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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 ...

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  • $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Jan 30, 2018 at 14:56
  • $\begingroup$ 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. $\endgroup$
    – jmaxx
    Feb 2, 2018 at 22:35
  • $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Feb 2, 2018 at 22:42
  • $\begingroup$ @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! $\endgroup$
    – jmaxx
    Feb 9, 2018 at 19:46

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I see your point. The so-called "closed form" of the CDF of a Gaussian, which in your case would be evaluated as part of this definite integral

$\Bigg[ \frac{1}{2} \Big[1 + \text{erf} \Big(\frac{t}{\sigma \sqrt{2}} \Big)\Big] \Bigg]_{t=y-0.5}^{t=y+0.5}$

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.

But it does not depend on any kind of sampling scheme (like Monte Carlo) to be evaluated, so in that sense it is 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.

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  • $\begingroup$ That's similar to my thoughts on the matter. Means it's unlikely I've overlooked something trivial, thanks a lot! $\endgroup$
    – jmaxx
    Jan 30, 2018 at 9:15

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