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I am working with lines both in polar ($\rho-\theta$) and Cartesian co-ordinates. Initially, the line is given by $(\rho_0, \theta_0)$ and the covariance in the parameter space is $$R=\left(\begin{array}{cc} A & B\\ B & C \end{array}\right).$$

I am able to plot the error covariance ellipse in the parameter space using Choleski decomposition as shown how to plot a 2D covariance error ellipse?, or using convectors and values (see ellipse equation from eigenvectors and eigenvalues).

To show the expression for the covariance ellipse using the eigen method, assume the eigenvectors to be $d_1=(a_{11}, a_{12})$ and $d_2=(a_{21}, a_{22})$ while the eigenvalues $v_1$ and $v_2$. Then the expression for the covariance ellipse is

$$ K_1\left(\rho-\rho_0\right)^2+K_2\left(\rho-\rho_0\right)\left(\theta-\theta_0\right)+K_3\left(\theta-\theta_0\right)^2=1. $$

Where $$ K_1 = \left(\frac{a_{11}^2}{v_1^2}+\frac{a_{12}^2}{v_2^2}\right),\\ K_2 = 2\left(\frac{a_{11}a_{21}}{v_1^2}+\frac{a_{12}a_{22}}{v_2^2}\right),\\ K_3 = \left(\frac{a_{21}^2}{v_1^2}+\frac{a_{22}^2}{v_2^2}\right). $$

The eigenvalues can be scaled to satisfy a different confidence level using Chi-squared table (I do not remember where I saw this).

My question is how can I transform this confidence interval into a cartesian space around the line. I need a closed form expression either using the eigenvectors and eigenvalues or the covaraince matrix.

Further information

Hoping an example might clarify things better. The shaded region in the following image is formed from lines that lie on the edge of a covariance ellipse.

enter image description here

The initial line is shown in dark. It seems the other lines are tangent to some hyperbola. The red hyperbola, in the figure, is an approximate solution I found using an implementation that I saw somewhere. Accordingly, if a point $P$ lies $m$ distance from the point $(\rho_0\cos(\theta_0), \rho_0\sin(\theta_0))$ on the line, the distance from this point to the hyperbola $d_n$ is given by:

$$d_n= \pm\sqrt{C*m^2+B*m+A}$$

Problem with this approximation:

  1. It is not perfectly aligned with the edge of the shaded region.

  2. I have no idea where this came from.

I am more comfortable with geometry than algebra, but I will appreciate any suggestion.

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    $\begingroup$ The key for understanding these kind of things is polarity with respect to a conic which transforms a point-defined curve, for example an ellipse (of incertitude) into the curve that is the envelope of the polar lines of this ellipse, which is... an hyperbola (of incertitude). I have not enough time spared this evening (CET 00:45) Tomorrow evening maybe. See (en.wikipedia.org/wiki/Dual_curve) in order to have a certain idea. $\endgroup$ – Jean Marie Nov 3 '16 at 23:47
  • $\begingroup$ I tried to look at the reference that you provide, I might need to study a little deeper than I thought. In the mean time if you can get some time it would be great. $\endgroup$ – hashmuke Nov 8 '16 at 10:19
  • $\begingroup$ What I have given you is a general framework. I have prefered not to go further because I didn't know if it is the kind of things you are on. Could you say me at first if it is more understandable. $\endgroup$ – Jean Marie Nov 8 '16 at 12:21
  • $\begingroup$ Is there a way to come up with an expression for the hyperbola given the ellipse or vise-versa? Where else might be the application for this type of transformation - just to look for any example. $\endgroup$ – hashmuke Nov 8 '16 at 12:47
  • $\begingroup$ I am going to make an Edit to my answer with an example. The formulas and information can be found on this French site (mathcurve.com/courbes2d/polaire/polaire.shtml), but probably you don't know French. $\endgroup$ – Jean Marie Nov 9 '16 at 11:24
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Your question is important. Permit me to reformulate it as I have met the issue some years ago.

Consider a family of straight lines with "normal equations":

$$x \cos(\theta) + y \sin(\theta) - p=0$$

for which we have fluctuations $\Delta \theta, \Delta p$ around a nominal value. Observation and/or simulation give rise to hyperbolic locii (plural of "locus"). Just as Gaussian behavior gives rise to incertitude ellipses. How to interpret a hyperbolic locus? How can we get certain characteristics out of it (for example, how its "narrowness" can be quantified in statistical terms ?...).

One good answer is through the duality "pole-polar". Have a look at the graphics below. I made it with Geogebra software that has a "Polar" tool (underlining its importance in geometry). All is with reference to circle centered in $0$ with radius $R=2$, but it could any other circle or even any other conic curve.

An example of the pole-polar association: point M has for its polar line the green line which has two properties: it is orthogonal to $OM$, and if $S$ denotes the intersection point with $OM$, one has $OM.OS=R^2.$ Moreover, as $M$ is external to the circle, there is another means for obtaining its polar line: take the two tangent lines issued from $M$ to the circle ; let U and V be their contact points ; then line $UV$ is the polar line of $M$.

The main thing to be observed is that the polar lines of points situated on the hyperbola $\frak{H}$ are enveloping a certain ellipse $\frak{E}.$ This process can be done in a dual way: had we taken points on ellipse $\frak{E}$, we would have obtained tangents to hyperbola $\frak{H}$. In particular, if we take the polar line of the center $C$ of $\frak{E}$, we obtain the main axis of $\frak{H}$.

With this possible transfer, one can take advantage of reliable methods and techniques developed for gaussian distributions.

Moreover, what has been said for 2D is at once extensible to 3D, because duality is not limited to 2D: instead of a point-line coupling, it is a point-plane coupling with respect to a sphere.

Edit : All this is accessible to simple analytic computations. Let us take an example. Consider the second image below. One can see a hyperbola and an ellipse with resp. parametric equations:

$$\frak{H}\cases{x=\tan(t)\\y=\dfrac{0.2}{\cos(t)}+3} \ \ \ \ \ \ \frak{E}\cases{X=\dfrac{-0.2\sin(t)R^2}{0.2 \cos(t)+3}\\Y=\dfrac{R^2}{0.2 \cos(t)+3}}$$

(where the radius of the circle is $R=2$).

How has been found the second system for $\frak{E}$ knowing the first system (for $\frak{H}$)?

By using transformation:

$$\cases{X=\dfrac{-R^2y'}{xy'-yx'}\\X=\dfrac{R^2x'}{xy'-yx'}}.$$

We refer to this French site for more explanation. I will look for an equivalent English speaking site.

enter image description here

enter image description here

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  • $\begingroup$ Is there a missing notation? Which O are you talking about when you say OM? How do you relate the points on the hyperbola with the ellipse? $\endgroup$ – hashmuke Nov 8 '16 at 12:23
  • $\begingroup$ O is the origin. $\endgroup$ – Jean Marie Nov 8 '16 at 12:40
  • $\begingroup$ As I remember, using polar equations, the correspondence between the equation $r=f(\theta)$ of the hyperbola and the equation $r=g(\theta)$ of the ellipse is not very complicated. Are you at ease with these notations ? $\endgroup$ – Jean Marie Nov 8 '16 at 12:52
  • $\begingroup$ Yes I am, then I can translate the $ r = f(\theta) $ into Cartesian form. $\endgroup$ – hashmuke Nov 8 '16 at 12:55
  • $\begingroup$ Edit done. In fact, it was simpler in this case to refer to a parametric representation instead of a polar one. $\endgroup$ – Jean Marie Nov 9 '16 at 11:47

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