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Is there a closed form expression for the covariance of a uniform distribution in a n-ball? I would like to develop a test for vector sums of points sampled from a uniform distribution in a n-ball. I need the covariance of the distribution to relate this to the CLT. I've taken a stab at it but I am not sure if I am correct.

Any pointers to statistics of uniform distributions in a n-ball would be very useful. Thanks.

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  • $\begingroup$ just to check I've understood the terminology: by an n-ball you mean all points in ${\mathbb R}^n$ whose distance from the origin is at most some fixed constant, right? $\endgroup$
    – Yemon Choi
    Commented Aug 11, 2010 at 21:06
  • $\begingroup$ (Not that there is anything wrong with your terminology, but I was momentarily uncertain whether you meant a $n$-polydisc.) $\endgroup$
    – Yemon Choi
    Commented Aug 11, 2010 at 21:07
  • $\begingroup$ I mean an n-dimensional sphere with points distributed uniformly within it. I don't mean a uniform distribution on the surface. $\endgroup$
    – drfrank
    Commented Aug 11, 2010 at 21:47
  • $\begingroup$ And sorry I wasn't clear earlier. :) $\endgroup$
    – drfrank
    Commented Aug 11, 2010 at 21:48

1 Answer 1

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The covariance is a multiple of the identity by simple symmetry considerations. For the constant, you just need, again by symmetry, and integration in spherical coordinates, $$ \mathbb{E} X_1^2 = \frac{1}{n} \mathbb{E} |X|^2 = \frac{c}{n} \int_0^R r^{n-1} r^2 dr = \frac{c}{n(n+2)}R^{n+2}, $$ where $R$ is the radius of your ball and $c$ is a constant depending on $n$ and $R$. To identify $c$, $$ 1 = c \int_0^R r^{n-1} dr = \frac{c}{n} R^n, $$ so $c = n/R^n$ and your covariance is $\frac{1}{n+2} R^2$ times the identity matrix. Hopefully I've included enough detail that if I've made an algebra mistake it will be easy for someone else to correct it, but I think I recognize that as the right answer.

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  • $\begingroup$ Sorry I don't why on the first line you have $\mathbb{E} X_1^2 = \frac{1}{n} \mathbb{E} |X|^2 $. What's $X_1$, and why the $1/n$? $\endgroup$ Commented Sep 27 at 14:25
  • $\begingroup$ @PinkCollins. $X_1$ is the first component of $X$, and $|X|^2 = X_1^2 + ... + X_n^2$ is the norm of $X$ squared. Since each of the components of $X$ has the same distribution, $\mathbb{E} |X|^2 = n \mathbb{E} X_1^2$. $\endgroup$ Commented Sep 27 at 15:56
  • $\begingroup$ Oh yes the norm, this is trivial I must confess. To be honest I was also confused by $|\cdot|$ usually used for absolute values. $\endgroup$ Commented Sep 29 at 6:17

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