When I'm reading this paper "On lattices, learning with errors, random linear codes, and cryptography" by O. Regev. I have trouble understanding the proof of claim 5.2.

"Hence, it is enough to show that $\left|\sum_{i=1}^k p x_i \bmod p\right|<p / 16$ with high probability. This condition is equivalent to the condition that $\left|\sum_{i=1}^k x_i \bmod 1\right|<1 / 16$. Since $\sum_{i=1}^k x_i \bmod 1$ is distributed as $\Psi_{\sqrt{k} \cdot \alpha}$, and $\sqrt{k} \cdot \alpha=o(1 / \sqrt{\log n})$, the probability that $\left|\sum_{i=1}^k x_i \bmod 1\right|<1 / 16$ is $1-\delta(n)$ for some negligible function $\delta(n)$."

Why can we obtain the probability that $\left|\sum_{i=1}^k x_i \bmod 1\right|<1 / 16$ is $1-\delta(n)$ and what is the detailed proof process? I would appreciate your response. Thank you very much!


1 Answer 1


$\Psi_\beta$ is the probability density function of the wrapped normal distribution so that per equation 7 of the paper $$\mathbb P\left(|\sum x_i|<1/16|\right)=\int_{-1/16}^{1/16}\Psi(r)dr$$ $$=\int_{-1/16}^{1/16}\sum_i\frac1{\sqrt k\alpha}\exp\left(-\pi\left(\frac{(r-i}{\sqrt k\alpha}\right)^2\right)dr\\ $$ Simply taking the $k=0$ term in the series we see that $$\mathbb P\left(|\sum x_i|<1/16|\right)>\int_{-1/16}^{1/16}\frac1{\sqrt k\alpha}\exp\left(-\pi\left(\frac r{\sqrt k\alpha}\right)^2\right)dr=\int_{-1/16}^{1/16}f(r)dr$$ where $f(r)$ is the density function for the normal distribution with mean 0 and standard deviation $\alpha\sqrt{k/2\pi}$.

Thus $1-\mathbb P\left(|\sum x_i|<1/16|\right)$ is less than twice the tail of the $\mathcal N(0,\alpha\sqrt{k/2\pi})$ distribution from 1/16 thus $$1-\mathbb P\left(|\sum x_i|<1/16|\right)<\frac{\exp(-(C\log n))}{\sqrt\log n}$$ for some constant $C$ which is $o(1/n)$ for large $n$.


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