이 글은 수학의 즐거움 채널에서 진행 되는 기하학적 통계학의 4번째 렉쳐의 핵심을 정리한 것입니다. 수업 중에 이루어진 디테일한 디스커션 및 자세한 설명들은 영상을 참고하십시오.
Setting.
Given $(\Omega, F, P)$ Probability space,
$$
\begin{aligned}
& X:(\Omega, F) \longrightarrow(S, B(S)) \text{ random variable } \\
& \text{ i.e., } \Leftrightarrow X^{-1}(B) \in F, \forall B \in B(S) \\
& \leadsto P_{X}: B(S) \longrightarrow[0,1], B \mid \longrightarrow P\left(X^{-1}(B)\right) \text {. } \\
& \stackrel{ \text { Radon-Nikodym theorem } }{\longrightarrow} \exists ! f \in L^{1}(\mu) \text { s.t } \\
& P_{x}(B)=\int_{B} f d \mu \text {. }
\end{aligned}
$$
Conditional Expectation
Given a probability space $(\Omega, F, P)$,
let $X \in L^{1}(P)$. i.e. $X: \Omega \rightarrow \mathbb{R}$ random variable and $\int_{\Omega}|X| d P<\infty$.
Let $G \subseteq F$ be a sub-field we say $E[X \mid G]$ is a condtional expectation of $X$ w.r.t $G$ if $\forall E \in G$,
$$
\int_{E} X d P=\int_{E} E[X \mid G] d P \text {. }
$$
Explanation:
Define the measure on $G$ by
$$
\widetilde{P}(\cdot)=\int_{(\cdot)} X d P: G \rightarrow \mathbb{R} .
$$
Then by Radon-Nikodym theorem, there exists unique
$$
\begin{gathered}
f \in L^{1}(P) \text { s.t } \\
\tilde{F}(.)=\int_{(.)} f d P \\
\text { i.e., } \forall E \in G, \widetilde{P}(E)=\int_{E} f d P .
\end{gathered}
$$
Remark.
$X, Y: \Omega \longrightarrow \mathbb{R}$ be $L^{1}$ functions.
$G$ is the smallest sigma field in $F$ containing $\left\{Y^{-1}(B) \in F \mid B \in P(\mathbb{R})\right\}$
Write the conditional expectation $\mathbb{E}[X \mid Y]=\mathbb{E}[X \mid G]$
$ \mathbb{E}[X \mid Y, Z]$ and $ \mathbb{E}[X, Y \mid Z] $ mean the conditional expectation of $X$ with respect to the smallest-sigma field from $Y,Z$ and the conditional expectation of joint random variable with respect to the smallest sigma field from $Z$ respectively.
Property 1.
$$
\begin{aligned}
& X, Y \in L^{1}(P) \\
& G \subseteq f \\
& X Y \in L^{1}(P), Y \text { is } \text { $G$-measurable. }
\end{aligned}
$$
Then
$$
E[X Y \mid G]=Y \cdot E[X \mid G]
$$
Tower property
$$
\begin{aligned}
& \int_{\Omega} X d P=\int_{\Omega} E[X \mid G] d P \\
& \Leftrightarrow \mathbb{E}[X]=\mathbb{E}[E[X \mid G]] \\
& \Rightarrow \mathbb{E}[\mathbb{E}[X Y \mid G]] \\
& =\mathbb{E}[Y \mathbb{E}[X \mid G]]
\end{aligned}
$$
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