Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that I have vector $\mathbf{x}$ that contains $n$ independently and identically distributed (i.i.d.) zero-mean Gaussian random variables $x_i\sim\mathcal{N}(0,\sigma^2)$.

Also suppose I have a uniform random rotation defined by matrix $\mathbf{R}$ that changes the angle of $\mathbf{x}$ with respect to each basis vector in the space $\mathbb{R}^n$ by some amount drawn uniformly at random from $[0,2\pi]$.

I am interested in the conditional distribution of $\mathbf{y}=\mathbf{Rx}$ given $\mathbf{x}$. It seems to me that $\mathbf{y}$ should contain i.i.d. zero-mean Gaussian random variables $y_i\sim\mathcal{N}(0,\sigma^2)$. Is that true? If so, how does one prove it?

I know that a vector of i.i.d. Gaussians is invariant to the rotation. I also know that any given rotation is a linear transformation, so if we know $\mathbf{R}$, then, obviously, $\mathbf{y}$ given $\mathbf{x}$ is deterministic and not random. I am wondering about the case when $\mathbf{R}$ is random. This question arises out of a study of a very strange interference channel in information theory. I appreciate any hints.

share|improve this question
1  
If R is uniform, then, conditionally on x, y=Rx is uniformly distributed on the sphere centered at 0 with radius |x|. –  Did Jul 6 '12 at 17:26
1  
Calling this $R$ uniformly random is rather misleading. In particular, for a fixed $x$, $Rx$ is not uniformly distributed over the surface of the sphere. (Think of choosing a point on the surface of the earth; your approach is to choose the latitude and longitude uniformly. But this means that with probability $1/2$, you will be in that band that's within $\pi/4$ of the equator, and this contains more than half the earth's area.) –  Nate Eldredge Jul 6 '12 at 18:58
    
Note that the conditional distribution of $\mathbf{y}$ given $\mathbf{x}$ cannot be normal, since $|\mathbf{y}| = |\mathbf{x}|$. –  Nate Eldredge Jul 6 '12 at 18:59
    
(I deleted my previous comment) @did, thanks now I understand what you meant, and yes, you are correct. –  M.B.M. Jul 6 '12 at 19:00
    
@NateEldredge When I was writing the question, I was (implicitly) using the wikipedia definition of uniform random matrices, where all the angles of the rotation w/r to the basis are uniformly random. In that setting, did's comment is correct, I think. –  M.B.M. Jul 6 '12 at 19:03

1 Answer 1

up vote 1 down vote accepted

If R is uniform, then, conditionally on x, y=Rx is uniformly distributed on the sphere centered at 0 with radius |x|.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.