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Niall
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 RANDOM POINTS ON SPHERE: statement correction   « on: Apr 17th, 2010, 10:06am » Quote Modify

New here, so apologies if this isn't the right approach. I'm hoping to change the phrasing of the problem below - not sure of the best way.

The problem: RANDOM POINTS ON SPHERE asks for the 'maximum likelihood estimate' of a a random variable (the random variable being the distance between two points uniformly drawn on the sphere).

This is a technical term and incorrect terminology here - it also sounds more complicated than needed, when we could use "most likely" (which is technically vague, but unambiguous here I think) as in the following suggested phrasing:

"RANDOM POINTS ON SPHERE
Contributor: Joc Koelman

9:00 PM 8/19/2004

Two points on the surface of a sphere are drawn uniformly at random. What is the most likely distance between these two points? The answer may be shocking at first. After getting the answer, try to explain it intuitively."

P.S. If people are worried about the interpretation of "expected distance" then perhaps by also asking for that would highlight the distinction?
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Niall
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 Re: RANDOM POINTS ON SPHERE: statement correction   « Reply #1 on: Apr 17th, 2010, 10:10am » Quote Modify

I guess it should also be made clear what "distance between these two points" means. Maybe:

"the distance between these two points across the surface of the sphere"

if this is what the author meant.
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towr
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 Re: RANDOM POINTS ON SPHERE: statement correction   « Reply #2 on: Apr 17th, 2010, 2:53pm » Quote Modify

Originally the question was asked simply as "What is the most likely distance between two randomly selected points on a sphere?"
But then there followed some discussion about what was precisely meant. You can read the topic about this puzzle here. (It's in the 12th post William suggest phrasing it as "What is the maximum likelihood estimate ..")

It would probably be a good idea to give both a simple, accessible phrasing and a more formal, precise one (in which case it would be important to also get the technical terms correctly, if they aren't. But I'm not familiar enough with them to judge).
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srussell
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 Re: RANDOM POINTS ON SPHERE: statement correction   « Reply #3 on: Nov 3rd, 2012, 12:27pm » Quote Modify

ok i am really disappointed by this whole thread.

Distance can mean one of two things: angular distance, measured in radians, and linear distance, measured as the difference between radial vectors that describe the position of the points. Since linear distance is a monotonic increasing function of anguar distance, between 0 and pi(), any result of maxima over that range will be the same either way you derive it.

But to go back to the initial problem, there is a major flaw in constructing the argument, that of compensating for spherical symmetry in developing the infinitessimal element.

If our first point is at the pole, the most probable position of the second point has to be at the equator, since the infinitessimal area at the equator is the largest of any angle. but its not easy to get this result from a standard workup.

The area a of the infinitessimal strip at an angular distance x from the pole can be found from the standard method of creating a derivative:

lim(delta(x)->0) of  Ksin(x)delta(x)
where K = (2pi()r^2)

which gives us
da(x) = Ksin(x)dx   ==>

a(x) = -Kcos(x)|(0 to x) = K(1+ cosx) + C

The probability P(x) that the point will be found within this strip is a/A, where A is the total surface area of the sphere. P(x) is obviously maximum at pi()/2.

But the problem with this is that da(pi()/2) is not equal to zero. which implies that the most likely angular distance is NOT at the distance of maximum probabilty

Right, you can see there is a problem. And it comes from the choice of delta(x). check it out:
for x < pi()/2, sin(x) + delta(x) > sin(x)
for x > pi()/2, sin(x) + delta(x) < sin(x)

To reduce it to a picture of Reimann sums (you may sketch these out yourself on a piece of paper), the rectangles on the near side of the equator approximate the area under the curve with an infinitessimal error ABOVE the curve, while the rectangles on the far side of the equator approximate the area under the curve with an infinetissimal error BELOW the curve. This gives a maximum somewhere just beyond an angular distance of 1.2 radians, on the near side of the equator. And its WRONG.

So how do we solve it?

Ans: the infinetissmal area has to be the average of an infinetissemal area that is centered around x, not in front or behind it.

(||d(x)||+||d(-x)||)/2)

Which seems more difficult than it is. The short answer (i leave the derivation up to you) is:

da(x) = K/2(sin(x)dx + sin(pi()-x)d(pi()-x))

and

a(x) = - K/2(-(1+cos(x))+(1+cos(pi()-x)))

so the probabilty is

P(x) = - 1/4(-(1+cos(x))+(1+cos(pi()-x)))

The probability of finding the second point at either pole is 0
P(0) = 0
P(pi()) = 0

And the most likely position of the point is at the equator.
dP(x=pi()/2)=0

The angular or linear distance between the pole and the equator is known relative to the radius of the sphere.

For completeness we note that the probability of finding the point someplace on the whole surface of the sphere is 1. (note we integrate from 0 to 2pi() and multiply by 2 because of symmetry issues again)

2 * int(P(x))|(x = 0 to x = pi()/2)=

-1/2(-(x+sin(x))+(x-sin(x)))|(0 to pi()/2) = 1

QED

. . . and let us never speak of this again.
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