A Spherically Distributed Random Vector Is Isotropic

Question

Definitions:

A random vector $X$ is spherically distributed if $X \sim \text{Unif}(\sqrt n S^{n-1})$ i.e. $X$ is uniformly distributed on the Euclidean sphere in $\mathbb{R}^n$ centered at origin with radius $\sqrt n$.

A random vector $X$ is isotropic if $\mathbb{E} XX^T=I_n$, where $I_n$ is the identity matrix in $\mathbb{R}^n$

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Question:

Show that the spherically distributed random vector $X$ is isotropic. Argue that the coordinates of $X$ are not independent.

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My Attempt:

To argue that coordinates of $X$ are not independent, if follows from $X$ is on $\sqrt n S^{n-1} $ then $\| X \|_2$ must equals to $\sqrt n$.

To show $X$ is isotropic, it suffices to show that for any $x\in \mathbb{R}^n, \ \mathbb{E}\left<X, x\right>^2=\|x \|_2^2.$ Then I tried to compute it directly by considering $\left<X, x\right>^2$ as the trace of a $1 \times 1$ matrix.

But it seems not working as we cannot get the form like $X^TXx^Tx$ or $x^TxX^TX$ using the cyclic property of trace. Even if we are able to do so, the expression would become $\|X\|_2^2 \|x\|_2^2 = n\|x \|_2^2$, which is not our desired result. May I ask what can I start with?

I found an answer here but I wonder if there is any other way to show it since I don't have any knowledge on rotation invariant of such distribution, I don't understand that solution at all. Any hint or help is appreciated.

Answer 1

Conditioning on the co-ordinates should give an answer. If $(X_1,X_2)$ are the first two co-ordinates, then $\mathbb{E}[X_1X_2] =\mathbb{E}[\mathbb{E}[X_1X_2|X_3,...X_n]]$. The conditional expecation will be zero, using say, polar co-ordinates. As for $\mathbb{E}[X_i^2]$, it must be same for all $i$ and their sum must be n.

Answer 2

I realize this is a 2.5 year old question, but I had trouble with this as well. Hopefully my answer can help others.

By definition, $X_1^2+\dots+X_n^2 = n$. Symmetry implies $E X_1^2 = \dots = E X_n^2 = 1$. Again, by symmetry, it suffices to show $E X_1X_2 = 0$. Now, $$X_2 = -\sqrt{n-\sum_{i\neq 2}X_i^2}\chi_{(-\infty,0)} + \sqrt{n-\sum_{i\neq 2}X_i^2}\chi_{[0,\infty)}$$ By symmetry, $P(X_2<0) = P(X_2\geq 0)$, and thus $EX_1X_2 = 0$.