# Why uniform distribution on sphere of radius $\sqrt n$ isotropic?

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?

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 $\mathbb{E}[\langle X, x \rangle^2] = \|x\|^2$.
• 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... – Ian Sep 12 '17 at 19:43