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I am trying to compute the Kolmogorov distance between two univariate gaussian distributions $\mathcal{N}(0,n)$ and $\mathcal{N}(0,2n)$ for large $n$. I have a feeling this should be simple but whatever I have tried so far doesn't work. Could anyone give me some hints?

By the Kolmogorov distance between two distributions $P$ and $Q$ I mean:

$$\displaystyle max_t | \Pr[P>t] - \Pr[Q>t] | $$


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1 Answer 1

up vote 2 down vote accepted

Because the Gaussians have the same mean and Gaussians are symmetric about the mean, the expression $\Pr[P>t] - \Pr[Q>t]$ should have a single maximum and a single minimum whose absolute values are the same. So you could just solve $\max_t (\Pr[P>t] - \Pr[Q>t])$. To do that, I would express $\Pr[P>t]$ as $$\int_t^{\infty} \frac{1}{\sqrt{2\pi n}} e^{-x^2/(2n)} dx,$$ and $\Pr[Q>t]$ similarly. Then take the derivative of the differences with respect to $t$ via the Fundamental Theorem of Calculus, set it to $0$, and solve for $t$. Since you don't want a full answer, I'll stop there.

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Thanks! I got the value of $t$ ($\sqrt{n \log 4}$) but now I have difficulty evaluating the expression $Pr[P >t]$. I understand there is no closed form known, are there approximations which might be useful in this case? –  Preyas Popat Nov 23 '10 at 6:47
@Preyas: Right, there's no closed form. The cdf of the Gaussian can be expressed in terms of the well-known error function erf$(x)$, though. See, for example, en.wikipedia.org/wiki/… –  Mike Spivey Nov 23 '10 at 6:51
Thanks again! I had not read the Taylor expansion of the error function –  Preyas Popat Nov 23 '10 at 7:04

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