Independence of random projection of Gaussian random vector If $v$, $y$ and $h$ are independent Gaussian random vectors, are $|v^H y|^2$ and $|v^H h|^2$ independent?
 A: Consider the case of 1-vectors where $v, h$, and $y$ have standard Normal distributions (whence their second moments equal $1$ and fourth moments equal $3$).  If $(vh)^2$ and $(vy)^2$ were independent, then their expectations would multiply:
$$\mathbb{E}((vh)^2(vy)^2) = \mathbb{E}((vh)^2)\mathbb{E}((vy)^2) = \mathbb{E}(v^2)\mathbb{E}(h^2)\mathbb{E}(v^2)\mathbb{E}(y^2) = (1)(1)(1)(1)=1.$$
(The second equality follows from the assumed independence of both $(v,h)$ and $(v,y)$.)
However, by virtue of the independence of $(v,h,y)$ we may easily compute
$$\mathbb{E}((vh)^2(vy)^2) = \mathbb{E}(v^4 h^2 y^2) = \mathbb{E}(v^4)\mathbb{E}(h^2)\mathbb{E}(y^2) = (3)(1)(1)=3.$$
Since $3\ne 1$, the independence assumption in the first calculation must be false: $(vh)^2$ cannot be independent of $(vy)^2$.
Independence fails for $n$-vectors in general, for much the same reason: when $|v|$ is large, both $|v^\prime h|$ and $|v^\prime y|$ will tend to be large; and when the former is small, both the latter expressions will tend to be small.  Although they are conditionally (on $v$) independent, they are not independent.
