Let $X\_1, X\_2, \ldots$ be independent and identically distributed random variables with mean $0$ and variance $1$ and let $S\_n = (X\_1 + \cdots + X\_n)/\sqrt{n}$ to be their normalized sum. Define $D\_n$ to the total variation (TV) distance between $S\_n$ and a standard Gaussian.


**Question:** Is $D\_n$ always monotonic decreasing?


This is similar to the well known [Shannon monotonicity of entropy](https://www.ams.org/journals/jams/2004-17-04/S0894-0347-04-00459-X/) problem which states that the entropy of $S\_n$ is monotonic increasing.


Intuitively, the conjecture might be true since visually, I cannot imagine a case where the TV distance ever decreases. Furthermore, the upper bound for TV distance from [Pinsker's inequality](https://en.wikipedia.org/wiki/Pinsker%27s_inequality) *does* derease monotonically (proof sketch: decompose KL divergence as a - entropy term + constant term and the first follows from the Shannon problem.)


Any references, counter examples, or discussions would be appreciated.



Is total variation distance of normalized sum of random variables to Gaussian monotonic decreasing?
Unless I made a mistake, this is not true.


Take the random variable $X$ to be $M$ with probability $\epsilon$, $-M$ with probability $-\epsilon$, and distributed as $N(0, \frac{ 1- 2 \epsilon M^2}{ 1-2\epsilon} )$ with probability $1-2\epsilon$. Here we take $M$ very large and $\epsilon M^2$ somewhat small, for instance $\epsilon = M^{-3}$.


The total variation distance is $$\int\_{-\infty}^{\infty} \max \left( \frac{ e^{ - x^2/2}}{\sqrt{2\pi}}- (1-2\epsilon)\frac{ e^{ - x^2 (1-2\epsilon) / (1-2 \epsilon M^2)} }{\sqrt{ 2\pi (1- 2 \epsilon m^2)/ (1-2\epsilon)}} , 0 \right) dx $$


since the spikes at $-M$ and $M$ don't contribute.


Now the distribution of $(X\_1 + X\_2)/\sqrt{2}$ is


* $N(0, \frac{ 1- 2 \epsilon M^2}{ 1-\epsilon} )$ with probability $(1-2 \epsilon)^2$
* $N(M/\sqrt{2}, \frac{ 1- 2 \epsilon M^2}{ 2 -2 \epsilon } )$, with probability $2 \epsilon (1-2\epsilon)$
* $N(-M/\sqrt{2}, \frac{ 1- 2 \epsilon M^2}{ 2 -2 \epsilon } )$
* $\sqrt{2} M$ with probability $\epsilon^2$
* $0$ with probaiblity $2\epsilon^2$
* $-\sqrt{2} M$ with probability $\epsilon^2$


The last three are peaks and don't contribute to the total variation distance. The two shifted Gaussians are supported around $M/\sqrt{2}$, with exponential error, and thus contribute an amount exponentially small in $M$ to the total variation distance. So the total variation distance is


The total variation distance is $$\int\_{-\infty}^{\infty} \max \left( \frac{ e^{ - x^2/2}}{\sqrt{2\pi}}- (1-2\epsilon)^2\frac{ e^{ - x^2 (1-2\epsilon) / (1-2 \epsilon M^2)} }{\sqrt{ 2\pi (1- 2 \epsilon m^2)/ (1-2\epsilon)}} , 0 \right) dx $$


plus a term exponentially small in $M$. This integral is greater than the integral above by a factor linear in $\epsilon$, which has size $M^{-3}$, which is much larger than the exponentially small factor factor, so the total variation distance is larger than the previous case.


