# 统计代写|随机过程代写stochastic process代考|STAT6540

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## 统计代写|随机过程代写stochastic process代考|Properties of the Conditional Expectation

The main rules that are useful in computing conditional expectations will be given once more, but this time in the general abstract framework.

Let $\mathcal{G}$ be a sub- $\sigma$-field of $\mathcal{F}$, and let $Y, Y_1, Y_2$ be integrable (resp. non-negative finite) random variables, $\lambda_1, \lambda_2 \in \mathbb{R}$ (resp. $\in \mathbb{R}_{+}$).
Theorem 3.3.19 Rule 1. (linearity)
$$\mathrm{E}^{\mathcal{G}}\left[\lambda_1 Y_1+\lambda_2 Y_2\right]=\lambda_1 \mathrm{E}^{\mathcal{G}}\left[Y_1\right]+\lambda_2 \mathrm{E}^{\mathcal{G}}\left[Y_2\right]$$
Proof. We consider the integrable case. We must check that $\lambda_1 \mathrm{EG}\left[X_1\right]+\lambda_2 \mathrm{E} \mathcal{G}\left[X_2\right]$ is $\mathcal{G}$-measurable (which is part of the definition of a conditional expectation with respect to $\mathcal{G}$ ) and that for all bounded $\mathcal{G}$-measurable random variable $U$
$$\mathrm{E}\left[\left(\lambda_1 \mathrm{EG}\left[X_1\right]+\lambda_2 \mathrm{E} \mathcal{G}\left[X_1\right]\right) U\right]=\mathrm{E}\left[\left(\lambda_1 Y_1+\lambda_2 Y_2\right) U\right]$$
This follows immediately from the definition of $\mathrm{E} \mathcal{G}\left[X_i\right]$, which says that $\mathrm{E}\left[\mathrm{E} \mathcal{G}\left[X_i\right] U\right]=$ $\mathrm{E}\leftY_i U\right$
Theorem 3.3.20 Rule 2. If $Y$ is independent of $\mathcal{G}$, then
$$\mathrm{E}^{\mathcal{G}}[Y]=\mathrm{E}[Y]$$
Proof. We consider the integrable case. First recall that the constant $\mathrm{E}[Y]$ is $\mathcal{G}$ measurable. It remains to prove that for all bounded $\mathcal{G}$-measurable random variables $U, \mathrm{E}[\mathrm{E}[Y] U]=\mathrm{E}[Y U]$. This is the case since $Y$ and $U$ are independent and therefore $\mathrm{E}[Y U]=\mathrm{E}[Y] \mathrm{E}[U]$
Theorem 3.3.21 Rule 3. If $Y$ is $\mathcal{G}$-measurable,
$$\mathrm{E}^{\mathcal{G}}[Y]=Y$$
Proof. We consider the integrable case. We must check that $Y$ is $\mathcal{G}$-measurable and that $\mathrm{E}[Y U]=\mathrm{E}[Y U]$.

## 统计代写|随机过程代写stochastic process代考|The Doubly Stochastic Framework

In imprecise but suggestive terms, the expression “doubly stochastic” refers to the situation where the samples are of the form $\omega=\left(\omega_1, \omega_2\right) \in \Omega_1 \times \Omega_2$ and are constructed in two steps. One draws $\omega_1$ according to a probability $P_1$ on some measurable space $\left(\Omega_1, \mathcal{F}_1\right)$, and then draws $\omega_2$ in another measurable space $\left(\Omega_2, \mathcal{F}_2\right)$ according to a probability “that depends on $\omega_1$ “. In order to formalize this notion, an extension of the Fubini-Tonelli theorem is needed, whose proof is a straightforward adaptation of the proof of the original result in Subsection 2.3.2.

Let $\left(\Omega_1, \mathcal{F}1\right)$ and $\left(\Omega_2, \mathcal{F}_2\right)$ be two measurable spaces, let $P_1$ be a probability measure on $\left(\Omega_1, \mathcal{F}_1\right)$ and let $P_2: \Omega_1 \times \mathcal{F}_2 \rightarrow[0,1]$ be a probability kernel from $\left(\Omega_1, \mathcal{F}_1\right)$ to $\left(\Omega_2, \mathcal{F}_2\right)$. By this we mean that (i) for all $A_2 \in \mathcal{F}_2$, the mapping from $\Omega_1$ to $[0,1]$ defined by $\omega_1 \rightarrow P_2\left(\omega_1, A_2\right)$ is measurable with respect to $\mathcal{F}_1$ and $\mathcal{B}([0,1])$, and (ii) for all $\omega_1 \in \Omega_1, P_2\left(\omega_1, \cdot\right)$ is a probability measure on $\left(\Omega_2, \mathcal{F}_2\right)$. Let $(\Omega, \mathcal{F}):=\left(\Omega_1 \times \Omega_2, \mathcal{F}_1 \otimes \mathcal{F}_2\right)$. Then, for any non-negative function $X:(\Omega, \mathcal{F}) \rightarrow \mathbb{R}$, the mapping $\omega_2 \rightarrow X\left(\omega_1, \omega_2\right)$ is $\mathcal{F}_2$-measurable for all $\omega_1 \in \Omega_1$ (this is Lemma 2.3.5) and the mapping $\omega_1 \rightarrow \int{\Omega_2} X\left(\omega_1, \omega_2\right) P_2\left(\omega_1, \mathrm{~d} \omega_2\right)$ is $\mathcal{F}_1$-measurable (the proof is an immediate adaptation of that of Lemma 2.3.6).

Also (this is the analog of Theorem 2.3.7) there exists a unique probability measure $P$ on $(\Omega, \mathcal{F})$ such that for all $A_1 \in \mathcal{F}1, A_2 \in \mathcal{F}_2$, $$P\left(A_1 \times A_2\right):=\int{A_1} P_2\left(\omega_1, A_2\right) P_1\left(\mathrm{~d} \omega_1\right) \text {. }$$
Finally (this is the analog of Theorem 2.3.9), if $X$ is a non-negative random variable defined on $(\Omega, \mathcal{F})$, then (Tonelli)
$$\int_{\Omega} X(\omega) P(\mathrm{~d} \omega)=\int_{\Omega_1}\left[\int_{\Omega_2} X\left(\omega_1, \omega_2\right) P_2\left(\omega_1, \mathrm{~d} \omega_2\right)\right] P_1\left(\mathrm{~d} \omega_1\right) .$$
The same is true for a random variable $X$ of arbitrary sign, provided $X$ is $P$-integrable (Fubini). (The integrability condition is usually checked by applying Tonelli’s theorem to the non-negative variable $|X|$.)

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Properties of the Conditional Expectation

$$\mathrm{E}^{\mathcal{G}}\left[\lambda_1 Y_1+\lambda_2 Y_2\right]=\lambda_1 \mathrm{E}^{\mathcal{G}}\left[Y_1\right]+\lambda_2 \mathrm{E}^{\mathcal{G}}\left[Y_2\right]$$

$\lambda_1 \mathrm{EG}\left[X_1\right]+\lambda_2 \mathrm{EG}\left[X_2\right]$ 是 $\mathcal{G}$-可衡量的（这是关于条 件期望的定义的一部分 $\mathcal{G}$ ) 并且对于所有有界的 $\mathcal{G}$-可测量 的随机变量 $U$
$$\mathrm{E}\left[\left(\lambda_1 \mathrm{EG}\left[X_1\right]+\lambda_2 \mathrm{EG}\left[X_1\right]\right) U\right]=\mathrm{E}\left[\left(\lambda_1 Y_1+\lambda_2 Y_2\right)\right.$$

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