Show that the vectors (ε1, ε2, . . . , εn−1) and (ε1ε2, ε2ε3, . . . , εn−1εn) have the same distribution.

Question

Let ε1, ε2, . . . , εn be independent random signs. Show that the vectors (ε1, ε2, . . . , εn−1) and (ε1ε2, ε2ε3, . . . , εn−1εn) have the same distribution. In particular, the variables ε1ε2, ε2ε3, . . . , εn−1εn are independent. Are the variables ε1ε2, ε2ε3, ε3ε1 independent?

Is this referring to the cross product? then shouldn't independence be obvious since cross product of pairs of vectors are independent of each other? I do not understand the distribution part, however. Please explain.

Answer 1

If $P(\epsilon_i=1)=0.5=P(\epsilon_i=-1)$ for $i=1,\dots,n$ then this is an answer.

$\begin{aligned}P\left(\epsilon_{1}\epsilon_{2}=1\right)=P\left(\epsilon_{1}=1=\epsilon_{2}\right)+P\left(\epsilon_{1}=-1=\epsilon_{2}\right)=P\left(\epsilon_{1}=1\right)P\left(\epsilon_{2}=1\right)+P\left(\epsilon_{1}=-1\right)P\left(\epsilon_{2}=-1\right)=0.5^{2}+0.5^{2}=0.5\end{aligned} $

and likewise $P\left(\epsilon_{2}\epsilon_{3}=1\right)=0.5$

$\begin{aligned}P\left(\epsilon_{1}\epsilon_{2}=1\wedge\epsilon_{2}\epsilon_{3}=1\right) & =P\left(\epsilon_{1}\epsilon_{2}=1\wedge\epsilon_{2}\epsilon_{3}=1\mid\epsilon_{2}=1\right)P\left(\epsilon_{2}=1\right)+P\left(\epsilon_{1}\epsilon_{2}=1\wedge\epsilon_{2}\epsilon_{3}=1\mid\epsilon_{2}=-1\right)P\left(\epsilon_{2}=-1\right)\\ & =P\left(\epsilon_{1}=1\wedge\epsilon_{3}=1\mid\epsilon_{2}=1\right)0.5+P\left(\epsilon_{1}=-1\wedge\epsilon_{3}=-1\mid\epsilon_{2}=1\right)0.5\\ & =P\left(\epsilon_{1}=1\wedge\epsilon_{3}=1\right)0.5+P\left(\epsilon_{1}=-1\wedge\epsilon_{3}=-1\right)0.5\\ & =P\left(\epsilon_{1}=1\right)P\left(\epsilon_{3}=1\right)0.5+P\left(\epsilon_{1}=-1\right)P\left(\epsilon_{3}=-1\right)0.5\\ & =0.5^{3}+0.5^{3}=0.25 \end{aligned} $

This gives $P\left(\epsilon_{1}\epsilon_{2}=1\wedge\epsilon_{2}\epsilon_{3}=1\right)=P\left(\epsilon_{1}\epsilon_{2}=1\right)P\left(\epsilon_{2}\epsilon_{3}=1\right)$ and because the rv's $\epsilon_{1}\epsilon_{2}$ and $\epsilon_{2}\epsilon_{3}$ can only take values is $\left\{ -1,1\right\} $ this proves that they are independent.

Further $\epsilon_{i}\epsilon_{i+1}$ and $\epsilon_{i}$ have equal distributions so the vectors $\left(\epsilon_{1},\dots,\epsilon_{n-1}\right)$ and $\left(\epsilon_{1}\epsilon_{2},\dots,\epsilon_{n-1}\epsilon_{n}\right)$ have equal distributions.