Exercise 2.32 (Independence of $\sigma$-algebras II)

Question

Let $X_1,X_2,\dots$ be a sequence of random variables. Show that $(X_n)_{n=1}^\infty$ are jointly independent if and only if $\sigma(X_{n+1})$ is independent of $\sigma(X_1,\dots,X_n)$ for all natural numbers $n$.

Elu Thingol · Accepted Answer

Notice the following identity
\begin{align*}
\sigma\left(X_1,\ldots,X_night)
&= \left\langle \left\{ (X_i)^{-1}(S_i): S_i \in {\mathcal B}_i, i \in A ight\} ightangle\
&= \left\langle \left\{ (X_1)^{-1}(S_1): S_1 \in {\mathcal B}_1 ight\} \cup \ldots \cup \left\{ (X_n)^{-1}(S_n): S_n \in {\mathcal B}_n ight\} ightangle\
&\subseteq \langle \sigma(X_1) \cup \ldots \cup \sigma(X_n) angle
\end{align*}

\begin{itemize}
\item[$\implies$] Suppose that $(X_n)_{n=1}^{\infty}$ are jointly independent. By the \href{./exercise_2-32}{previous exercise}, the $\sigma$-algebras $\sigma(X_n)_{n=1}^{\infty}$ are independent as well. In particular, $\sigma(X_{n+1})$ is independent of $\sigma(X_1),\ldots,\sigma(X_n)$. From this, we conclude that 
$\sigma(X_{n+1})$ is independent of $\sigma(X_1) \cup \ldots \cup \sigma(X_n)$. To conclude from the generating set to the $\sigma$-algebra generated by it, we use \href{exercise_0-11}{measurable induction}. Let $X$ be an arbitrary $\sigma\left(X_1,\ldots,X_night)$-measurable random variable.
\begin{enumerate}
\item[$\sigma_1$] We have ${\bf P}(X_{n+1} \in S_{n+1} \wedge X \in \emptyset) = {\bf P}(X_{n+1}\in S_{n+1}) \cdot {\bf P}(X \in \emptyset) = 0$.
\item[$\sigma_2$] Let $E$ be a set such that ${\bf P}(X_{n+1} \in S_{n+1} \wedge X \in E) = {\bf P}(X_{n+1}\in S_{n+1}) \cdot {\bf P}(X \in E)$. Then for $E^c$ we have
\begin{align*}
{\bf P}\left(X_{n+1} \in S_{n+1}ight) \cdot {\bf P}\left(X \in E^cight) &= {\bf P}\left(X_{n+1} \in S_{n+1}ight) \cdot \left[1 - {\bf P}\left(X \in Eight)ight]\
&= {\bf P}\left(X_{n+1} \in S_{n+1}ight) - {\bf P}\left(X_{n+1} \in S_{n+1}ight)\cdot {\bf P}\left(X \in Eight)\
&= {\bf P}\left(X_{n+1} \in S_{n+1}ight) - {\bf P}\left(X_{n+1} \in S_{n+1},\, X \in Eight)\
&= {\bf P}\left(X_{n+1} \in S_{n+1},\, X 
ot\in Eight)
\end{align*}
\item[$\sigma_3$] Let $E_1,E_2,\ldots$ be a sequence of sets in $\mathcal{B}$. Then $\bigcup_{n=1}^{\infty} E_n \in \mathcal{B}$.
\begin{align*}
{\bf P}\left(X_{n+1} \in S_{n+1},\, X \in \bigcup_{n=1}^{\infty} E_night) &= {\bf P}\left(X_{n+1} \in S_{n+1},\, Y \in \bigcup_{n=1}^{\infty} E_n \setminus \bigcup_{i=1}^{n-1} E_i ight)\
&= \sum_{n=1}^{\infty} {\bf P}\left(X_{n+1} \in S_{n+1},\, X \in E_n \setminus \bigcup_{i=1}^{n-1} E_i ight)\
&= \sum_{n=1}^{\infty} {\bf P}\left(X_{n+1} \in S_{n+1}ight) \cdot {\bf P}\left(X \in E_n \setminus \bigcup_{i=1}^{n-1} E_i ight)\
&= {\bf P}\left(X_{n+1} \in S_{n+1}ight) \cdot \sum_{n=1}^{\infty} {\bf P}\left(X \in E_n \setminus \bigcup_{i=1}^{n-1} E_i ight)\
&= {\bf P}\left(X_{n+1} \in S_{n+1}ight) \cdot {\bf P}\left(X \in \bigcup_{n=1}^{\infty}E_n \setminus \bigcup_{i=1}^{n-1} E_i ight)\
&= {\bf P}\left(X_{n+1} \in S_{n+1}ight) \cdot {\bf P}\left(X \in \bigcup_{n=1}^{\infty}E_night)
\end{align*}
\end{enumerate}
Thus, independence of $\sigma(X_{n+1})$ from $\sigma(X_1)\cup\ldots\cup\sigma(X_{n})$ is a $\sigma$-algebra property, and is thus true on all $\sigma\left(X_1,\ldots,X_night)$.

\item[$\impliedby$] Suppose that $\sigma(X_{n+1})$ is independent of $\sigma(X_1,\dots,X_n)$ for all natural numbers $n$. By definition, $\sigma(X_{n+1})$ must be independent of each $\sigma(X_1),\ldots,\sigma(X_n) \subseteq \sigma(X_1,\dots,X_n)$ separately. Since this is true for all $n\in\mathbf{N}$, we conclude the joint independence.
\end{itemize}

Exercise 2.32 (Independence of $\sigma$-algebras II)

Answers

Comments

Exercise 2.32 (Independence of $\sigma$-algebras II)

Answers

Comments

Add answer