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Is there some analogous of the Central limit theorem for discrete uniforms and discrete normal distributions?

To be more specific, let's say we have identical and independent random random variables $U_i$ uniform on a given integer interval $[0, k]\cap \mathbb{Z}$, for some $k \in \mathbb{N}$, $k \ge 2$.

Then, we can define the distribution $S_n := \sum_{i=1}^n U_i$.

Let $f_{c, s}(x) = \exp(-(x-c)/(2s^2))$, then the discrete Gaussian distribution centered on $c$ with variance $s^2$, which I will denote by $G_{c, s^2}$, assigns to each $x \in \mathbb{Z}$ the probability $$Pr[G_{c,s} = x] = \frac{f_{c, s}(x)}{\sum_{\forall y \in \mathbb{Z}}f_{c, s}(y)}.$$ That is, the "mass" of $x$ normalized by the "mass" of the whole integers.

I would like to know if there is some result that relates $S_n$ with the discrete normal distribution. For example, will $S_n$ converge to $G_{c, s^2}$ as $n$ increases?

I know that the mean of each $U$ is $\mu := k/2$ and the variance is $\sigma^2 := (k+1)^2 / 12$, so I made some tests using $G_{c, s^2}$ with $c = n\mu$ and $s = \sqrt{n}\sigma$, but it seems that $S_n$ does not converge. But it is hard to tell by the graphs...

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  • $\begingroup$ The CLT asserts that the standardized version of $S_n$ converges to a standard Normal distribution. Since that's not the same as a discrete Gaussian, the answer to your question must be in the negative. Perhaps some further study of the CLT would help you, or maybe a detailed analysis of $S_n.$ $\endgroup$ – whuber Jul 17 at 15:06
  • $\begingroup$ Yes, but if I divide by $n$, then I get rational values, and in this case there is no chance for $S_n$ to converge to the discrete Gaussian over the integers, since the supports will be diferent. Isn't there a version of CLT for non-standardized sums? Thank you! $\endgroup$ – Hilder Vítor Lima Pereira Jul 17 at 16:03
  • $\begingroup$ I think you missed one of the main points about the CLT: it concerns the (cumulative) distribution function, not the probability mass function or probability density function. Because the CLT governs sums, the only possible different conclusion--whether the sums are standardized or not--occurs when the underlying distribution has infinite variance. That's not the case here. $\endgroup$ – whuber Jul 17 at 16:50
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Partially answered in comments:

The CLT asserts that the standardized version of $S_n$ converges to a standard Normal distribution. Since that's not the same as a discrete Gaussian, the answer to your question must be in the negative. Perhaps some further study of the CLT would help you, or maybe a detailed analysis of $S_n$.

– whuber

( Yes, but if I divide by $n$, then I get rational values, and in this case there is no chance for $S_n$ to converge to the discrete Gaussian over the integers, since the supports will be diferent. Isn't there a version of CLT for non-standardized sums? – Hilder Vítor Lima Pereira )

I think you missed one of the main points about the CLT: it concerns the (cumulative) distribution function, not the probability mass function or probability density function. Because the CLT governs sums, the only possible different conclusion--whether the sums are standardized or not--occurs when the underlying distribution has infinite variance. That's not the case here.

– whuber

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