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Definition. A random vector $X = (X_1, \cdots, X_n)$ is said to be isotropic if $\mathbb{E}[XX^T] = I$.

Now let $X \sim \text{Unif}(\sqrt n S^{n-1})$ where $S^{n-1}$ denotes unit sphere (surface of unit ball) in $\mathbb{R}^n$. I want to show that $X$ is isotropic. Any help?

My Intuition: It can be proved that $X$ is isotropic iff $\mathbb{E}\left[\left<X, x\right>^2\right] = \|x\|^2$ for all $x \in \mathbb{R}^n$ which intuitively means all marginal projections of $X$ have unit variance. This intuition helps to get a sence of why uniform distribution on the surface of sphere should be isotropic but how to show that rigorously?

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By the rotational invariance of the distribution of $X$, we know that $\mathbb{E}[\langle X, x \rangle^2] $ depends only on the length of $x$. So if $\{ x = x_1, \cdots, x_n \}$ is an orthogonal system with $\|x_i\| = \|x\|$ for all $i$, then

$$n\mathbb{E}[\langle X, x \rangle^2] = \mathbb{E}\left[ \sum_{i=1}^{n} \langle X, x_i \rangle^2 \right] = \mathbb{E}[ \|X\|^2 \|x\|^2 ] = n\|x\|^2 $$

and hence $\mathbb{E}[\langle X, x \rangle^2] = \|x\|^2$.

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  • $\begingroup$ I think that first step that you take for granted is probably where the difficulty lies. After that it's just a short linear algebra calculation as you show here. But then even checking that is also a fairly short linear algebra calculation: $\langle X,x \rangle \sim \langle QX,x \rangle = \langle X,Q^T x \rangle$, for an arbitrary orthogonal matrix $Q$, where $\sim$ denotes "has the same distribution as". That first step, strictly speaking, requires a little bit of elbow grease with change of variables. Now just rotate $x$ appropriately... $\endgroup$ – Ian Sep 12 '17 at 19:43
  • $\begingroup$ @Ian, That's true. But in order to establish that step, we need to open the gut of the definition of uniform distribution on the sphere and any further details would depend on it. (A convenient definition in this context would be that it is the unique normalized Haar measure on the sphere w.r.t. the group of rotations.) Probably I need OP's input on this matter to expand my answer further. $\endgroup$ – Sangchul Lee Sep 12 '17 at 19:51

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