# Does bounding moments make distributions close in total variation distance?

Let $W\sim\mathcal{N}(0,\sigma^2)$ be a "reference" Gaussian random variable.

Suppose I have a set of distributions, $\mathcal{W}$, where $W_a\in\mathcal{W}$ if it satisfies the following criteria:

1. Symmetry: $E[W_a^{2k-1}]=0$ for all $k=1,2,\ldots$
2. Moments close to reference: $E[W_a^{2k}]\in\left[E[W^{2k}]/\rho^k,\rho^kE[W^{2k}]\right]$ for all $k=1,2,\ldots$ and some $\rho>1$.

I am wondering if this has implications for the total variation distance between the distributions in $\mathcal{W}$, that is do the above constraints yield an upper bound on $\frac{1}{2}\|W-W_a\|_1$?

This question arose as I was reading this paper (an early pre-print is available here for free). In particular, the authors seem to claim (in the proof of Theorem 1) that the distributions in $\mathcal{W}$ cannot be separated via a hypothesis test even given an infinite number of samples. However, their mathematics are rather sloppy, and, thus, I am a little skeptical of their claim.

• Great point -- this completely makes sense. Now I wonder what are these other conditions on the distributions in $\mathcal{W}$ that would result in the total variation distance bound. Obviously, if $\mathcal{W}$ contains a set of slightly disturbed $W$'s, the bound should trivially hold. I wonder if the set is larger... – Bullmoose Jan 16 '15 at 21:16