If two Gaussians disagree on one moment, it seems like this should imply that they have a large variation distance--equivalently, if two Gaussians are close in variation distance it's hard for their moments to differ very much. But I'm having trouble proving this without getting into a terrible mess and ending up with weak bounds. I'm hoping someone here can offer a slick proof or point me to one. 


Suppose that $\Sigma\_1$ and $\Sigma\_2$ are two positive-semidefinite matrices, with diagonal entries bounded by 1, and that $\Sigma\_1$ and $\Sigma\_2$ differ by at least $\delta$ in one of their entries.


Can we lower-bound the total variation distance between the multivariate gaussians with covariance matrices $\Sigma\_1$ and $\Sigma\_2$? In particular, is this distance at least $\Theta(\delta)$? It seems like it ought to be, since the gaussians are pretty well concentrated. But trying to formalize the obvious approach seems to create a bit of a mess.


Edit: Douglas Zare and guest point out that you can trivially reduce to the 2 dimensional case by projecting onto the relevant subspace. So we can restrict attention to that case.



Are gaussians with different moments far in total variation distance?
Letting $\mu\_{a,\Sigma}$ be the Gaussian measure with covariance matrix $\Sigma$ and mean $a$. Then (double) the variation distance can be written as
$$
\left\lVert\mu\_{a\_1,\Sigma\_1}-\mu\_{a\_2,\Sigma\_2}\right\rVert\_1 = \max\_f\left\lvert\mu\_{a\_1,\Sigma\_1}(f)-\mu\_{a\_2,\Sigma\_2}(f)\right\rvert,
$$
where the maximum is taken over functions $f\colon\mathbb{R}^n\to\mathbb{C}$ with $\lvert f\rvert\le1$. Taking $f(x)=\exp(ix\cdot u)$ for a fixed $u\in\mathbb{R}^n$, plugging in the characteristic function for Gaussian distributions,
$$
\begin{align}
\left\lVert\mu\_{a\_1,\Sigma\_1}-\mu\_{a\_2,\Sigma\_2}\right\rVert\_1&\ge\max\_{u\in\mathbb{R}^n}\left\lvert e^{ia\_1\cdot u-\frac12 u^t\Sigma\_1u}-e^{ia\_2\cdot u-\frac12 u^t\Sigma\_2u}\right\rvert\\
&\ge\max\_{u\in\mathbb{R}^n}\left\lvert e^{-\frac12 u^t\Sigma\_1u}-e^{-\frac12 u^t\Sigma\_2u}\right\rvert\\
&=\max\_{u\in\mathbb{R}^n,\lVert u\rVert\_1=1}\max\_{\lambda\ge0}\left\lvert e^{-\frac12 \lambda^2u^t\Sigma\_1u}-e^{-\frac12 \lambda^2u^t\Sigma\_2u}\right\rvert.
\end{align}
$$
Note that, for any $1\ge\alpha > \beta > 0$, we can consider $\lambda=1/\sqrt{\alpha}$ to get the following bound,
$$
\max\_{\lambda\ge0}\left(e^{-\lambda^2\beta}-e^{-\lambda^2\alpha}\right)\ge e^{-\frac\beta\alpha}-e^{-1}=\Theta\left(1-\frac\beta\alpha\right)=\Theta(\alpha-\beta).
$$
Hence, as we are assuming $\Sigma\_i$ are positive semidefinite with diagonal entries bounded by 1, we have $1\ge u^t\Sigma\_iu\ge0$ for $u\in\mathbb{R}^n$ with $\lVert u\rVert\_1=1$ and,
$$
\left\lVert\mu\_{a\_1,\Sigma\_1}-\mu\_{a\_2,\Sigma\_2}\right\rVert\_1=\Theta\left(\max\_{\lVert u\rVert\_1=1}\left\lvert u^t(\Sigma\_1-\Sigma\_2)u\right\rvert\right)
$$
For any symmetric matrix $M$ we have $2M\_{ij}=(e\_i+e\_j)^tM(e\_i+e\_j)-e\_i^tMe\_i-e\_j^tMe\_j$ where $e\_i$ is the unit vector along coordinate dimension $i$, from which we get
$$
\left\lVert\mu\_{a\_1,\Sigma\_1}-\mu\_{a\_2,\Sigma\_2}\right\rVert\_1=\Theta\left(\lVert\Sigma\_1-\Sigma\_2\rVert\_\infty\right)
$$


