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Say that a random variable $X$ follows the Gaussian distribution $\mathcal{N}(\mu, \sigma)$. Then will the ceiling $\lceil X\rceil$, the floor $\lfloor X\rfloor$, and the rounding $\lfloor X\rceil$ follow some discrete Gaussian distributions? Could you prove or disprove it? And if they do follow, what are the means and the standard deviations?

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It depends on what you mean by "discrete Gaussian distribution". For example, if $\sigma \ll 1,$ the floor will be uniformly distributed between the values $0$ and $-1.$ Is that "discrete gaussian"? – Igor Rivin Mar 16 '12 at 19:16

2 Answers 2

This type of discretized normal distribution occurs in some practical problems. For example, the median of the vertex degrees of an Erdős-Renyi random graph with fixed edge probability has such a distribution with bounded variance. (This is a result of myself and Wormald that we still haven't published more than 10 years later.)

I don't recall having seen a name for this distribution, and I don't know of a useful expression for its mean or variance. Any information about those things would be appreciated.

Regarding the question, if $\sigma\to\infty$ the distribution is asymptotically normal in some senses (such as total variation distance), while if $\sigma=O(1)$ it isn't.

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I don't know what you mean by "discrete Gaussian distribution". Explicitly, the probability mass function of $\lceil X \rceil$ is $P(\lceil X \rceil = k) = P(k-1 < X \le k) = F(k) - F(k-1)$, where $F$ is the cumulative distribution function of $X$, and similarly the pmf's of $\lfloor X \rfloor$ and $\lfloor X \rceil$ can be written in terms of $F$.

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