# Estimating the Variance of a Discrete Normal Distribution

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 cumulative distribution 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 $\sigma^2$?

• That's the probability mass function, not the cdf. – Robert Israel Aug 20 '14 at 21:44

$$\sigma^2 + \dfrac{1}{12} + \sum_{k=1}^\infty (-1)^k e^{-2\sigma^2 k^2 \pi^2} (4 \sigma^2 + 1/(\pi^2 k^2))$$
If $\sigma$ is not too small, the series converges quite rapidly.
If X is the continuous normal random variable and you write the discretized random variable Z as X + Y, then Var(Z) = Var(X) + Var(Y) + 2 Cov(X, Y). When sigma is large enough that Y becomes almost decorrelated from X and becomes almost uniformly distributed on (-1/2, 1/2), the variance of the discretized distribution approaches Var(X) + Var(Unif(-1/2, 1/2)) = $\sigma^2 + \frac{1}{12}$ which is the zeroth order term (and large $\sigma$ asymptotic) of the formula already provided by a more thorough answer.
To detail Robert Israel's answer (I can't comment), you have to compute the Fourier transform of the function $x^2F(x;\sigma)$ and then apply the Poisson summation formula. To compute the Fourier transform, you can use the usual tricks, i.e., replace product by convolutions of Fourier transform and get rid of integral by the integration property and the translation property of FTs.