# Closed form of Gaussian CDF for integers

I'm reviewing a paper that contains the following expression:

"$...=\int_{y-0.5}^{y+0.5}\mathcal{N}(t|0,\sigma)dt$ , which can be evaluated in closed form"

As far as I know, the CDF of the normal distribution has in general no closed-form expression. In this case, $\sigma$ is not known ahead of time and y is known to be an integer. However, this still shouldn't make it closed-form evaluable?! It's written with such naturalness that I'm wondering whether I'm overlooking something ...

• What does "$\mathcal N$" refer to? Would it be the density of a Normal distribution of variance $\sigma^2$, as is usual (and would be expected when appearing in such an integral)? If so, then its evaluation is simply $\Phi((y+1/2)\sigma)-\Phi((y-1/2)/\sigma)$ where $\Phi$ is the standard Normal cdf (and that looks like a "closed form" to me). Otherwise, what is it?
– whuber
Jan 30, 2018 at 14:56
• Dear @whuber: Yes, it's written in closed form, but that's not the same as evaluation in closed form as erf() requires numerical evaluation, that was the point I tried to make. Feb 2, 2018 at 22:35
• Considering that elementary functions like $\exp$ (which is involved in all non-integral powers) and $\cos$ require "numerical evaluation," too, might provide some useful context. $\operatorname{erf}$ is no more difficult or time-consuming to evaluate than any of those. I'm still trying to figure out whether you're asking about integrating the density or the CDF itself: could you address that question about the meaning of your notation?
– whuber
Feb 2, 2018 at 22:42
• @whuber I'm asking about the CDF, i.e. the result of the integration. I see your point and actually, it's sort of valid! Feb 9, 2018 at 19:46

$\Bigg[ \frac{1}{2} \Big[1 + \text{erf} \Big(\frac{t}{\sigma \sqrt{2}} \Big)\Big] \Bigg]_{t=y-0.5}^{t=y+0.5}$
does indeed depend on the error function $\text{erf(x)}=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-x^2} dx$, which must be computed using numerical methods. So in that sense it is not really closed form.
As for the standard deviation parameter $\sigma$, I can't see how you would get around evaluating the CDF without knowing it. Indeed, the conditional form the author wrote the PDF expression in - $\mathcal{N}(t|0,\sigma)$ - does suggest it is a dependency.