Independence of 2 events involving inner products of random vector

Question

We have $4$ fixed $n$-dimensional binary vectors $U,V,A,B$. We have another random $n$-dimensional binary vector $Z$ where each element is Ber$(p)$. So $P$ is a event where $\left<U,Z\right>\equiv\left<V,Z\right>\pmod 2$ (inner product) and $Q$ is an event where $\left<A,Z\right>\equiv\left<B,Z\right>\pmod 2$. So are $P$ and $Q$ independent or not? Note that all calculations are done modulo $2$.

Answer 1

It depends on the vectors.

$P\equiv\{\omega: \sum_i (U_i-V_i)\,Z_i(\omega)=0\}$

$Q\equiv\{\omega: \sum_i (A_i-B_i)\,Z_i(\omega)=0\}$

Take $U=(1,0,0,0), V=(0,1,0,0), A=(0,0,1,0), B=(0,0,0,1)$

Then $P=\{\omega: (Z_1-Z_2)(\omega)=0\}$ and $\mathsf P(P)=p^2+(1-p)^2$

And $Q=\{\omega: (Z_3-Z_4)(\omega)=0\}$ and $\mathsf P(P)=p^2+(1-p)^2$

And $P\cap Q=\{\omega: (Z_1-Z_2)(\omega)=0~\wedge~(Z_3-Z_4)(\omega)=0\}$ and $\mathsf P(P\cap Q)= p^4+2p^2(1-p)^2+(1-p)^4$

Thus the events are independent.

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Take $U=(1,0), V=(0,1), A=(1,1), B=(0,0)$

Then $P=\{\omega:(Z_1-Z_2)(\omega)=0\}$ and $\mathsf P(P)=p^2+(1-p)^2$

But $Q=\{\omega: (Z_1+Z_2)(\omega)=0\}$ and $\mathsf P(Q)=(1-p)^2$

And $P\cap Q=\{\omega: Z_1(\omega)=0\wedge Z_2(\omega)=0\}$ and $\mathsf P(P\cap Q)=(1-p)^2$

$P(P\cap Q) \neq P(P)P(Q)$. So the events are not independent.