Radial axis transformation in polar kernel density estimate Consider a kernel density estimate of a continuous, non-negative random variable defined over the unit circle with no discontinuity between 360 and 0 degrees.
Unlike in the most common KDE implementations that use a Gaussian distribution (such as Seaborn's kdeplot), apparently distributions of a polar nature should use the von Mises distribution for their kernel.
If such a KDE is shown on a polar plot, I think there will be a kind of visual distortion introduced. The KDE's area-under-curve is significant because the integral of a KDE should be 1. In the polar plot, looking at two sectors where

*

*the sector angle spans are the same, but

*the first sector radius is double the second sector radius,

the first sector will not have double the area; it will have more than double. The effect will be a visual bias where higher densities are over-emphasized when compared with lower densities, especially if the graph is drawn to be filled under the curve.
I imagine that a way to correct for this would be a non-linear radius dimension where lower values are more spaced-out than higher-values. I have searched and cannot find example images where this has been done. My questions are:

*

*Is this kind of visual bias commonly corrected-for when showing rendered polar plots?

*I believe the expression that defines the radial corrective transformation is simply $r_i = \sqrt{i}$ . Does this seem correct?

*Would this corrective transformation be valid in the context of a von Mises KDE?

 A: Consider any density $f$ for the circular parameter $\theta.$  The relevant integrals are of the form $$\Pr(\mathcal A) = \int_\mathcal{A}f(\theta)\,\mathrm d\theta$$ where $\mathcal A\subset[0,2\pi)$ is any circular event. Ordinarily we would plot them in Cartesian coordinates, as in this example:

Now, if you wish to represent these integrals as circular areas, perhaps you are thinking of plotting the graph of some related functions $g$ and $h$ in polar coordinates, given by the region
$$\{(\theta, r)\mid g(\theta)\le r \le h(\theta);\ 0\le \theta\lt 2\pi\}.$$
The area on the plot itself therefore is
$$\int_\mathcal{A}\int_{g(\theta)}^{h(\theta)} r\,\mathrm dr\,\mathrm d\theta = \int_{\mathcal A}\frac{h(\theta)^2 - g(\theta)^2}{2}\,\mathrm d\theta.$$
Consequently, if you pick any nonnegative functions for which $h(\theta)^2 - g(\theta)^2 = f(\theta)$ the right side works out to the desired probability.
Two natural choices are
$$(g(\theta), h(\theta)) = (0, \sqrt{2 f(\theta)}),$$
the "filled" version

and
$$(g(\theta), h(\theta)) = (\sqrt{f(\theta)}/\lambda, \lambda\sqrt{f(\theta)})$$
where $\lambda = \sqrt{1 + \sqrt{2}},$ the "symmetric" version.

Other choices are possible.  For instance, you could enclose everything within a disk provided $f$ is bounded.
