is the sign of symmetry distribution projected still keep symmetry?

Question

Given a real random vector $\vec{r} \sim N(0, I^{d \times d})$ and a real matrix $B^{n\times d}$

I would obtain a new random vector $\vec{S} = sign(B \cdot \vec{r})$.

Would this hold $P(\vec{S}=all+)$ == $P(\vec{S}= all -)$ ?

Why?

Answer 1

Yes, we can do this using the Jacobian matrix and the transformation $C^1$ diffeomorphism $\Phi(x)=-x$. Then for any Borel measurable set $E$ we have that $$P(-E)=P(\Phi(E))=\int_E |\det D\Phi(x)|P(dx) =\int_E P(dx) = P(E)$$