\left(Y^{\prime}>Y\right)=0$. Since we can prove $P\left(Y>Y^{\prime}\right)=0$ similarly, we obtain $Y=Y^{\prime}$ a.s.

数学代写|随机分析作业代写stochastic analysis代考|progressively measurable processes

Corollary 1.3.1 Let G be a sub- $\sigma$-algebra, and $X$ be an integrable random variable. Then there exists an integrable random variable $Y$ satisfying the following.
(1) $Y$ is G-measurable.
(2) $E[Y, B]=E[X, B]$ for any $B \in G$.
Moreover, if $Y^{\prime}$ is another integrable random variable satisfying Conditions (1) and (2), then $Y^{\prime}=Y$ a.s.

Proof Let $X_{+}$and $X_{-}$be non-negative random variables given by $X_{+}=\max {X, 0}$ and $X_{-}=\max {-X, 0}$. Then we see that $E\left[X_{+}\right]<\infty$, and $E\left[X_{-}\right]<\infty$. Since
$$E\left[E\left[X_{+} \mid \mathcal{B}\right]+E\left[X_{-} \mid \mathcal{B}\right]\right]=E\left[E\left[X_{+} \mid \mathcal{B}\right]+E\left[X_{-} \mid \mathcal{B}\right], \Omega\right]=E\left[X_{+}+X_{-}, \Omega\right]<\infty,$$
we see that $E\left[X_{+} \mid \mathcal{B}\right]<\infty$ and $E\left[X_{-} \mid \mathcal{B}\right]<\infty$ a.s. Therefore letting $Y=E\left[X_{+} \mid \mathcal{B}\right]$ $-E\left[X_{-} \mid \mathcal{B}\right]$, we see that $Y$ is an integrable random variable satisfying Conditions (1) and (2).

Suppose that $Y^{\prime}$ is an integrable random variable satisfying Conditions (1) and (2). Then we see that $\left{Y>Y^{\prime}\right} \in \mathcal{G}$, and that

\begin{aligned} E\left[Y-Y^{\prime},\left{Y>Y^{\prime}\right}\right] &=E\left[Y,\left{Y>Y^{\prime}\right}\right]-E\left[Y^{\prime},\left{Y>Y^{\prime}\right}\right] \ &=E\left[X,\left{Y>Y^{\prime}\right}\right]-E\left[X,\left{Y>Y^{\prime}\right}\right]=0 \end{aligned}
This implies that $P\left(Y>Y^{\prime}\right)=0$. Similarly, we see that $P\left(Y^{\prime}>Y\right)=0$. So we obtain $P\left(Y=Y^{\prime}\right)=1$.

We denote also by $E[X \mid \mathcal{G}]$ the integrable random variable $Y$ in Corollary 1.3.1. Note that $E[X \mid \mathcal{G}]$ is determined only almost surely in either case that $X$ is a non-negative random variable or $X$ is an integrable random variable.
We call $E[X \mid \mathcal{G}]$ the conditional expectation of $X$ given a sub- $\sigma$-algebra $\mathcal{G}$.
For any $A \in \mathcal{F}$, we denote $E\left[1_{A} \mid \mathcal{G}\right]$ by $P(A \mid \mathcal{G})$ and we call this the conditional probability of an event $A$ given a sub- $\sigma$-algebra $\mathcal{G}$.

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|NON-NEGATIVE RANDOM VARIABLE

1 是是 G 可测量的。
2 和[是,乙]=和[X,乙]对于任何乙∈G.

1和2， 然后是′=是作为
Proof Let $X_{n}=X \wedge n, n=0,1,2, \ldots$ Since $E\left[X_{n}^{2}\right]<\infty$, by Proposition $1.2 .3$ there are $Y_{n} \in \mathcal{L}{\mathcal{G}}^{2}, n=1,2, \ldots$, such that $E\left[Y{n}, B\right]=E\left[X_{n}, B\right]$ for any $B \in \mathcal{G}$. Let $Y_{0}=0$. Then $Y_{0} \in \mathcal{L}{G}^{2}$ and $E\left[Y{0}, B\right]=E\left[X_{0}, B\right]=0$ for any $B \in \mathcal{G}$. Let $A_{n, m}=\left{Y_{n}-Y_{m}<0\right}$ for $n>m \geqq 0$. Then we see that $A_{n, m} \in \mathcal{G}$. Since $X_{n} \geqq X_{m}$, we see that
$$E\left[Y_{n}-Y_{m}, A_{n, m}\right]=E\left[X_{n}-X_{m}, A_{n, m}\right] \geqq 0$$

We call $E[X \mid \mathcal{G}]$ the conditional expectation of $X$ given a sub- $\sigma$-algebra $\mathcal{G}$.
For any $A \in \mathcal{F}$, we denote $E\left[1_{A} \mid \mathcal{G}\right]$ by $P(A \mid \mathcal{G})$

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|PROGRESSIVELY MEASURABLE PROCESSES

1 是是 G 可测量的。
2 和[是,乙]=和[X,乙]对于任何乙∈G.