# Does the sum of discrete uniforms coverge to a discrete Gaussian?

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...

• 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 Jul 17 at 15:06
• 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! – Hilder Vítor Lima Pereira Jul 17 at 16:03
• 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 Jul 17 at 16:50

## 1 Answer

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