Orthogonal projection of a random vector on the sphere

Lucky Hans · **Joined:** Thu, 7 Jun 2012 11:13:33 UTC **Posts:** 4

Hey there,

I have a fixed orthogonal projection P on R^d and a uniformly distributed random vector X on the unit sphere in R^d. Now I am interested in the distribution of the norm (or squared norm) of the orthogonal projection of X, i.e., I am looking for the distribution of \|P(X)\| or \|P(X)\|^2.

If each entry of X is independent and standard normally distributed, then \|P(X)\|^2 is chi-square distributed. Moreover, if I normalized such X, i.e., use Y=X/\|X\|, then Y is uniformly distributed on the sphere. However, I do not know how this may change \|P(X)\|^2.

Any comments???

outermeasure · **Posted:** Thu, 7 Jun 2012 12:27:12 UTC

Lucky Hans wrote:

Hey there,

I have a fixed orthogonal projection P on R^d and a uniformly distributed random vector X on the unit sphere in R^d. Now I am interested in the distribution of the norm (or squared norm) of the orthogonal projection of X, i.e., I am looking for the distribution of \|P(X)\| or \|P(X)\|^2.

If each entry of X is independent and standard normally distributed, then \|P(X)\|^2 is chi-square distributed. Moreover, if I normalized such X, i.e., use Y=X/\|X\|, then Y is uniformly distributed on the sphere. However, I do not know how this may change \|P(X)\|^2.

Any comments???

It changes dramatically. Instead, think about volume (or area) element.

Lucky Hans · **Joined:** Thu, 7 Jun 2012 11:13:33 UTC **Posts:** 4

well, I am sure that is changes a lot. In fact, I know that the expectation is k/d, where d is the dimension of the ambient space and k the rank of the orthogonal projection. I also have some deviation estimates that hold with high probability. Nevertheless, I am interested in the actual distribution and was wondering if it is possible to determine it analytically.

outermeasure · **Posted:** Thu, 7 Jun 2012 13:57:02 UTC

Lucky Hans wrote:

well, I am sure that is changes a lot. In fact, I know that the expectation is k/d, where d is the dimension of the ambient space and k the rank of the orthogonal projection. I also have some deviation estimates that hold with high probability. Nevertheless, I am interested in the actual distribution and was wondering if it is possible to determine it analytically.

You can.

Without loss of generality, we can assume P is projection ontothe first k components of $\mathbb{R}^d$ . So changing to spherical polars by
$\begin{aligned} x^1&=r\cos\theta^1&r&=\sqrt{(x^1)^2+\dots+(x^n)^2}\\ x^2&=r\sin\theta^1\cos\theta^2&\theta^1&=\cot^{-1}\frac{x^1}{\sqrt{(x^2)^2+\dots+(x^n)^2}}\\ &\vdots&\qquad\qquad&\vdots\\ x^{n-1}&=r\sin\theta^1\cdots\sin\theta^{n-2}\cos\theta^{n-1}&\theta^{n-2}&=\cot^{-1}\frac{x^{n-2}}{\sqrt{(x^{n-1})^2+(x^n)^2}}\\ x^n&=r\sin\theta^1\cdots\sin\theta^{n-2}\sin\theta^{n-1}&\theta^{n-1}&=2\cot^{-1}\frac{\sqrt{(x^n)^2+(x^{n-1})^2}+x^{n-1}}{x^n} \end{aligned}$
Since r=1, the area element is $i_{\partial_r}\mathrm{d}x^1\cdots\mathrm{d}x^n=\sin^{n-2}(\theta^1)\sin^{n-3}(\theta^2)\cdots\sin(\theta^{n-2})\,\mathrm{d}\theta^1\,\mathrm{d}\theta^2\cdots\mathrm{d}\theta^{n-1}$ , now integrate the $\theta^{k+1},\dots,\theta^{n-1}$ coordinates, remembering $\theta^i\in(0,\pi)$ except $\theta^{n-1}\in(0,2\pi)$ parametrises almost every point on the unit sphere. Finally, change coordinates back.

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Orthogonal projection of a random vector on the sphere

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