Is there a counterexample to the Thin Shell Conjecture for sub-exponential distributions?

Question

The thin shell conjecture states that there exist universal constants $C,c>0$ such that every logconcave isotropic random vector $X$ in every Euclidean space $\mathbb{R}^n$ satisfies

$$\mathbb{P}\Big\{\big|\|X\|_2-\sqrt{n}\big|\geq t\Big\}\leq Ce^{-ct}\qquad \forall t\geq0.$$

That is, most of the probability mass resides in a constant-thickened sphere of radius $\sqrt{n}$, i.e, a "thin shell." A random vector $X$ in $\mathbb{R}^n$ is said to be sub-exponential if there exist constants $c_1,c_2>0$ such that

$$\mathbb{P}\{|\langle X,v\rangle|\geq t\}\leq c_1e^{-c_2t} \qquad \forall v\in S^{n-1}, ~t\geq0.$$

One may verify that every logconcave isotropic distribution is sub-exponential with $c_1=1$ and $c_2$ being the reciprocal of Paouris' constant.

Question: Does there exist a sequence of random vectors of increasing dimension that are isotropic and sub-exponential with fixed constants $c_1,c_2>0$, but fail to reside in thin shells?

While I'm having difficulty producing such a counterexample, I never read about the thin shell conjecture beyond logconcave distributions, so perhaps a counterexample is known. Or maybe any sub-exponential counterexample can be transformed into a logconcave counterexample?

Answer 1

Ahh, just take $X$ to be $0$ with probability $1/2$ and otherwise draw uniformly from the sphere of radius $\sqrt{2n}$.

If the support of the distribution were convex, then it wouldn't be able to "skip" the thin shell. I guess this is why logconcavity is important.