Covariance matrix of uniform spherical distribution I need to figure out the covariance matrix of a uniform spherical distribution.  But there I can't even find a closed form of the distribution.  This link says it is $\frac{1}{n}\mathbf{I}$, where $\mathbf{I}$ is the $n \times n$ identity matrix.  But can someone please help me derive it?  Or even point to literature on how to?
Thanks!
p.s. By uniform spherical distribution, yes, I meant the unit sphere on $\mathbb{R}^n$, that is $S^{n-1} = \{ s \in \mathbb{R}^n : |s| = 1 \}$.
 A: According to @whuber's answer posted here, the spherical distribution is best seen as
$$ \left( Y_1 = \frac{X_1}{\sqrt{X_1^2+...+X_n^2}}, ... , Y_n = \frac{X_n}{\sqrt{X_1^2+...+X_n^2}}\right)$$
where all the $X_i$ are independent Gaussian $(0,1)$.
If $(Y_1, ..., Y_i, ... Y_n)$ is uniform on the unit sphere, then so is $(Y_1, ..., -Y_i, ... Y_n)$, so they have the same distribution. In particular this implies that $E(Y_i)=-E(Y_i)$ and also that $E(Y_iY_j) = - E(Y_iY_j)$ for all $j \neq i$. Therefore the means and the covariance terms are equal to 0, as mentions @whuber in the comments.
For the variance, notice that
$$E \left( Y_1^2 \right) + ... + E \left( Y_n^2 \right) = 
E \left( Y_1^2 + ... + Y_n^2 \right) = 1.$$
For reasons of symmetry, the $Y_i$ are obviously exchangeable (but not independent), so that $E \left( Y_1^2 \right) = ... = E \left( Y_n^2 \right)$ and thus each of them is equal to $1/n$.
Im summary, the variance terms are equal to $1/n$ and the covariance terms are equal to $0$, so the covariance matrix is $\frac{1}{n} \mathbf{I}$. This is a great example of uncorrelated dependent variables (for example if $Y_1 = 1$ then all other values have to be $0$).
