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 $X,Y$.


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 $\sigma_i \to \sigma$.

  • 1
    $\begingroup$ Is $\sigma$ the variance or the standard deviation? $\endgroup$ – Michael R. Chernick Jun 25 '17 at 0:45
  • $\begingroup$ $\sigma$ is the standard deviation, I will edit the question, thanks! $\endgroup$ – vkonton Jun 25 '17 at 0:46

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}


\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}


\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 $k$.

  • $\begingroup$ Should something similar work to bound also the difference of the variances? $\endgroup$ – vkonton Jun 25 '17 at 9:27
  • 1
    $\begingroup$ @vkonton Yes indeed. The idea is the same, but LMK if you need something. $\endgroup$ – Ami Tavory Jun 25 '17 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.