# Uniform distribution on $\ell_1$ ball

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.

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

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.