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Topic: random points on a sphere (Read 9949 times) |
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Rejeev
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see the attached diagram. A' and A are same.
cosA' is perpenticular component of area and
sin2A is radius [in my previous posting i was wrongly used sinA; in fact i was thinking for a while sin2A = 2 sinA  ].
Hence probability for a purticular distance is maximun at: where cosAsin2A is maximum (not cosAsinA).
Is this matching with Eigemray's solution?
How to paste a image (jpg) in reply ( so i have attached it)
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ccornchip
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Re: random points on a sphere
« Reply #26 on: Oct 6th, 2011, 1:15pm » |
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Ok, so this is an ancient thread: I send my standard bumping apologies.
The answer, 2R, bothers me. It goes all the way back to Leo Broukhis's question: "How do you define the notion of maximum likelyhood in order to arrive at your intended answer?" which was answered by "The distance d for which the probability distribution P(x) peaks at it's (sic) maximum value."
Meaning, "let f(x) be the probability density function (p.d.f.) of the Euclidean distance between two points on a sphere. Find the x that maximises f(x)." And the answer to that is indeed x=2R.
The reason this doesn't sit right with many people (myself included) is that the number is meaningless and does not relate to the "most likely distance."
As a physicist, I ask a similar question, "given that on average, there is one star every cubic megaparsec of space in the universe, what is the 'most likely distance' than one can travel before encountering a star"
... the proposed solution would give zero since the p.d.f. goes as exp(-x).
I interpret "most likely distance" as
"Given a ball with two dots placed at random on its surface, what is the (Euclidean) distance between the dots that you would place a bet on finding, given that the closest punter wins."
This becomes an expected value problem, which has an answer not yet mentioned on this thread at all.
It's not 2R, nor is it R*sqrt(2), it's:
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4R/3.
Take the p.d.f. f(x) by william wu, multiply by x then integrate.
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towr
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Re: random points on a sphere
« Reply #27 on: Oct 6th, 2011, 10:59pm » |
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on Oct 6th, 2011, 1:15pm, ccornchip wrote:I interpret "most likely distance" as
"Given a ball with two dots placed at random on its surface, what is the (Euclidean) distance between the dots that you would place a bet on finding, given that the closest punter wins."
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Running a quick simulation* supports that you'd be about twice as likely to win against 2R.
However if I picked ever so slightly higher than your 4/3R, I seem to win more than 55% of the time. So it does not seem like an unambiguous best bet.
(Which come to think of it isn't surprising, because the expected value isn't necessarily the median, and in a one on one bet the median is the better bet if the closest one wins.)
* by picking random x,y,z in space, keeping ones that fall in a small range of distance from the center of the sphere, rather than exactly on a surface.
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| « Last Edit: Oct 6th, 2011, 11:01pm by towr » |
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TenaliRaman
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Re: random points on a sphere
« Reply #28 on: Oct 13th, 2011, 11:06pm » |
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Choosing the random points on the unit sphere in the following way,
1] Choose x,y,z from a normal distribution
2] Normalize the vector using the norm
Now, computing distance between various points and computing the histogram, I get the largest bucket to be the range 1.9 - 2.0. This supports the original assertion of 2R.
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Grimbal
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Re: random points on a sphere
« Reply #29 on: Oct 25th, 2011, 2:04am » |
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I did a quick simulation with 1000 buckets and 1000000 samples. It shows that P(x) is proportional to x.
Method:
each sample:
draw x, y and z uniformly in [-1,1]
d = |(x,y,z)|
discard samples that don't have 0.1<d<1
divide (x,y,z) by d
compute d = distance between (x,y,z) and (0,0,-1)
increment bucket[d/2*1000]
at the end:
plot bucket[n] vs n
Graphically it shows a straight line starting at (0,0).
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