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A random vector $X \in \mathbb{R}^n$ is isotropic if $\mathbb{E}XX^T = I_n$. However isotropic random vectors don't have the property of isotropy. See 1. So why are they called isotropic?

Similarly a random vector $X$ is anisotropic if $X = AY$ where $Y$ is isotropic. Although this might align more with the meaning of anisotropy, I'm not sure why these words are chosen.

Edit: Definition of isotropy

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    $\begingroup$ It's hard to make sense of your question. You assert that an isotropic object doesn't have the property of isotropy, without explaining what you mean by "property of isotropy". $\endgroup$
    – Michael Engelhardt
    Commented Nov 22, 2020 at 20:08
  • $\begingroup$ I added a link. If a random vector was to some amount uniform in different directions, it should at the least have expectation $0$. (If it wasn't rotate the vector to get a contradiction). $\endgroup$
    – lamlame
    Commented Nov 22, 2020 at 20:23
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    $\begingroup$ Well, I think they mean "isotropic" as "the same in all directions", roughly speaking, but with respect to the origin only. In the other post, the OP asked whether translation preserves isotropy and it turns out the answer is no. But $\mathbb{R}^n$ is not just an affine space, it also has a "special point", namely the origin. Your question was about terminology, so I hope my somewhat vague comment helps. $\endgroup$
    – Malkoun
    Commented Nov 22, 2020 at 20:30
  • $\begingroup$ Okay thank you. In the other post the OP asked if the specific translation X-EX is isotropic. This would be true if EX = 0, but since the answer was this translation did not preserve isotropy presumably EX=/=0. So there isn't "sameness in all directions", even with respect to the origin, unless their answer was wrong. $\endgroup$
    – lamlame
    Commented Nov 22, 2020 at 21:11
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    $\begingroup$ I still don't get why you assert that "isotropic random vectors don't have the property of isotropy". The contradiction doesn't seem to be between these notions of isotropy, but between the given definition of an isotropic random vector and a mistaken interpretation of that definition in a post on math.stackexchange.com. $\endgroup$
    – Michael Engelhardt
    Commented Nov 22, 2020 at 22:49

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This is a contamination quite common in probability when properties of distributions are instead attributed to the associated random objects. Strictly speaking one should talk about isotropic (i.e., rotation invariant) measures or distributions rather than vectors.

Yet another more recent example of this contamination is provided by so called "invariant random subgroups" which are in reality probability measures on the space of subgroups invariant with respect to the group action by conjugations. I wouldn't be surprised if one day probabilists begin to talk about "invariant random points" instead of invariant measures of a dynamical system.

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  • $\begingroup$ If the definition of isotropic random vectors implies the existence of such rotational invariances then this makes sense. I didn't know that. If X has components x_i, where P(x_1 = 1)=1 and P(x_i = 1)=P(x_i = -1) = 1/2 for i=2,...n then X satisfies E XX^T = I_n. The distribution has rotational symmetries, but all such rotations leave the 1st component fixed. This seems allowed with your definition of isotropy, but I'm not sure what norm $\mu$ to choose to make it isotropic for Carlo Beenakker's answer. $\endgroup$
    – lamlame
    Commented Nov 23, 2020 at 0:38
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A random vector $\mathbf{x}$ is called isotropic with respect to a norm $\mu$ (more generally, a quasinorm) if the equiprobability curves are given by $\mu(\mathbf{x})=\text{constant}$. If $\mu$ is the Euclidean norm, the equiprobability curves are rotationally invariant, hence the name "isotropic". This is a special case, more generally the equiprobability curves are unbounded, but the name is kept.

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