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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 $$P(X_i=1)=P(X_i=-1)=\frac{1}{2}\quad\tag{*}$$ 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 $Z_i=X_1X_2\cdots X_i$. How can I show that $(Z_i)_{i=1}^{m}$ are independent for any finite $m\in{\Bbb N}$?


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.

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Let $\epsilon_i=-1$ or $1$. We have $$P\left(\bigcap_{j=1}^n\{Z_j=\varepsilon_j\}\right)=P\left(\bigcap_{j=1}^n\{Z_j=\varepsilon_j\}\mid \{X_n=-1\}\right)P(X_n=-1)+P\left(\bigcap_{j=1}^n\{Z_j=\varepsilon_j\}\mid \{X_n=1\}\right)P(X_n=1),,$$ hence $$P\left(\bigcap_{j=1}^n\{Z_j=\varepsilon_i\}\right)=\frac 12P\left(\bigcap_{j=1}^{n-1}\{Z_j=\varepsilon_i\}\cap\{Z_{n-1}=-\epsilon_n\}\right)+\frac 12P\left(\bigcap_{j=1}^{n-1}\{Z_j=\varepsilon_i\}\cap\{Z_{n-1}=\epsilon_n\}\right).$$ We just deal with the case $\epsilon_{n-1}=\epsilon_n$. Then $$P\left(\bigcap_{j=1}^n\{Z_j=\varepsilon_j\}\right)=\frac 12P\left(\bigcap_{j=1}^{n-1}\{Z_j=\varepsilon_j\}\right)=\frac 12\prod_{j=1}^{n-1}P\left(\{Z_j=\varepsilon_j\}\right),$$ which is what we want. The other case is similar.

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  • $\begingroup$ But you've already have $P(X_n=-1)$, how did you come up with $\frac 12$? $\endgroup$ – Jack Oct 17 '12 at 15:37
  • $\begingroup$ I don't understand: this probability is $1/2$. $\endgroup$ – Davide Giraudo Oct 17 '12 at 15:41
  • $\begingroup$ What i'm saying is that $$ P\left(\bigcap_{j=1}^n\{Z_j=\varepsilon_j\}\mid X_n=-1\right)P(X_n=-1)=P\left(\bigcap_{j=1}^{n-1}\{Z_j=\varepsilon_i\}\cap\{Z_{n-1}=-\varepsilon_n\}\right).$$ Why do you need $\frac 12$? $\endgroup$ – Jack Oct 17 '12 at 15:45
  • $\begingroup$ Also, why is $\varepsilon_i=0,1$ instead of $\varepsilon_i=-1,1$? $\endgroup$ – Jack Oct 17 '12 at 15:51
  • $\begingroup$ I go back to the definition of conditional probability, using independence of $X_n$ with respect to $(Y_1,\dots,Y_{n-1})$. $\endgroup$ – Davide Giraudo Oct 17 '12 at 19:45
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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) =\frac{1}{2} $ 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 $(Z_i)_{i=1}^{m}$.

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