On the one hand we toss $n$ times a fair coin, and we sum the outcomes (+1 for heads, -1 for tails). Let $f:\mathbb{N}\to\mathbb{R}$ describe the probability distribution of the outcome.

On the other hand we toss $m$ times a fair die with $2k$ sides, and we sum the outcomes (to avoid parity issues, assume outcomes in $\{\pm 1,\pm 3,\dots,\pm (2k-3),\pm (2k-1)\}$). Let $g:\mathbb{N}\to\mathbb{R}$ describe the probability distribution of the outcome.

What can we say about the total variation distance between both experiments (i.e., 1-distance between $f$ and $g$), as a function of $n$, $m$ and $k$?

BACKGROUND: I am interested in this question since simulations indicate that throwing $m$ times a die of $O(\sqrt{n/m})$ sides allows for $f$ and $g$ to get $1/\sqrt{m}$-close in TV distance (for $n\gg m$). I.e., we could approximate (up to constant distance) an $n$-fold coin toss experiment by a constant number of tosses of a $O(\sqrt{n})$ sided die.

It is easy to find $m$ and $k$ such that $f$ and $g$ converge weakly. And the Berry-Esseen theorem shows that the cumulative probability distributions converge pointwise as $O(1/\sqrt{m})$, if we choose $\tau\in O(\sqrt{n/m})$. This is however not sufficient to prove anything about the TV distance.

I have also tried to work with local limit theorems, which show that the probability distributions converge pointwise as $O(1/m)$ if $\tau\in O(\sqrt{n/m})$. But seeing that the support of the distributions will be $\gg \sqrt{m}$, this also seems insufficient to bound TV distance.

Any other ideas?

(This question is a duplicate from a question on math.stackexchange: link)

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