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 $\mathbb{E} e^{X^2/C^2} \leq 2$.  
 A: 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.
