Uniform sampling of points on the curve of an ellipse. Jacobian of transformation.

Question

I have been wondering on how to properly uniformly samples points on the curve of an ellipse using well-explored, given workflow. That is, consider the following parametric representation of an ellipse: $$ x = a r \cos(t) $$ $$ y = b r \sin(t) $$

, where $t\in[0, 2\pi) $

Note that parameter $r$ simply shrinks or enlarges the ellipse, and $a$ and $b$ are constants (axes).

Now, given the probability density in Cartesian coordinates p(x,y) (which is, as far as I am concerned, simply 1 over circumference of an ellipse (I originally said area, which is wrong, of course)) probability density in the new coordinate system can be found as follows: $$ p(r, t) = |J_t| p(x,y) $$ where $Jt=a b r$ is the Jacobian of transformation

At this point I tried to see if this new pdf integrates to 1, that is: $$ \int_0^{2\pi}p(r, t)dt $$

Note once again that I set $r$ to 1 and simply try to integrate over the domain of parameter $t$. Unfortunately, it just doesn't add up to 1. Consequently, having wrong pdf I can not proceed and apply inversion method to sample the points. Could you please point out the flaw in my reasoning? How is that my pdf does not integrate to 1? Thank you.

The method described here: How to generate points uniformly distributed on the surface of an ellipsoid? works for arbitrary dimensions, I believe. It certainly works in two. (I just tried) — Linas, Aug 09 '20 at 23:42

Ng Chung Tak · Accepted Answer · 2016-12-15T11:29:58.807

2

It needs to be parametrized by arc length:

\begin{align*} ds &= r\sqrt{a^2\sin^2 t+b^2\cos^2 t} \, dt \\ S_\text{total} &= r\int_{0}^{2\pi} \sqrt{a^2\sin^2 t+b^2\cos^2 t} \, dt \\ &= 4raE\left( \sqrt{1-\frac{b^2}{a^2}} \right) \\ f(s) &= \frac{1}{4raE\left( \sqrt{1-\frac{b^2}{a^2}} \right)} \\ p_r(t) &= f(s) \times \frac{ds}{dt} \\ &= \frac{\sqrt{a^2\sin^2 t+b^2\cos^2 t}} {4aE\left( \sqrt{1-\frac{b^2}{a^2}} \right)} \end{align*}

where $E(\cdot)$ is complete elliptic integral of the second kind.

Caution

For partial arc length with $a>b$, $(a\sin \theta, b\cos \theta)$ or clockwise convention is used instead.

$$s(\theta)=a\int_{0}^{\theta} \sqrt{1-\left( 1-\frac{b^2}{a^2} \right)\sin^2\theta} \, d\theta = aE\left( \theta, \sqrt{1-\frac{b^2}{a^2}} \right)$$ where $E(\cdot, \cdot)$ is an incomplete elliptic integral of the second kind.

edited Dec 15 '16 at 11:29

answered Dec 15 '16 at 11:10

Ng Chung Tak

19,693

You might want to explain the complete elliptic integral of the second kind for the kids at home. ;) – J. M. ain't a mathematician Dec 15 '16 at 11:16
@J.M. Explanation added, thanks for your remind. – Ng Chung Tak Dec 15 '16 at 11:17
Almost there; you'll want to include the full definition you're using, since there are a number of different conventions for the argument; your preference might not be the same as the OP's. – J. M. ain't a mathematician Dec 15 '16 at 11:18
@J.M. Links of references are added with precaution. – Ng Chung Tak Dec 15 '16 at 11:26
@NgChungTak A fine answer, much appreciated! Now I can indeed proceed to sample parameter t. Might I also ask why does the approach described by me fails? It was described in a certain text book on the examples of shapes such as circles and spheres and worked just fine. – MutomboDikey Dec 15 '16 at 12:12
Your original approach was considering $2$D space similar to polar coordinates up to linear scaling (either $a$, $b$ or $r$). However, in your case is a closed curve and hence line integral is the way. Due to non-linearity of arc-length, the pdf is invariant under the scaling of $r$ only. In terms of eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$, $$p(t)=\frac{\sqrt{1-e^2\cos^2 t^2}}{4E(e)}$$ – Ng Chung Tak Dec 15 '16 at 13:06

Uniform sampling of points on the curve of an ellipse. Jacobian of transformation.

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