Counter-proof of orthogonality of random points in a higher-dimensional unit sphere
$begingroup$
I seek to provide a counter-proof against the following statement about the unit sphere in a N-dimensional space, with a large value for N.
Statement: Two randomly selected points on the surface of the sphere are nearly orthogonal.
Counter-proof using a random arc length
Create an arc of a unit circle with a length of random value in the interval of (0, π) and drop it on the surface of the N-dimensional sphere at a random place with a random orientation in the (N-1) dimensional surface. The two ends of the arc locate the two random points. However, the length of the arc does not have a high probability of being nearly π/2.
Causes for the orthogonality found in theory and in computations
I think the pole and equator method is only one of the multiple possible methods to estimate the probability in this case. Random selections in a multi-dimensional continuum might have indeterministic probabilities.
Question: What are the flaws in my reasoning?
probability probability-theory random-variables analytic-geometry dimensional-analysis
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
$endgroup$
add a comment |
$begingroup$
I seek to provide a counter-proof against the following statement about the unit sphere in a N-dimensional space, with a large value for N.
Statement: Two randomly selected points on the surface of the sphere are nearly orthogonal.
Counter-proof using a random arc length
Create an arc of a unit circle with a length of random value in the interval of (0, π) and drop it on the surface of the N-dimensional sphere at a random place with a random orientation in the (N-1) dimensional surface. The two ends of the arc locate the two random points. However, the length of the arc does not have a high probability of being nearly π/2.
Causes for the orthogonality found in theory and in computations
I think the pole and equator method is only one of the multiple possible methods to estimate the probability in this case. Random selections in a multi-dimensional continuum might have indeterministic probabilities.
Question: What are the flaws in my reasoning?
probability probability-theory random-variables analytic-geometry dimensional-analysis
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
$endgroup$
1
$begingroup$
The Jacobian for your process does not agree with what the world calls uniform distribution on the sphere. If $x$ and $y$ are two random points you should be able to work out the expectation of $|x-y|^2$ for your way and for the usual way.
$endgroup$
– kimchi lover
Jan 8 at 23:50
$begingroup$
The mid points of the arc have a standard uniform distribution over the sphere's surface. The lengths and orientations have a standard uniform distribution. Any hints on why the end points would not have a uniform distribution?
$endgroup$
– Venkata Pagadala
Jan 9 at 5:43
add a comment |
$begingroup$
I seek to provide a counter-proof against the following statement about the unit sphere in a N-dimensional space, with a large value for N.
Statement: Two randomly selected points on the surface of the sphere are nearly orthogonal.
Counter-proof using a random arc length
Create an arc of a unit circle with a length of random value in the interval of (0, π) and drop it on the surface of the N-dimensional sphere at a random place with a random orientation in the (N-1) dimensional surface. The two ends of the arc locate the two random points. However, the length of the arc does not have a high probability of being nearly π/2.
Causes for the orthogonality found in theory and in computations
I think the pole and equator method is only one of the multiple possible methods to estimate the probability in this case. Random selections in a multi-dimensional continuum might have indeterministic probabilities.
Question: What are the flaws in my reasoning?
probability probability-theory random-variables analytic-geometry dimensional-analysis
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
$endgroup$
I seek to provide a counter-proof against the following statement about the unit sphere in a N-dimensional space, with a large value for N.
Statement: Two randomly selected points on the surface of the sphere are nearly orthogonal.
Counter-proof using a random arc length
Create an arc of a unit circle with a length of random value in the interval of (0, π) and drop it on the surface of the N-dimensional sphere at a random place with a random orientation in the (N-1) dimensional surface. The two ends of the arc locate the two random points. However, the length of the arc does not have a high probability of being nearly π/2.
Causes for the orthogonality found in theory and in computations
I think the pole and equator method is only one of the multiple possible methods to estimate the probability in this case. Random selections in a multi-dimensional continuum might have indeterministic probabilities.
Question: What are the flaws in my reasoning?
probability probability-theory random-variables analytic-geometry dimensional-analysis
probability probability-theory random-variables analytic-geometry dimensional-analysis
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
asked Jan 8 at 23:33
Venkata PagadalaVenkata Pagadala
164
164
1
$begingroup$
The Jacobian for your process does not agree with what the world calls uniform distribution on the sphere. If $x$ and $y$ are two random points you should be able to work out the expectation of $|x-y|^2$ for your way and for the usual way.
$endgroup$
– kimchi lover
Jan 8 at 23:50
$begingroup$
The mid points of the arc have a standard uniform distribution over the sphere's surface. The lengths and orientations have a standard uniform distribution. Any hints on why the end points would not have a uniform distribution?
$endgroup$
– Venkata Pagadala
Jan 9 at 5:43
add a comment |
1
$begingroup$
The Jacobian for your process does not agree with what the world calls uniform distribution on the sphere. If $x$ and $y$ are two random points you should be able to work out the expectation of $|x-y|^2$ for your way and for the usual way.
$endgroup$
– kimchi lover
Jan 8 at 23:50
$begingroup$
The mid points of the arc have a standard uniform distribution over the sphere's surface. The lengths and orientations have a standard uniform distribution. Any hints on why the end points would not have a uniform distribution?
$endgroup$
– Venkata Pagadala
Jan 9 at 5:43
1
1
$begingroup$
The Jacobian for your process does not agree with what the world calls uniform distribution on the sphere. If $x$ and $y$ are two random points you should be able to work out the expectation of $|x-y|^2$ for your way and for the usual way.
$endgroup$
– kimchi lover
Jan 8 at 23:50
$begingroup$
The Jacobian for your process does not agree with what the world calls uniform distribution on the sphere. If $x$ and $y$ are two random points you should be able to work out the expectation of $|x-y|^2$ for your way and for the usual way.
$endgroup$
– kimchi lover
Jan 8 at 23:50
$begingroup$
The mid points of the arc have a standard uniform distribution over the sphere's surface. The lengths and orientations have a standard uniform distribution. Any hints on why the end points would not have a uniform distribution?
$endgroup$
– Venkata Pagadala
Jan 9 at 5:43
$begingroup$
The mid points of the arc have a standard uniform distribution over the sphere's surface. The lengths and orientations have a standard uniform distribution. Any hints on why the end points would not have a uniform distribution?
$endgroup$
– Venkata Pagadala
Jan 9 at 5:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not a counterproof.
Implicit in the original statement ("two randomly selected points on the surface of the sphere are nearly orthogonal") is that the randomly selected points are (separately) chosen uniformly. The fact that the conclusion doesn't hold for some other distribution you propose just shows that your distribution (on pairs of points) isn't the uniform one.
This is, by the way, the same phenomenon illustrated by the Bertrand Paradox. You might find it instructive to study that paradox, which is (at least dimensionally) simpler than the situation you describe.
answered Jan 10 at 7:22
TravisTravis
60.6k767147
$endgroup$
$begingroup$
Of course, my distribution is not the same as the one used by the original statement, but I don't think one can claim that one of these distributions is uniform and the other isn't. For example, the distribution used by the original statement isn't uniform as well, as it selects the random points one after the other, imposing an artificial order among the elements which is not related to the uniformity of the distribution.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:55
$begingroup$
And actually Bertrand Paradox supports my counter proof, as I have already stated, by indicating that there could be multiple ways of constructing randomness, leading to paradoxical truths.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:58
$begingroup$
The term uniform has a precise meaning, so one certainly can claim that one distribution is uniform and the other is not.
$endgroup$
– Travis
Jan 10 at 8:05
$begingroup$
For any distribution on pairs of points we can define a new distribution simply by reversing the orders of the pairs of points. The ordering of the points isn't an issue for uniforming in the sense that if we apply this operation to the uniform distribution, we get the uniform distribution back.
$endgroup$
– Travis
Jan 10 at 8:09
$begingroup$
In any case, I don't see where you've mentioned the Bertrand Paradox. It's true that there are "multiple ways of constructing randomness" in the sense that there's more than one distribution on the space of pairs of points on the $N$-sphere, but it's a little grandiose to describe this situation as "leading to paradoxical truths". What's going on here is just that (1) different distributions give different results, and (2) the particular result ("...nearly orthogonal") you mention implicitly refers to a particular distribution, nothing more.
$endgroup$
– Travis
Jan 10 at 8:13
|
show 6 more comments
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$begingroup$
This is not a counterproof.
Implicit in the original statement ("two randomly selected points on the surface of the sphere are nearly orthogonal") is that the randomly selected points are (separately) chosen uniformly. The fact that the conclusion doesn't hold for some other distribution you propose just shows that your distribution (on pairs of points) isn't the uniform one.
This is, by the way, the same phenomenon illustrated by the Bertrand Paradox. You might find it instructive to study that paradox, which is (at least dimensionally) simpler than the situation you describe.
answered Jan 10 at 7:22
TravisTravis
60.6k767147
$endgroup$
$begingroup$
Of course, my distribution is not the same as the one used by the original statement, but I don't think one can claim that one of these distributions is uniform and the other isn't. For example, the distribution used by the original statement isn't uniform as well, as it selects the random points one after the other, imposing an artificial order among the elements which is not related to the uniformity of the distribution.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:55
$begingroup$
And actually Bertrand Paradox supports my counter proof, as I have already stated, by indicating that there could be multiple ways of constructing randomness, leading to paradoxical truths.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:58
$begingroup$
The term uniform has a precise meaning, so one certainly can claim that one distribution is uniform and the other is not.
$endgroup$
– Travis
Jan 10 at 8:05
$begingroup$
For any distribution on pairs of points we can define a new distribution simply by reversing the orders of the pairs of points. The ordering of the points isn't an issue for uniforming in the sense that if we apply this operation to the uniform distribution, we get the uniform distribution back.
$endgroup$
– Travis
Jan 10 at 8:09
$begingroup$
In any case, I don't see where you've mentioned the Bertrand Paradox. It's true that there are "multiple ways of constructing randomness" in the sense that there's more than one distribution on the space of pairs of points on the $N$-sphere, but it's a little grandiose to describe this situation as "leading to paradoxical truths". What's going on here is just that (1) different distributions give different results, and (2) the particular result ("...nearly orthogonal") you mention implicitly refers to a particular distribution, nothing more.
$endgroup$
– Travis
Jan 10 at 8:13
|
show 6 more comments
$begingroup$
This is not a counterproof.
Implicit in the original statement ("two randomly selected points on the surface of the sphere are nearly orthogonal") is that the randomly selected points are (separately) chosen uniformly. The fact that the conclusion doesn't hold for some other distribution you propose just shows that your distribution (on pairs of points) isn't the uniform one.
This is, by the way, the same phenomenon illustrated by the Bertrand Paradox. You might find it instructive to study that paradox, which is (at least dimensionally) simpler than the situation you describe.
answered Jan 10 at 7:22
TravisTravis
60.6k767147
$endgroup$
$begingroup$
Of course, my distribution is not the same as the one used by the original statement, but I don't think one can claim that one of these distributions is uniform and the other isn't. For example, the distribution used by the original statement isn't uniform as well, as it selects the random points one after the other, imposing an artificial order among the elements which is not related to the uniformity of the distribution.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:55
$begingroup$
And actually Bertrand Paradox supports my counter proof, as I have already stated, by indicating that there could be multiple ways of constructing randomness, leading to paradoxical truths.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:58
$begingroup$
The term uniform has a precise meaning, so one certainly can claim that one distribution is uniform and the other is not.
$endgroup$
– Travis
Jan 10 at 8:05
$begingroup$
For any distribution on pairs of points we can define a new distribution simply by reversing the orders of the pairs of points. The ordering of the points isn't an issue for uniforming in the sense that if we apply this operation to the uniform distribution, we get the uniform distribution back.
$endgroup$
– Travis
Jan 10 at 8:09
$begingroup$
In any case, I don't see where you've mentioned the Bertrand Paradox. It's true that there are "multiple ways of constructing randomness" in the sense that there's more than one distribution on the space of pairs of points on the $N$-sphere, but it's a little grandiose to describe this situation as "leading to paradoxical truths". What's going on here is just that (1) different distributions give different results, and (2) the particular result ("...nearly orthogonal") you mention implicitly refers to a particular distribution, nothing more.
$endgroup$
– Travis
Jan 10 at 8:13
|
show 6 more comments
$begingroup$
This is not a counterproof.
Implicit in the original statement ("two randomly selected points on the surface of the sphere are nearly orthogonal") is that the randomly selected points are (separately) chosen uniformly. The fact that the conclusion doesn't hold for some other distribution you propose just shows that your distribution (on pairs of points) isn't the uniform one.
This is, by the way, the same phenomenon illustrated by the Bertrand Paradox. You might find it instructive to study that paradox, which is (at least dimensionally) simpler than the situation you describe.
answered Jan 10 at 7:22
TravisTravis
60.6k767147
$endgroup$
This is not a counterproof.
Implicit in the original statement ("two randomly selected points on the surface of the sphere are nearly orthogonal") is that the randomly selected points are (separately) chosen uniformly. The fact that the conclusion doesn't hold for some other distribution you propose just shows that your distribution (on pairs of points) isn't the uniform one.
This is, by the way, the same phenomenon illustrated by the Bertrand Paradox. You might find it instructive to study that paradox, which is (at least dimensionally) simpler than the situation you describe.
answered Jan 10 at 7:22
TravisTravis
60.6k767147
answered Jan 10 at 7:22
TravisTravis
60.6k767147
answered Jan 10 at 7:22
TravisTravis
60.6k767147
answered Jan 10 at 7:22
TravisTravis
60.6k767147
60.6k767147
$begingroup$
Of course, my distribution is not the same as the one used by the original statement, but I don't think one can claim that one of these distributions is uniform and the other isn't. For example, the distribution used by the original statement isn't uniform as well, as it selects the random points one after the other, imposing an artificial order among the elements which is not related to the uniformity of the distribution.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:55
$begingroup$
And actually Bertrand Paradox supports my counter proof, as I have already stated, by indicating that there could be multiple ways of constructing randomness, leading to paradoxical truths.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:58
$begingroup$
The term uniform has a precise meaning, so one certainly can claim that one distribution is uniform and the other is not.
$endgroup$
– Travis
Jan 10 at 8:05
$begingroup$
For any distribution on pairs of points we can define a new distribution simply by reversing the orders of the pairs of points. The ordering of the points isn't an issue for uniforming in the sense that if we apply this operation to the uniform distribution, we get the uniform distribution back.
$endgroup$
– Travis
Jan 10 at 8:09
$begingroup$
In any case, I don't see where you've mentioned the Bertrand Paradox. It's true that there are "multiple ways of constructing randomness" in the sense that there's more than one distribution on the space of pairs of points on the $N$-sphere, but it's a little grandiose to describe this situation as "leading to paradoxical truths". What's going on here is just that (1) different distributions give different results, and (2) the particular result ("...nearly orthogonal") you mention implicitly refers to a particular distribution, nothing more.
$endgroup$
– Travis
Jan 10 at 8:13
|
show 6 more comments
$begingroup$
Of course, my distribution is not the same as the one used by the original statement, but I don't think one can claim that one of these distributions is uniform and the other isn't. For example, the distribution used by the original statement isn't uniform as well, as it selects the random points one after the other, imposing an artificial order among the elements which is not related to the uniformity of the distribution.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:55
$begingroup$
And actually Bertrand Paradox supports my counter proof, as I have already stated, by indicating that there could be multiple ways of constructing randomness, leading to paradoxical truths.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:58
$begingroup$
The term uniform has a precise meaning, so one certainly can claim that one distribution is uniform and the other is not.
$endgroup$
– Travis
Jan 10 at 8:05
$begingroup$
For any distribution on pairs of points we can define a new distribution simply by reversing the orders of the pairs of points. The ordering of the points isn't an issue for uniforming in the sense that if we apply this operation to the uniform distribution, we get the uniform distribution back.
$endgroup$
– Travis
Jan 10 at 8:09
$begingroup$
In any case, I don't see where you've mentioned the Bertrand Paradox. It's true that there are "multiple ways of constructing randomness" in the sense that there's more than one distribution on the space of pairs of points on the $N$-sphere, but it's a little grandiose to describe this situation as "leading to paradoxical truths". What's going on here is just that (1) different distributions give different results, and (2) the particular result ("...nearly orthogonal") you mention implicitly refers to a particular distribution, nothing more.
$endgroup$
– Travis
Jan 10 at 8:13
$begingroup$
Of course, my distribution is not the same as the one used by the original statement, but I don't think one can claim that one of these distributions is uniform and the other isn't. For example, the distribution used by the original statement isn't uniform as well, as it selects the random points one after the other, imposing an artificial order among the elements which is not related to the uniformity of the distribution.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:55
$begingroup$
Of course, my distribution is not the same as the one used by the original statement, but I don't think one can claim that one of these distributions is uniform and the other isn't. For example, the distribution used by the original statement isn't uniform as well, as it selects the random points one after the other, imposing an artificial order among the elements which is not related to the uniformity of the distribution.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:55
$begingroup$
And actually Bertrand Paradox supports my counter proof, as I have already stated, by indicating that there could be multiple ways of constructing randomness, leading to paradoxical truths.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:58
$begingroup$
And actually Bertrand Paradox supports my counter proof, as I have already stated, by indicating that there could be multiple ways of constructing randomness, leading to paradoxical truths.
$endgroup$
– Venkata Pagadala
Jan 10 at 7:58
$begingroup$
The term uniform has a precise meaning, so one certainly can claim that one distribution is uniform and the other is not.
$endgroup$
– Travis
Jan 10 at 8:05
$begingroup$
The term uniform has a precise meaning, so one certainly can claim that one distribution is uniform and the other is not.
$endgroup$
– Travis
Jan 10 at 8:05
$begingroup$
For any distribution on pairs of points we can define a new distribution simply by reversing the orders of the pairs of points. The ordering of the points isn't an issue for uniforming in the sense that if we apply this operation to the uniform distribution, we get the uniform distribution back.
$endgroup$
– Travis
Jan 10 at 8:09
$begingroup$
For any distribution on pairs of points we can define a new distribution simply by reversing the orders of the pairs of points. The ordering of the points isn't an issue for uniforming in the sense that if we apply this operation to the uniform distribution, we get the uniform distribution back.
$endgroup$
– Travis
Jan 10 at 8:09
$begingroup$
In any case, I don't see where you've mentioned the Bertrand Paradox. It's true that there are "multiple ways of constructing randomness" in the sense that there's more than one distribution on the space of pairs of points on the $N$-sphere, but it's a little grandiose to describe this situation as "leading to paradoxical truths". What's going on here is just that (1) different distributions give different results, and (2) the particular result ("...nearly orthogonal") you mention implicitly refers to a particular distribution, nothing more.
$endgroup$
– Travis
Jan 10 at 8:13
$begingroup$
In any case, I don't see where you've mentioned the Bertrand Paradox. It's true that there are "multiple ways of constructing randomness" in the sense that there's more than one distribution on the space of pairs of points on the $N$-sphere, but it's a little grandiose to describe this situation as "leading to paradoxical truths". What's going on here is just that (1) different distributions give different results, and (2) the particular result ("...nearly orthogonal") you mention implicitly refers to a particular distribution, nothing more.
$endgroup$
– Travis
Jan 10 at 8:13
|
show 6 more comments
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$begingroup$
The Jacobian for your process does not agree with what the world calls uniform distribution on the sphere. If $x$ and $y$ are two random points you should be able to work out the expectation of $|x-y|^2$ for your way and for the usual way.
$endgroup$
– kimchi lover
Jan 8 at 23:50
$begingroup$
The mid points of the arc have a standard uniform distribution over the sphere's surface. The lengths and orientations have a standard uniform distribution. Any hints on why the end points would not have a uniform distribution?
$endgroup$
– Venkata Pagadala
Jan 9 at 5:43