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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.

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    If R is uniform, then, conditionally on x, y=Rx is uniformly distributed on the sphere centered at 0 with radius |x|.2012-07-06
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    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.)2012-07-06
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    Note that the conditional distribution of $\mathbf{y}$ given $\mathbf{x}$ cannot be normal, since $|\mathbf{y}| = |\mathbf{x}|$.2012-07-06
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    (I deleted my previous comment) @did, thanks now I understand what you meant, and yes, you are correct.2012-07-06
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    @NateEldredge When I was writing the question, I was (implicitly) using the [wikipedia definition](http://en.wikipedia.org/wiki/Rotation_matrix#Uniform_random_rotation_matrices) 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.2012-07-06
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    @M.B.M.: If you read the Wikipedia passage carefully, it says that the rotation angles are *not* uniformly distributed.2012-07-07
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    @NateEldredge Thanks for your comment. Yes, you are right, the angles aren't distributed uniformy if $\mathbf{R}$ moves $\mathbf{x}$ to some random point on the sphere with radius $|\mathbf{X}|$ centered at 0. So, the $\mathbf{R}$ defined in the question clearly does not do that. Again, thank you for the insight.2012-07-08

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