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Let $X_0,X_1,X_2,...$ be independent random variables with $$P(X_n =1)=P(X_n =-1)={1\over 2},\forall n.$$ Let $$Z_n = \prod_{i=0}^n X_i .$$

Show that $Z_1,Z_2,...$ are independent.


An intuition is to prove it by induction. By definition a sequence of r.v.s $(X_n)_{n\ge1}$ are independent if the sigma algebra generated by them are independent. Here, I try to prove $P(Z_1=1,Z_2=-1)=P(Z_1 =1)P(Z_2 =-1)$ and this is easy to do, but when includes like $n$ R.V.s should I write all the possible combination of events? I am not so sure how to write this proof.

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For any finite $S\subset\left\{ 0,1,2,\ldots\right\} $ define $Z_{S}:=\prod_{s\in S}X_{s}$. Note that here $Z_{n}$ and $Z_{\left\{ 0,\ldots,n\right\} }$ denote the same random variable.

By induction it can be shown that $Z_{n}$ has exactly the same distribution as - let's pick one out - $X_{0}$ . It uses the apparant independency of $Z_{n-1}$ and $X_{n}$. Note that $Z_{n}$ can only take values in $\left\{ -1,1\right\} $ so that it is enough to prove that $P\left\{ Z_{n}=1\right\} =\dfrac{1}{2}$.

$P\left\{ Z_{n}=1\right\} =P\left\{ Z_{n-1}=1\wedge X_{n}=1\right\} +P\left\{ Z_{n-1}=-1\wedge X_{n}=-1\right\} =P\left\{ Z_{n-1}=1\right\} P\left\{ X_{n}=1\right\} +P\left\{ Z_{n-1}=-1\right\} P\left\{ X_{n}=-1\right\} =\dfrac{1}{2}\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{2}$.

In fact this means that $Z_{S}$ has the same distribution as $X_{0}$ for any finite $S\subset\left\{ 0,1,2,\ldots\right\} $.

Let $n,k$ be integers with $k>0$ and let $S:=\left\{ n+1,\cdots,n+k\right\} $. Then $Z_{n}$ and $Z_{S}$ are independent and for $\varepsilon,\delta\in\left\{ -1,1\right\} $ we find:

$P\left\{ Z_{n}=\varepsilon\wedge Z_{n+k}=\delta\right\} =P\left\{ Z_{n}=\varepsilon\wedge Z_{S}=\delta\varepsilon^{-1}\right\} =P\left\{ Z_{n}=\varepsilon\right\} P\left\{ Z_{S}=\delta\varepsilon^{-1}\right\} =\dfrac{1}{2}\times\dfrac{1}{2}=P\left\{ Z_{n}=\varepsilon\right\} P\left\{ Z_{n+k}=\delta\right\} $.

This proves that $Z_n$ and $Z_{n+k}$ are independent.

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Let $n$ be a fixed integer and $\left(\varepsilon_i\right)_{i=0}^n\in\left\{-1,1\right\}^{n+1}$. Then the following equality holds: $$ \bigcap_{j=0}^n\left\{Z_j=\varepsilon_j \right\}=\bigcap_{j=0}^n\left\{X_j=\prod_{i=0}^j\varepsilon_i \right\}. $$ Consequently, using independence and the fact that the $\varepsilon_i$ are identically distributed,
$$\tag{*} \mathbb P\left(\bigcap_{j=0}^n\left\{Z_j=\varepsilon_j \right\}\right)=2^{-n-1}. $$ Moreover, for all $0\leqslant j\leqslant n$, $X_j$ is independent of $Z_{j-1}$ hence
\begin{align} \mathbb P\left\{Z_j=\varepsilon_j \right\}&=\mathbb P\left(\left\{Z_j=\varepsilon_j \right\}\cap \left\{X_j=1 \right\} \right)+\mathbb P\left(\left\{Z_j=\varepsilon_j \right\}\cap \left\{X_j=-1 \right\} \right)\\ &=\mathbb P\left(\left\{Z_{j-1}=\varepsilon_j \right\}\cap \left\{X_j=1 \right\} \right)+\mathbb P\left(\left\{Z_{j-1}=-\varepsilon_j \right\}\cap \left\{X_j=-1 \right\} \right)\\ &=\frac 12\left(\mathbb P\left\{Z_{j-1}=\varepsilon_j \right\}+\mathbb P\left\{Z_{j-1}=-\varepsilon_j \right\}\right) \end{align} and by induction on $j$ it follows that $Z_j$ takes the values $1$ and $-1$ with probability $1/2$. In view of (*), independent of the sequence $\left(Z_j\right)_{j\geqslant 1}$ follows.

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