# Asymptotic bound on the total variation distance between a standard multivariate normal and a simple mixture

Let $$P = N(\vec{0}, I^d)$$ be a standard multivariate Gaussian distribution in $$d$$ dimensions. Let $$Q$$ be distributed the same as $$P$$, except that samples from $$Q$$ have one of their coordinates, chosen uniformly at random from $$1$$ to $$d$$, distributed as $$N(\mu, \sigma^2)$$ instead.

That is, $$Q$$ is a mixture of $$d$$ multivariate Gaussians, each with weight $$\frac1d$$. The $$i$$th mixture component has mean $$\vec{0} + \vec{e}_i \cdot \mu$$ and variance with diagonal $$\vec{1} + \vec{e}_i \cdot (\sigma^2 - 1)$$.

I'd like to bound the total variation distance between $$P$$ and $$Q$$, as a function of $$d$$ (when $$\mu$$ and $$\sigma$$ are constants that do not depend on $$d$$).

My intuition is that $$Q$$ is very close to a multivariate Gaussian $$N(\frac{\vec{\mu}}{d}, I^d \cdot (1+\frac{\sigma^2}{d}+ o(\frac1d))$$, which would give a Kullback–Leibler divergence of $$D_{KL}(P || Q) = O(\frac1d)$$ and thus a total variation distance of $$O(\frac{1}{\sqrt d})$$.

The case where $$\sigma^2=1$$ might be easier and sufficient for what I need. Any suggestions on how I could bound the TV distance between P and Q?

## 1 Answer


• Very nice! I'll try to go through a similar approach for the general case with $\sigma \neq 1$ (or does your approach crucially rely on $\sigma=1$ somewhere?). It also seems that the Berry--Esseen inequality should let us get an explicit (rather than asymptotic) result involving $\mu, \sigma$ and $1/\sqrt{n}$, right? – Florian Tramèr Mar 15 at 0:27
• I think this should work for $\sigma\ne1$ too, but I have not gone through the details. – Iosif Pinelis Mar 15 at 2:16