On Jun 21, 11:35*am, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 20 Jun, 23:44, Tim Wescott <t...@seemywebsite.com> wrote:
>
>
>
>
>
> > On Sat, 20 Jun 2009 11:37:27 -0700, Rune Allnor wrote:
> > > On 20 Jun, 20:14, Tim Wescott <t...@seemywebsite.com> wrote:
> > >> On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
> > >> > On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>
> > >> >> My choice of oddball integrals was intentional, as I want to go on
> > >> >> to calculating various moments for the probability distributions of
> > >> >> the surface of the hypersphere when 3D probability distributions are
> > >> >> mapped onto it. *Clearly if I map a tight Gaussian distributiononto
> > >> >> the hypersphere with a standard deviation that's much smaller than
> > >> >> the hypersphere radius the resulting probability distribution will
> > >> >> be easy; it's figuring out what happens as that probability
> > >> >> distribution opens up that's making my brain cramp.
>
> > >> > I'm a bit curious about what kind of problem leads you out in such
> > >> > kinds of calculations?
>
> > >> Unscented transformations for quaternion PDFs used to represent angles
> > >> in a (hopefully soon-to-be) unscented Kalman filter.
>
> > > The expected value is a 4-vector pointing in some desired direction and
> > > the PDF represents the probability distribution of actual directions?
>
> > Yup. *Given a PDF I'd _like_ to be able to solve for the expected value
> > of the four vector elements, as well as their cross-correlation. *I'd
> > like this to be in a form that's tractable enough that I can code it into
> > an algorithm without either making people's heads explode when they read
> > it and without making the processor bog down.
>
> > But for now I'll just settle with being able to get the element means and
> > variances out of the thing for a variety of variances of the Gaussians,
> > or a clear indication that if I insist on using Gaussians the math is
> > going to be hopelessly intractable.
>
> I did have a go at this for a Gaussian on the unit circle .
> Note that I did this at 4AM (no intoxication though), arithmetics
> has never been a force of mine, my pencil broke and I generally
> am not able to read my own handwriting.
>
> In other words, don't trust the details in what follows:
>
> ================================================== ===========
> Leaving out the scaling factors, the Gaussian PDF on the
> circle becomes (view with fixed-width font)
>
> * * * * * *inf
> g(phi) = * sum * * exp-((phi - nu)/sigma + 2 n pi)^2 *[1]
> * * * * *n = -inf
>
> since there is a possibility that the angle error in reality
> includes any number of revolutions around the unit circle.
>
> For simplicity, define
>
> a = (phi - nu)/sigma * * * * * * * * * * * * * * * * *[2a]
> b = 2pi * * * * * * * * * * * * * * * ** * * * * * * [2b]
>
> and substitute into [1]:
>
> * * * * * *inf
> g(phi) = * sum * *exp-(a + bn)^2 * * * * * * * * * * *[3]
> * * * * *n = -inf
>
> * * * * * *inf
> * * * *= * sum * *exp-(a^2+2abn+b^2n^2) * * * * * * * [4]
> * * * * *n = -inf
>
> * * * * * * * * * * *inf
> * * * *= *exp(a^2) * sum * * exp(-2abn)exp(-b^2 n^2) *[5]
> * * * * * * * * * *n = -inf
>
> Next, convert from a double-sided infinite sum to a
> one-sided infinite sum, using
>
> * inf * * * * * * * * * * * *inf
> * sum * * exp(-x) * *= 1 + 2 sum cosh(x). * * * * ** [6]
> n = -inf * * * * * * * * * * n=1
>
> Apply [6] to the first exponential inside the summation
> in [5] to find
>
> * * * * * * * * * * * * inf
> g(phi) = exp(a^2){1 + 2 sum cosh(2abn)exp(-b^2 n^2)} *[7]
> * * * * * * * * * * * * n=1
>
> To get the end result, substitute [2a-b] into [7] and insert
> the missing scaling factors.
> ================================================== ===========
>
> That's as far as I can get. Because of the n^2 fector in
> the exponent inside the sum, the summation formula for
> geometric series can't be used (at least that's how I
> understand it).
>
> Now, the form of [7] seems to be rather nice: It's a basic
> Gaussian (the outer term) and a lot of correction terms.
> Since the exponential correction term is dominated by n^2,
> you probably don't need to many correction terms to get an
> impression of the PDF. Note also that the cosh term contains
> the (phi-nu)/sigma correction, so let sigma -> inf and
> investigate what hapepns when the PDF 'opens up' towards
> uniform. Maybe one is able to work this out further,
> using power series etc.
>
> Or you can mage a worst-case assumption about the series
> being a sum of cosh(2abn)exp(-b^2 n) terms (substituted
> n for n^2 in the exponent) and use the formula for a
> geometric series to proceed.
>
> Now, you do introduce an error, but you push the PDF
> towards something 'worse' in the sense
>
> * * * * * *g(phi)_n > *g(phi)_n^2,
>
> where the subscript indicates what n term you used in
> the exponential.
>
> Whatever blunders and errors I might have made in the
> above, here is the maths challenge:
>
> Now, this excercise (or the corrected / amended version
> of it...) works well when working on a unit circle in
> 2D space. Even generalizing it to the spheric shell in
> 3D space is beyond me: The above works for the azimuth
> angle in the spherical coordinate system. I have no idea
> how to handle the elevation angle: It covers only half
> the sphere (from 'pole' to 'pole') and once it extends
> beyond +/- pi, it couples back in on the azimuth angle,
> which flips by pi.
>
> I don't see how to handle this for the problem at hand.
>
> But, as somebody made a point of earlier this week:
> That's what mathematicians are for - handle the
> technicalities in the theory we mere mortal engineers
> have to use.
>
> Any takers among the mathematicians? How does one
> extend the 'infintely many wrap-arounds' on the
> circle in 2D space to a sphere in 3D space?
>
> Rune

You extend it via the geodesics that start at the mean and wrap round
and round the sphere, using distance along the geodesic as a
coordinate. This corresponds to identifying circular sets of points in
R^{n}, where n is the dimensionality of the sphere (embedded in R^{n +
1}), just as the circle case (n = 1) corresponds to identifying all
the odd integers and all the even integers. But one has to be careful
about the underlying measure also. This has to be the measure derived
from the constant curvature metric on the sphere, i.e. that induced by
its embedding in Euclidean space.

But the Gaussian used in this way doesn't make much sense, by which I
mean there is no compelling argument for using it. There are other
distributions that reduce to a Gaussian for small variances (i.e. when
one can approximate the situation by the tangent space at the mean),
e.g. the von Mises distribution, and that are better defined on a
periodic domain. The general area is known as directional statistics.

illywhacker;