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Hi,

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?

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up vote 3 down vote accepted

$X/|X|$ 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.

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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 –  Jessica Feb 19 at 12:52
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