# Gaussian function in lattices

Probability density function of gaussian distribution is $$1/{\sqrt{2 \pi} \sigma} \times {e^{{(x-c)^2/ 2{\sigma}^2 }}}$$ in lattices we assume $$\sigma =s/\sqrt{2 \pi}$$so the gaussian function becomes For $$X \in R^n$$ $$\rho_s(X)=1/s^n \times e^{-{\pi}||(x-c||)^2/s^2}$$ but it is given as $$\rho_s(X)= e^{-{\pi}||(x-c||)^2/s^2}$$ can you explain why is this difference?

• This isn't my area, but if I had to guess it's because the $\frac{1}{s^n}$ is a scaling factor that doesn't affect sampling from the distribution? – pg1989 Feb 24 '16 at 1:29
• Is it really 1 over $s^n$ or it should be $s^2$ ? – Hilder Vítor Lima Pereira Feb 24 '16 at 9:38

The Gaussian function over a lattice, if you define it with or without the scaling factor $1/s^n$, is not a probability distribution, since it does not sum to 1. In order to construct a probability distribution, that samples points proportionally to the Gaussian function, one has to rescale anyway by dividing by the sum of the Gaussian over all lattice points (even though this is an infinite sum, it is well defined and one can compute sharp bounds on it). So any scaling factor you apply to the Gaussian function will be cancelled out by this normalization anyway, so you might as well define the Gaussian without it.
$\rho_s(X)$ for a lattice with dimension $n$ is $(1/s)^ne^{-\pi||X||^2/s^2}$
As we know it is Spherically Symmetric. So The probability of $X$ only depends on its length. We can consider $s=1$ in this case.