Let $\rho_s(x) = e^{-\pi x^2/s^2}$ be the Gaussian measures, then the discrete Gaussian distribution on $\mathbb{Z}$ could be defined as $D_{\mathbb{Z},s}(x) = \rho_s(x)/\sum_{n\in \mathbb{Z}}\rho_s(n)$.

In [Regev05], the distribution $\Psi_s(r)$ on torus $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ is defined :

$$\Psi_s(r)=\sum_{k=-\infty}^{+\infty}\frac1s\cdot exp\left(-\pi \left(\frac{r-k}{s}\right)^2 \right), r\in \mathbb{T}$$

then, it can be discretized as a distribution on $\mathbb{Z}_p$ :

$$\overline{\Psi}_s(i) = \int_{(i-1/2)/p}^{(i+1/2)/p}\Psi_s(x)dx, \ \ i\in \mathbb{Z}_p$$

What I want to know is: if the $D_{\mathbb{Z},s} \mod p$ is the same as the $\overline{\Psi}_s$ on $\mathbb{Z}_p$ ?

  • $\begingroup$ Doesn't your discritization method "ignore the tails" of $\Psi_s$? In particular I see no reason why the $\overline{\Psi}$ should have total probability 1, especially if one chooses very large $s$ (say $s \gg p$) so a relatively large amount of probability mass will be contained in the tails. $\endgroup$
    – Mark Schultz-Wu
    Apr 15, 2021 at 16:33
  • $\begingroup$ the $\Psi_s$ is the distribution on torus, it looks like the Gaussian density function mod 1. Because the function $1/s exp(-\pi r^2/s^2)$ is density function on real R, so the $\Psi$ should be a density function on torus T. $\endgroup$
    – Bob
    Apr 16, 2021 at 5:45

1 Answer 1


There are two ways to discretize: rounding and conditioning.

The first discretization you use to define $D_{\mathbb Z,s}$ from $\rho_s$ is using conditionning.

The second one defining $\bar \Psi_s$ from $\Psi_s$ is using rounding (the density at integer $x$ of the former distribution is given the total density of the interval $[x-\frac 1 2, x+ \frac 12]$ of the latter distribution).

So, because of this, the two distributions are not the same. However, for large $s$, they would be very close to each other, because they would both be very close to uniform.

To summarize: the moding (ie going from R to the torus) and discretizing do commute, but only if you use the same discretization on both path.

  • $\begingroup$ Thank you! According to you say, In [Regev05], the author defined the discretized LWE, the error term is the distribution $\overline{\Psi}_s$ on $\mathbb{Z}_p$, so it is just a rounding, not the discrete Gaussian distribution $D_{Z,s}$ ? $\endgroup$
    – Bob
    Apr 23, 2021 at 8:30
  • $\begingroup$ I find many open libraries which implement the LWE or RLWE module, the way they generate the error is just sampling from a continuous Gaussian distribution and then round it to nearest number. But in [GPV08], the author points out that rounding a continuous Gaussian is not the $D_{Z,s}$. $\endgroup$
    – Bob
    Apr 23, 2021 at 8:44
  • $\begingroup$ Sampling exactly from $D_{Z,s}$ is critical for trapdoor inversion (such as in the GPV signature scheme) but not for LWE/RLWE. $\endgroup$
    – LeoDucas
    Apr 26, 2021 at 15:59

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