Let $f(x; \sigma) = \frac{1}{\sigma\sqrt{2\pi}}\cdot e^{-\frac{x^2}{2\sigma^2}}$ be the probability density function of a normal distribution $\mathcal{N}(0, \sigma^2)$. We consider a *discrete* normal distribution over $\mathbb{Z}$, obtained by sampling from $\mathcal{N}(0, \sigma^2)$ and rounding the result to the nearest integer.
The probability mass function of this discrete normal is then given by
\begin{equation\*}
F(x;\sigma) = \int\limits\_{x-\frac{1}{2}}^{x+\frac{1}{2}} f(\theta; \sigma)\ \mathrm{d}\theta.
\end{equation\*}


From the symmetry of the normal around $0$, we can easily deduce that the expected value of our discrete normal is $0$ as in the continuous case.


My question however, is how one could estimate the variance of this discrete distribution, which is thus given by
\begin{equation\*}
\sum\limits\_{x=-\infty}^{\infty} F(x; \sigma)\cdot x^2 \;.
\end{equation\*}
From some numerical simulations, it would seem that this value is very close to $\sigma^2$, the variance of the continuous distribution. Is there a way to either prove what the exact value is, or at least give some tight bound with respect to $\sigma^2$?



Estimating the variance of a discrete normal distribution
Using the Poisson summation formula, I find that the variance is


$$ \sigma^2 + \dfrac{1}{12} + \sum\_{k=1}^\infty (-1)^k e^{-2\sigma^2 k^2 \pi^2} (4 \sigma^2 + 1/(\pi^2 k^2)) $$


If $\sigma$ is not too small, the series converges quite rapidly.


