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I am trying to show that a random vector that is uniformly distributed on the $\ell_1$ ball (or a scaled version of the $\ell_1$ ball) in $\mathbb{R}^n$ is isotropic (i.e. $\mathbb{E}X X^T = Id$). I can show that the off-diagonal terms are zero since $(x_1, \cdots, x_i, \dots, x_n)$ and $(x_1, \cdots, -x_i, \dots, x_n)$ are identically distributed so $\mathbb{E} x_i x_j = - \mathbb{E} x_i x_j$ implies that $\mathbb{E} x_i x_j = 0$ for $i \neq j$.

Additionally, I would like to show that $X$ is not subgaussian where we define subgaussian to mean there exists a constant $C$ such that $\mathbb{E} e^{X^2/C^2} \leq 2$.

Source : Vershynin, High Dimensional Probability, Exercise 3.4.9.

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  • $\begingroup$ What do you mean by uniformly distributed on the $\ell^1$ ball? $\endgroup$
    – Jason
    Oct 19, 2017 at 1:36
  • $\begingroup$ The density is supported on the $\ell_1$ ball and is constant. $\endgroup$
    – JohnKnoxV
    Oct 19, 2017 at 13:18
  • $\begingroup$ Density with respect to what measure? The distribution you are talking about does not exist. $\endgroup$
    – Jason
    Oct 19, 2017 at 15:56
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    $\begingroup$ The Lebesgue measure. The density is simply the reciprocal of the volume. What measure are you talking about? What doesn't exist? $\endgroup$
    – JohnKnoxV
    Oct 19, 2017 at 19:24
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    $\begingroup$ $X$ is bounded, so of course it is sub-Gaussian. $\endgroup$
    – zhoraster
    Oct 20, 2017 at 7:16

1 Answer 1

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Since $\mathbb{E}X_1^2 = \cdots = \mathbb{E}X_n^2$, it is clear that one can make an $\ell_1$ ball isotropic by scaling it. To find the scaling factor, the key boils down to finding the marginal distribution of $p_{X_1}(x_1)$.

Let $B^n(r)$ denote the $n$-dimensional $\ell_1$ ball of radius $r$, that is, $B^n(r) := \{x\in\mathbb{R}^n: \|x\|_1\leq r\}$. Then $$ \Pr(X_1 \leq x) = \int_{-1}^x \frac{\operatorname{vol}(B^{n-1}(1-|t|))}{\operatorname{vol}(B^n(1))} dt = \int_{-1}^x (1-|t|)^{n-1} dt \cdot \frac{\operatorname{vol}(B^{n-1}(1))}{\operatorname{vol}(B^n(1))}. $$ Plugging in $x=1$ and since $\Pr(X_1\leq 1) = 1$, we know that $$ \frac{2}{n}\cdot \frac{\operatorname{vol}(B^{n-1}(1))}{\operatorname{vol}(B^n(1))} = 1, $$ that is, $$ \frac{\operatorname{vol}(B^{n-1}(1))}{\operatorname{vol}(B^n(1))} = \frac{n}{2}. $$ Therefore the marginal density $p_{X_1}(x_1) = \frac{n}{2}(1-|x_1|)^{n-1}$. Now we can compute $$ \mathbb{E} X_1^2 = \frac{n}{2}\int_{-1}^1 x_1^2(1-|x_1|)^{n-1} dx_1 = \frac{n}{2}\cdot 2\cdot \frac{2! (n-1)!}{(n+2)!} = \frac{2}{(n+1)(n+2)}, $$ which means that we should rescale the $\ell_1$ ball by a factor of $C_n \approx n$ to make it isotropic.

The marginal density of $X_1$ behaves like $(1-|x_1|/C_n)^{n-1} \approx e^{-c|x_1|}$, and this is an exponential decay instead of a subgaussian decay.

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