Convolution "square root" for kernels

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Convolution "square root" for kernels

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MKBHD

#1 h Apr 17, 2025, 3:09 AM

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Given a kernel function (typically even, unimodal and nonnegative), $f(x)$ how can I find it's "square root", $k(x)$ that is a function so that $k(x)*k(x)=f(x)$ ?

This is pretty easy when $f(x)$ is a normal distribution (you just halve the variance to find $k(x)$ since the sum of two normal RVs is another normal RV), but what about when $f(x)$ is slightly different from a normal distribution (e.g. logistic or cosine)?

Logistic is applicable for Elo ratings, where we know the link function ( $\frac{1}{1+10^{-R_D/400}}$ )based on the difference in rating but want to find the kernel for the corresponding "TrueSkill" (range of performances) -

around the 2:00 mark here: https://www.youtube.com/watch?v=AsYfbmp0To0&pp=ygUFIzJlbG8%3D

I know a normal gives a pretty good approximation, but I want the exact function.

I think there's a way to get an explicit form of the answer by taking the fourier transform into the frequency domain and square rooting and then translating back, but it's so complicated, I just want to see the graph plotted and how it differs from normal, I don't care about the explicit formula if it's not intuitive.

I'm more interested in the case of the "square root" of cosine from $-\pi/2$ to $\pi/2$ .

Here's a relaxed set of constraints for the problem that I truly care about the most:

Find a kernel function, $k(x)$ , that meets the following criteria (similar to $k(x)*k(x)= f(x) = \frac{\cos\left(\frac{\pi x}{6}\right)\pi}{3}$ )

- define the cdf: $F(x) = \int_{-3}^x f(x) dx$
- $F(0) = \sin(\pi/4)^2 = 0.5$ - trivial when $k(x)$ is even
- $F(1) = \sin(\pi/3)^2 = 0.75$
- $F(2) = \sin(5\pi/12)^2 \approx 0.933$
- $F(3) = \sin(\pi/2)^2 = 1$ - this can be achieved by limiting the support of $k(x)$ to $(-1.5,1.5)$

The constraints for $F(0)=0$ and $F(3)=1$ just means that $k(x) = 0$ for $|x|>1.5$ and that $k(-x)=k(x)$ and $\int_{0}^{1.5} k(x) = 0.5$ (normalization).

So we just have 2 "difficult constraints" for $F(1) = 0.75$ and $F(2) =\frac{2+\sqrt{3}}{4}\approx0.933$

This can easily be satisfied by creating a set of functions with 2 degrees of freedom and then fitting the two constraints.

Additional criteria:
- I want $k(x)$ to be unimodal at $x=0$
- Ideally $k(x)$ should be continuous
- Ideally $k(x)$ is differentiable on $(-1.5, 1.5)$
--

Maybe this is related: https://en.wikipedia.org/wiki/Infinite_divisibility_(probability)

I tried asking the latest ChatGPT models which are near-perfect on AIME (o3 and o4-mini) and the answers are so subpar, so I'm back on AoPS.

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greenturtle3141

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greenturtle3141

#2 h Apr 17, 2025, 6:49 AM

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You are insulting every mathematician (and artist!) by depending on chatgpt whatsoever. Don't do that.

I don't know why you are phrasing this in probabilistic terms since the problem of finding a "convolution square root" of $g$ is perfectly well-defined for, say, any $g \in L^1(\mathbb{R})$ .

The answer is Fourier analysis, whether you like it or not. It gives you all possible answers immediately and moreover gives the simplest possible form for such answers: If $f \star f = g$ then $(\hat{f})^2 = \hat{g}$ , and $\hat{f} = \sqrt{\hat{g}}$ where the expression $\sqrt{\hat{g}}$ is to be interpreted as some function which gives $\hat{g}$ when squared, which certainly may not be unique. So the explicit form is
$f(x) = \int_{-\infty}^\infty \sqrt{\int_{-\infty}^\infty g(z)e^{-2\pi i yz}}e^{2\pi xy}\,dy.$ Fourier analysis is the tool to help you here, you are not going to find anything better. (Either that, or prove me wrong!) You say you don't like this because you just want to plot the graph, but this is perfectly plottable with some sufficient numerical computation.

We can attempt to find the convolution square root of cosine restricted to an interval as you wish. Say $g = \cos(2\pi x)1_{(-1/2,1/2)}$ for simplicity. Then $\hat{g} = (\delta_1 + \delta_{-1})/2 \star \hat{1}_{(-1/2,1/2)} = \frac12\left(\hat{1}_{(-1/2,1/2)}(\xi-1) + \hat{1}_{(-1/2,1/2)}(\xi+1)\right).$ But
$\hat{1}_{(-1/2,1/2)}(\xi) = \int_{-1/2}^{1/2} e^{-2\pi ix\xi}\,dx = \frac{1}{-\pi i \xi} \sin(\pi \xi)$ so
$\hat{g} = \frac{1}{-2\pi i \xi}\left( \sin(\pi \xi-\pi) + \sin(\pi\xi + \pi)\right) = \frac{1}{\pi i \xi}\sin(\pi \xi).$ Now
$\sqrt{\hat{g}} = \sqrt{\frac{\sin(\pi \xi)}{\pi i \xi}},$ where again "square root" should be interpreted carefully, and so the convolution square root, $f$ , is the inverse Fourier of this, or
$f(x) = \int_{-\infty}^\infty \sqrt{\frac{\sin(\pi \xi)}{\pi i \xi}}e^{2\pi i x \xi}\,d\xi.$ Now let's pick a square root. One sees that the expression under the square root is pure imaginary for all $\xi$ and so we may take the principal square root. In this case, this means that
$\sqrt{\frac{\sin(\pi \xi)}{\pi i \xi}} = \sqrt{\left|\frac{\sin(\pi\xi)}{\pi\xi}\right|}e^{i\pi/4 \cdot \varepsilon_\xi}$ where
$\varepsilon_\xi = \begin{cases}1, & \frac{\sin(\pi \xi)}{\pi\xi} < 0 \\ -1, & \frac{\sin(\pi \xi)}{\pi\xi} > 0\end{cases}.$ So
$f(x) = \int_{-\infty}^\infty \sqrt{\left|\frac{\sin(\pi\xi)}{\pi\xi}\right|}e^{2\pi i\left(x \xi + \frac{\varepsilon_\xi}{8}\right)}\,d\xi.$ As far as I can tell this is not real, so if you'd like a real answer you may instead consider
$f(x) = \int_{-\infty}^\infty \sqrt{\left|\frac{\sin(\pi\xi)}{\pi\xi}\right|}e^{2\pi i\left(x \xi + \frac{\varepsilon_\xi}{8}\right)} h(\xi)\,d\xi$ where $h:\mathbb{R} \to \{-1,1\}$ is carefully chosen. Anyways, it's more or less evident that a closed form for this is quite unlikely, so I highly doubt any alternative approach could be any cleaner.

This post has been edited 1 time. Last edited by greenturtle3141, Apr 17, 2025, 6:50 AM

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