Vtyjkyuk

The following is a classic example that pairwise independent does not necessarily imply mutually independent:

Let $X_1$ and $X_2$ be independent r.v.'s with distributions $$P(X_i=1)=P(X_i=-1)=frac12quadtag*$$ for $i=1,2.$ Let $Z=X_1X_2$. Then $X_1,X_2,Z$ are pairwise independent but they are not mutually independent.

In this case $X_1$ and $Z=X_1X_2$ are independent. Now let $(X_i)_i=1^infty$ be a family of independent variables which satisfy $(*)$ and let $Z_i=X_1X_2cdots X_i$. How can I show that $(Z_i)_i=1^m$ are independent for any finite $minBbb N$?

When $m=2$, it is done. If one needs induction, then a key step is to show that the $n-1$ dimensional random vector $(Z_1,Z_2,dots,Z_n-1)$ and $Z_n$ are independent. This where I have no idea how to go on after writing down the definition.

Let $epsilon_i=-1$ or $1$. We have
$$Pleft(bigcap_j=1^nZ_j=varepsilon_jright)=Pleft(bigcap_j=1^nZ_j=varepsilon_jmid X_n=-1right)P(X_n=-1)+Pleft(bigcap_j=1^nZ_j=varepsilon_jmid X_n=1right)P(X_n=1),,$$
hence
$$Pleft(bigcap_j=1^nZ_j=varepsilon_iright)=frac 12Pleft(bigcap_j=1^n-1Z_j=varepsilon_icapZ_n-1=-epsilon_nright)+frac 12Pleft(bigcap_j=1^n-1Z_j=varepsilon_icapZ_n-1=epsilon_nright).$$
We just deal with the case $epsilon_n-1=epsilon_n$. Then
$$Pleft(bigcap_j=1^nZ_j=varepsilon_jright)=frac 12Pleft(bigcap_j=1^n-1Z_j=varepsilon_jright)=frac 12prod_j=1^n-1Pleft(Z_j=varepsilon_jright),$$
which is what we want. The other case is similar.

For example, $P(Z_4 | Z_3 Z_2 Z_1) = P(Z_4 | Z_3)$ (because $Z_4 = X_4 Z_3 $, with $X_4$ independent from $Z_3$; hence, if $Z_3$ is known, $Z_2$, $Z_1$ do not give more information; in other words, $Z_i$ is a first-order Markov sequence)

But, say $P(Z_4 = 1| Z_3=1)= P(X_4=1) =frac12 $ and the same happens in all cases, hence $P(Z_4 | Z_3 Z_2 Z_1) = P(Z_4 | Z_3) = P(Z_4) = $ and $Z_4$ is independent of previous values; applying induction, you get the independence of the full set $(Z_i)_i=1^m$.