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Let $Y_0,Y_1,...$ be independent random variables with $P(Y_n=1)=P(Y_n=-1)=\frac{1}{2}$ and let $X_n=\prod_{i=0}^nY_i$.

Let $\mathcal{Y}=\sigma(Y_1,Y_2,...)$ , $\mathcal{T}_n=\sigma(X_r, r>n)$, $\mathcal{L}=\bigcap_n\sigma(\mathcal{Y}, \mathcal{T}_n)$ and $\mathcal{R}=\sigma(\mathcal{Y}, \bigcap_n\mathcal{T}_n)$.

1) Prove $X_n$ are independent.

2) Prove $\mathcal{L}\neq \mathcal{R}$.

This is my attempt for the first part.

$(X_n=1)$ happens when there is an even number of $Y$s equal to $-1$, so

$$P(X_n=1)=\left({n+1 \choose 0}+{n+1 \choose 2}+...+{n+1 \choose 2[\frac{n+1}{2}]}\right)\frac{1}{2^{n+1}}.$$ I computed that the sum is equal to $2^n$, so $P(X_n=1)=P(X_n=-1)=\frac{1}{2}$. Now the event $(X_n=1)\cap(X_{n+k}=1)$ consists of the number of cases when in the first $n+1$ terms we have an even number of $-1$ (which is $2^n$ by the previuos computation) times the number of cases where $\prod_{i=n+1}^kY_{i}=1$, which is $2^{k-1}$ because it's the same number of cases that would give $X_{k-1}=1$. Therefore $$P((X_n=1)\cap(X_{n+k}=1))=\frac{2^n2^k}{2^{1+n+k}}=2^{-2}$$ which is $P(X_n=1)P(X_{n+k}=1)$. Similarly I can prove that the $X_n$ are independent because when I intersect an $n_1$ number of such events I get probability $2^{-n_1}$.

Is this correct? Are there simpler ways to do it?

As for the second part I don't know how to proceed.

Any help would be appreciated, thank you.

bibo
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2 Answers2

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1) $X_n$'s are independent: here

2)

  1. $\sigma(Y_0) \subseteq \mathscr{L}$

  2. $\sigma(Y_0)$ and $\mathscr{R}$ are independent

Now assume on the contrary that $\mathscr{L} = \mathscr{R}$.

If $\sigma(Y_0) \subseteq \mathscr{L}$, then $\sigma(Y_0) \subseteq \mathscr{R}$.

Since $\sigma(Y_0)$ and $\mathscr{R}$ are independent, $\sigma(Y_0)$ is independent of itself.

This means $\forall F \in \sigma(Y_0), P(F \cap F) = P(F)P(F) \ \to \ P(F) \in \{0,1\}$.

Choose $F = (Y_0 = 1)$ or $F = (Y_0 = -1)$. We have $P(F) = 1/2 \notin \{0,1\}$

Community
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BCLC
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0

You can use induction.

For any cardinal $k$ then $P(X_k=1) = 1/2$.

By reason of $P(X_1=1) = 1/2 \quad\wedge\quad P(X_k)=1/2 \to P(X_{k+1})=1/2$

Likewise, $P(X_k=-1)=1/2$ for any cardinal $k$

Then show that for any $k , h$ then $X_k$ is independent of $X_{k+h}$, because : for all $(\alpha, \beta) \in \{-1, 1\}^2$

$$\begin{align} P(X_{k+h} = \beta\mid X_k= \alpha) & = P(\prod_{i=1}^k Y_k=\alpha, \prod_{j=1}^h Y_{k+j}=\beta/\alpha\mid \prod_{i=1}^k Y_k=\alpha) \\ & = P(\prod_{j=1}^h Y_{k+j}=\beta/\alpha) \\ & =1/2 \\[2ex]\therefore P(X_{k+h}=\beta \mid X_k=\alpha) & = P(X_{k+h}=\beta) & =1/2 \end{align}$$

Graham Kemp
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