A part of this answer says that the inner product $\left( \frac{X}{|X|} \right)^T Y$, where X and Y are vectors with i.i.d. zero-mean Gaussian elements, is independent on X and is again Gaussian distributed. I have found the same claim here, but no reasoning, and my skills are apparently not sufficient to see it. Where should I look? Is $\frac{X}{|X|}$, a vector with N complex elements, somehow just a rotation?


$X/|X|$ is almost surely a uniformly random element on the unit sphere of dimension $n-1$; this is not the same thing as a rotation, which is a matrix. Since a multivariate standard Gaussian vector is invariant in law under a fixed rotation, its law is certainly invariant under a random rotation as well: thus if $A$ is a uniformly random orthogonal matrix, then $A Y$ is again standard Gaussian. Now let $Z = (X/|X|)^T Y$. Then $Z$ has the same law as the first component of $AY$, which is clearly univariate standard Guassian. To verify its independence with $X$: if you rotate $X$ by an orthogonal matrix $A$, you can absorb $A$ into $Y$, whose law is invariant under rotation. So conditional law of $Z$ under two different $X$ values differing by a rotation stays the same. More obviously, if you scale $X$, the conditional law of $Z$ remains the same.

Another thing I noticed is that $Z$ and $Y$ are not jointly normal! Notice that conditioning on $Y$ clearly has an effect on $Z$. Now assuming they are jointly normal, we can show $E Z v^T Y = 0$ for all vector $v$. In fact we can show $E (X/|X|)^T u v^T u = 0$ for fixed $u,v$. In fact, this follows simply from $E (X/|X|)^T u = E (-X/|-X|)^T u = 0$. This is a contradiction. This is similar to the situation $B X$ and $X$, where $X$ is standard 1d gaussian and $B$ is an independent Bernoulli $\pm 1$ variable. They are not jointly normal either.

  • $\begingroup$ About your answer, I still have several questions: 1.Is $\mathbf{A}$ must be the uniformly distributed matrix? What about the unitary matrix with some other distribution? 2.If $\mathbf{Y}$ is not the standard Gaussian vector, i.e. its mean is not zero or its variance is not $1$, is $\mathbf{Y}$ still invariant under a rotation? Looking forward to your reply $\endgroup$ – user47169 Feb 19 '14 at 12:52
  • $\begingroup$ @user47169 1. There is no uniformly distributed matrix because the space of such matrices is unbounded, hence cannot be normalized to have unit total mass under Lebesgue measure. The space of rotation matrices, however, is bounded and can be given an induced Lebesgue measure and normalized to 1. 2. If Y is not standard Gaussian, no guarantee exists about rotation invariance. $\endgroup$ – John Jiang Feb 16 '19 at 23:10

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