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Isotropic random vectors crop up with some frequency when talking about random matrices. A random vector $X$ is defined to be isotropic if, in expectation, its outer product is the identity matrix: $\mathbb{E}[XX^T] = I$.

This is all well and good, and there appear to be many results reasoning about what you can say when working with isotropic random vectors (e.g. here).

My problem is that I'm not sure when a random vector is isotropic. For example, if I draw a vector $X$ uniformly at random from the unit ball, is $X$ isotropic?

Thanks!

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From your definition, an isotropic random vector is a vector such that, with $n$ the size of the vector space, $$ \mathbb{E} [X_iX_j] = 0 \qquad \forall i,j \in \{1, \dots n \}, \;i\ne j \\ \mathbb{E} [X_iX_i] = 1 \qquad \forall i \in \{1, \dots n \} $$ Thus if $X_i$ are i.i.d. centered gaussian variables of variance 1, the random vector is isotropic.

Considering a ball of radius $R$, for uniform vectors, in dimension 2, with $x = rcos(\theta)$ and $y = r cos (\theta)$: $$ \mathbb{E} [X^2] = \frac{1}{\pi R^2} \iint_{r=0, \theta = 0}^{r=R, \theta = 2\pi} (r\cos(\theta))^2 r dr d\theta = \frac{R}{3} $$ So for $R=1$ the vectors can certainly not be isotropic.

You can show for centered vectors that if they belong to a sphere they can not be isotropic.

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