Bounding Mean and Variance of Normal Assuming Total Variation Upper Bound

Question

Let $\mathcal{N}(\mu,\sigma)$ denote the Normal Distribution with mean $\mu$ and variance $\sigma^2$, let $d_{tv}(X,Y)$ be the total variation distance of $X,Y$.

Question

Assuming $d_{tv}(\mathcal{N}(\mu_1, \sigma_1), \mathcal{N}(\mu_2, \sigma_2)) < \epsilon$, where $\epsilon$ is a positive constant what could we say about the distance of the means and variances?

I think that assuming total variation distance at most $\epsilon$ should imply something like $| \mu_1 - \mu_2 | = O( \epsilon (\sigma_1 + \sigma_2))$ and $|\sigma_1^2 - \sigma_2^2 | = O( \epsilon (\sigma_1^2 + \sigma^2_2))$. A result quantifying this intuition should exist because $\mathcal{N}(\mu_i, \sigma_i) \to \mathcal{N}(\mu, \sigma)$ iff $\mu_i \to \mu$ and $\sigma_i \to \sigma$.

Answer 1

I think the following is just a more formalized version of your reasoning in the question, but here goes.

By definition,

\begin{equation} \mu_1 - \mu_2 = \int_X X f_X(X) dX - \int_X X f_Y(X) dX = \int_X X [f_X(X) -f_Y(X)] dX. \end{equation}

Fix some $k > 0$, and define

\begin{equation} C_k = [\text{min}(\mu_1, \mu_2) - k (\sigma_1 + \sigma_2), \text{max}(\mu_1, \mu_2) + k(\sigma_1 + \sigma_2)], \end{equation}

and \begin{equation} C'_k = [- \infty, \infty] \setminus C_k. \end{equation}

Then

\begin{equation} \mu_1 - \mu_2 = \int_{X | X \in C_k} X [f_X(X) -f_Y(X)] dX + \int_{X | X \in C'_k} X [f_X(X) -f_Y(X)] dX. \end{equation}

Because of the known bound on total variation \begin{equation} \int_{X | X \in C_k} \left| X [f_X(X) -f_Y(X)] \right| dX = O(k \epsilon (\sigma_1 + \sigma_2)^2). \end{equation}

Also

\begin{equation} \int_{X | X \in C'_k} X [f_X(X) -f_Y(X)] dX \end{equation}

is bounded by the error function, taking parameters determined by $\mu_1, \mu_2, \sigma_1, \sigma_2$. Formally, to complete this, you can optimize over $k$.