Let $Y_0, Y_1, \ldots$ be independent random variables with

$P(Y_n = 1) = P(Y_n = -1) = 1/2$ for $n = 0, 1, 2, \ldots$

Define $X_n = Y_0Y_1Y_2\cdots Y_n = \prod_{i=0}^n Y_i$ for $n = 0, 1, 2, \ldots$

It can be shown that $X_0, X_1, X_2, \ldots$ are independent.

Define $\mathscr{Y} \doteq \sigma(Y_1, Y_2, \ldots)$

$\mathscr{T_n} \doteq \sigma(X_r \mid r > n) = \sigma(X_{n+1}, X_{n+2}, \ldots)$

$\mathscr{L} \doteq \bigcap_n \sigma(\mathscr{Y}, \mathscr{T_n})$

Prove $Y_0$ is $\mathscr{L}$-measurable $\iff$ $\sigma(Y_0) \subseteq \mathscr{L}$

What I tried:

\begin{align} & \forall n \in \mathbb{N}, \\[8pt] & \sigma(\mathscr{Y}, \mathscr{T_n}) = \sigma(\sigma(Y_1, Y_2, \ldots), \sigma(X_r \mid r > n)) \\[8pt] = {} & \sigma(\sigma(Y_1, Y_2, \ldots), \sigma(X_{n+1}, X_{n+2}, \ldots)) \\[8pt] = {} & \sigma(\sigma(Y_1, Y_2, \ldots), \sigma\left(\prod_{i=0}^{n+1} Y_i, \prod_{i=0}^{n+2} Y_i, \ldots\right)) \\[8pt] = {} & \sigma(\sigma(Y_1, Y_2, \ldots), \sigma\left(\prod_{i=0}^{n+1} Y_i, \prod_{i=0}^{n+2} Y_i, \ldots\right)) \\[8pt] = {} & \sigma(\sigma(Y_1, Y_2, \ldots), \sigma\left(Y_0\prod_{i=1}^{n+1} Y_i, Y_0\prod_{i=1}^{n+2} Y_i, \ldots\right)) \end{align}

It seems that $\sigma(Y_0) \subseteq \sigma(\sigma(Y_1, Y_2, \ldots), \sigma\left(\color{red}{Y_0}\prod_{i=1}^{n+1} Y_i, \color{red}{Y_0}\prod_{i=1}^{n+2} Y_i, \ldots\right))$? How do I prove that exactly?


For brevity of notation set $$\mathcal{A}_1 := \sigma(Y_1,Y_2,\dots) \qquad \quad \mathcal{A}_2 := \sigma(Y_0 \prod_{i=1}^{n+1} Y_i, Y_0 \prod_{i=1}^{n+2} Y_i,\dots).$$

Since $Y_j \neq 0$ almost surely for all $j \in \mathbb{N}$, we can write

$$Y_0 = \underbrace{ \frac{1}{\prod_{i=1}^{n+1} Y_i}}_{\text{$\mathcal{A}_1$-measurable}} \cdot \underbrace{Y_0 \prod_{i=1}^{n+1} Y_i}_{\text{$\mathcal{A}_2$-measurable}}$$

for any $n \in \mathbb{N}$. This shows that $Y_0$ is measurable with respect to

$$\sigma(\mathcal{A}_1 \cup \mathcal{A}_2) = \sigma( \sigma(Y_1,Y_2,\dots), \sigma(Y_0 \prod_{i=1}^{n+1} Y_i, Y_0 \prod_{i=1}^{n+2} Y_i,\dots)).$$

Since $\sigma(Y_0)$ is the smallest $\sigma$-algebra such that $Y_0$ is measurable, it follows that

$$\sigma(Y_0) \subseteq \sigma(\mathcal{A}_1 \cup \mathcal{A}_2) = \sigma( \sigma(Y_1,Y_2,\dots), \sigma(Y_0 \prod_{i=1}^{n+1} Y_i, Y_0 \prod_{i=1}^{n+2} Y_i,\dots)).$$

  • 1
    $\begingroup$ Yes, exactly, if $X=Y \cdot Z$, then $\sigma(X) \subseteq \sigma(Y,Z)$. $\endgroup$ – saz Sep 21 '15 at 14:18
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    $\begingroup$ why does $X=YZ$ mean $X$ is $Y-$meas, and $Z-$meas? $\endgroup$ – nomadicmathematician Jun 15 '16 at 15:55
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    $\begingroup$ @takecare What exactly do you mean by $Y$-measurable? $X$ is measurable with respect to the $\sigma$-algebra generated by $Y$ and $Z$. $\endgroup$ – saz Jun 15 '16 at 16:10
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    $\begingroup$ @saz oh you're right, I was wondering about what BCLC wrote in the first comment. I don't think we can say $X$ is $\sigma(Y)$ measurable can we? $\endgroup$ – nomadicmathematician Jun 15 '16 at 16:16
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    $\begingroup$ @takecare No, in general $X$ is not $\sigma(Y)$-measurable. $\endgroup$ – saz Jun 15 '16 at 17:06

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