Probability of two random points being orthogonal in higher-dimensional unit sphere
I understand that most points will be close to surface due to volume concentration. Also I also understand the concentration of volume near the equator, relative to any specific point (North pole).
However, I can't understand why the North pole concept should influence the angle between the two points.
The argument seems to that, once the first point is selected randomly, then considering the first point as the North pole, the second point is more likely to be found near the equator.
I don't understand this argument because I think the orientation of second point has no bearing with the location of the first point. Given both are drawn randomly, all locations on the sphere are equally probable candidates for both points. The angle should have an equal probability for all values, (from zero to 180) instead of having more probability for orthogonality.
What if both points are selected at the same time (no first and second), or selected on two separate spheres? How does the North pole effect the outcomes?
[edit: add another counter-argument] Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
Why was the North pole concept considered to influence the location for the second point?
Any thoughts?
probability geometry geometric-probability dimensional-analysis topological-data-analysis
asked Jan 2 at 17:47
Pagadala
163
|
show 1 more comment
I understand that most points will be close to surface due to volume concentration. Also I also understand the concentration of volume near the equator, relative to any specific point (North pole).
However, I can't understand why the North pole concept should influence the angle between the two points.
The argument seems to that, once the first point is selected randomly, then considering the first point as the North pole, the second point is more likely to be found near the equator.
I don't understand this argument because I think the orientation of second point has no bearing with the location of the first point. Given both are drawn randomly, all locations on the sphere are equally probable candidates for both points. The angle should have an equal probability for all values, (from zero to 180) instead of having more probability for orthogonality.
What if both points are selected at the same time (no first and second), or selected on two separate spheres? How does the North pole effect the outcomes?
[edit: add another counter-argument] Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
Why was the North pole concept considered to influence the location for the second point?
Any thoughts?
probability geometry geometric-probability dimensional-analysis topological-data-analysis
asked Jan 2 at 17:47
Pagadala
163
Pick the first point; rotate the hypersphere so this point is on the north pole, see that other points are more likely to be close to orthogonal to this North Pole than at a very narrow angle or a very wide angle; rotate back to the original orientation and the second point remains more likely to be close to orthogonal. Or just run a simulation
– Henry
Jan 2 at 17:53
I think about it this way: . Let X and Y be two Gaussian random vectors (normalized to lie on the sphere - note this yields to points uniformly distributed on the sphere). The expected value of $Xcdot Y$ is 0, since each component in the sum is the product of two independent Gaussians with mean 0. Intuitively, if the first point is fixed, then each coordinate of the second point is equally likely to agree/disagree with the respective coordinate of the second point. Hence, Our best guess for the inner product is 0.
– mm8511
Jan 2 at 17:53
1
The north pole is a reference point. It is a trick to ease calculation. It works because rotations do not change the angles between pairs of points. You can do the same thing in the plane. Imagine you want to calculate the distance between two points. Distance is preserved under translation so you can move the plane so that one of the points is at the origin.
– John Douma
Jan 2 at 18:00
How about this? Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
– Pagadala
Jan 2 at 18:01
In fact, you can learn the dot product's distribution here.
– J.G.
Jan 2 at 18:04
|
show 1 more comment
I understand that most points will be close to surface due to volume concentration. Also I also understand the concentration of volume near the equator, relative to any specific point (North pole).
However, I can't understand why the North pole concept should influence the angle between the two points.
The argument seems to that, once the first point is selected randomly, then considering the first point as the North pole, the second point is more likely to be found near the equator.
I don't understand this argument because I think the orientation of second point has no bearing with the location of the first point. Given both are drawn randomly, all locations on the sphere are equally probable candidates for both points. The angle should have an equal probability for all values, (from zero to 180) instead of having more probability for orthogonality.
What if both points are selected at the same time (no first and second), or selected on two separate spheres? How does the North pole effect the outcomes?
[edit: add another counter-argument] Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
Why was the North pole concept considered to influence the location for the second point?
Any thoughts?
probability geometry geometric-probability dimensional-analysis topological-data-analysis
asked Jan 2 at 17:47
Pagadala
163
I understand that most points will be close to surface due to volume concentration. Also I also understand the concentration of volume near the equator, relative to any specific point (North pole).
However, I can't understand why the North pole concept should influence the angle between the two points.
The argument seems to that, once the first point is selected randomly, then considering the first point as the North pole, the second point is more likely to be found near the equator.
I don't understand this argument because I think the orientation of second point has no bearing with the location of the first point. Given both are drawn randomly, all locations on the sphere are equally probable candidates for both points. The angle should have an equal probability for all values, (from zero to 180) instead of having more probability for orthogonality.
What if both points are selected at the same time (no first and second), or selected on two separate spheres? How does the North pole effect the outcomes?
[edit: add another counter-argument] Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
Why was the North pole concept considered to influence the location for the second point?
Any thoughts?
probability geometry geometric-probability dimensional-analysis topological-data-analysis
probability geometry geometric-probability dimensional-analysis topological-data-analysis
asked Jan 2 at 17:47
Pagadala
163
asked Jan 2 at 17:47
Pagadala
163
edited Jan 2 at 18:06
asked Jan 2 at 17:47
Pagadala
163
asked Jan 2 at 17:47
Pagadala
163
asked Jan 2 at 17:47
Pagadala
163
163
Pick the first point; rotate the hypersphere so this point is on the north pole, see that other points are more likely to be close to orthogonal to this North Pole than at a very narrow angle or a very wide angle; rotate back to the original orientation and the second point remains more likely to be close to orthogonal. Or just run a simulation
– Henry
Jan 2 at 17:53
I think about it this way: . Let X and Y be two Gaussian random vectors (normalized to lie on the sphere - note this yields to points uniformly distributed on the sphere). The expected value of $Xcdot Y$ is 0, since each component in the sum is the product of two independent Gaussians with mean 0. Intuitively, if the first point is fixed, then each coordinate of the second point is equally likely to agree/disagree with the respective coordinate of the second point. Hence, Our best guess for the inner product is 0.
– mm8511
Jan 2 at 17:53
1
The north pole is a reference point. It is a trick to ease calculation. It works because rotations do not change the angles between pairs of points. You can do the same thing in the plane. Imagine you want to calculate the distance between two points. Distance is preserved under translation so you can move the plane so that one of the points is at the origin.
– John Douma
Jan 2 at 18:00
How about this? Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
– Pagadala
Jan 2 at 18:01
In fact, you can learn the dot product's distribution here.
– J.G.
Jan 2 at 18:04
|
show 1 more comment
Pick the first point; rotate the hypersphere so this point is on the north pole, see that other points are more likely to be close to orthogonal to this North Pole than at a very narrow angle or a very wide angle; rotate back to the original orientation and the second point remains more likely to be close to orthogonal. Or just run a simulation
– Henry
Jan 2 at 17:53
I think about it this way: . Let X and Y be two Gaussian random vectors (normalized to lie on the sphere - note this yields to points uniformly distributed on the sphere). The expected value of $Xcdot Y$ is 0, since each component in the sum is the product of two independent Gaussians with mean 0. Intuitively, if the first point is fixed, then each coordinate of the second point is equally likely to agree/disagree with the respective coordinate of the second point. Hence, Our best guess for the inner product is 0.
– mm8511
Jan 2 at 17:53
1
The north pole is a reference point. It is a trick to ease calculation. It works because rotations do not change the angles between pairs of points. You can do the same thing in the plane. Imagine you want to calculate the distance between two points. Distance is preserved under translation so you can move the plane so that one of the points is at the origin.
– John Douma
Jan 2 at 18:00
How about this? Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
– Pagadala
Jan 2 at 18:01
In fact, you can learn the dot product's distribution here.
– J.G.
Jan 2 at 18:04
Pick the first point; rotate the hypersphere so this point is on the north pole, see that other points are more likely to be close to orthogonal to this North Pole than at a very narrow angle or a very wide angle; rotate back to the original orientation and the second point remains more likely to be close to orthogonal. Or just run a simulation
– Henry
Jan 2 at 17:53
Pick the first point; rotate the hypersphere so this point is on the north pole, see that other points are more likely to be close to orthogonal to this North Pole than at a very narrow angle or a very wide angle; rotate back to the original orientation and the second point remains more likely to be close to orthogonal. Or just run a simulation
– Henry
Jan 2 at 17:53
I think about it this way: . Let X and Y be two Gaussian random vectors (normalized to lie on the sphere - note this yields to points uniformly distributed on the sphere). The expected value of $Xcdot Y$ is 0, since each component in the sum is the product of two independent Gaussians with mean 0. Intuitively, if the first point is fixed, then each coordinate of the second point is equally likely to agree/disagree with the respective coordinate of the second point. Hence, Our best guess for the inner product is 0.
– mm8511
Jan 2 at 17:53
I think about it this way: . Let X and Y be two Gaussian random vectors (normalized to lie on the sphere - note this yields to points uniformly distributed on the sphere). The expected value of $Xcdot Y$ is 0, since each component in the sum is the product of two independent Gaussians with mean 0. Intuitively, if the first point is fixed, then each coordinate of the second point is equally likely to agree/disagree with the respective coordinate of the second point. Hence, Our best guess for the inner product is 0.
– mm8511
Jan 2 at 17:53
1
1
The north pole is a reference point. It is a trick to ease calculation. It works because rotations do not change the angles between pairs of points. You can do the same thing in the plane. Imagine you want to calculate the distance between two points. Distance is preserved under translation so you can move the plane so that one of the points is at the origin.
– John Douma
Jan 2 at 18:00
The north pole is a reference point. It is a trick to ease calculation. It works because rotations do not change the angles between pairs of points. You can do the same thing in the plane. Imagine you want to calculate the distance between two points. Distance is preserved under translation so you can move the plane so that one of the points is at the origin.
– John Douma
Jan 2 at 18:00
How about this? Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
– Pagadala
Jan 2 at 18:01
How about this? Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
– Pagadala
Jan 2 at 18:01
In fact, you can learn the dot product's distribution here.
– J.G.
Jan 2 at 18:04
In fact, you can learn the dot product's distribution here.
– J.G.
Jan 2 at 18:04
|
show 1 more comment
1 Answer
1
active
oldest
votes
I'm assuming a uniform continuous distribution, assuming the probability distributions of the two points are independent, and assuming you mean "close to orthogonal" rather than strictly orthogonal.
It is true that "all locations on the sphere are equally probable candidates for both points." Assuming independent distributions,
all locations are equally probable for the second point even given a known position for the first point.
The point is that the part of the sphere close to the "equator" that you randomly selected when you placed the first point
is much larger than the part that is close to that point or to the antipodal point.
If you go to a high enough dimension, most of the sphere will be "close" to the "equator," for whatever your definition of "close" is.
The remarks above, of course, amount merely to a restatement of the claim in slightly different terms.
Better would be an actual proof.
Suppose we have a hypersphere of $n$ dimensions.
To simplify the calculations, let the radius of the sphere be $1.$
The total volume inside this sphere is therefore
$$
V_n(1) = frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}.
$$
We assume a uniform distribution over the volume inside the sphere. That is, when we pick a random point according to this distribution, the probability that the point lands inside a particular region of hyperspace inside the sphere is proportional to the volume of that region.
Now arbitrarily pick a number $q$ with $0 < q leq 1$ and consider the probability that the distance $r$ between the center of the hypersphere and the random point will be less than $q.$
This probability is the volume inside the hypersphere of radius $q$ divided by the total volume inside the hypersphere, namely,
$$
P(r leq q) = frac{V_n(q)}{V_n(1)}
= frac{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)} q^n}
{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}}
= q^n.
$$
Now suppose $n$ is very large, for example, $n = 10000,$ and let's see what is the probability that the random point lands within a distance $q = 0.999$ from the center of the hypersphere. This is
$$
P(r leq 0.999) = 0.999^{10000} approx 0.000045173.
$$
So the random point will land in this region less than $0.005%$ of the time.
The other $99.995%$ of the time, the point lands in the thin hyperspherical shell within just $0.001$ units from the surface.
This doesn't happen because the point was somehow attracted to the surface of the hypersphere or because we skewed the distribution to something non-uniform to make it close to the surface. This is a distribution that is strictly uniform by volume, with no other influences. But it almost always lands very close to the surface simply because that's where almost all the volume is.
High dimensions are weird!
Now let's randomly choose two points and measure the angle between them.
In order to measure the angle, after the points have already been chosen
(so that we can no longer unduly influence the random placement of the points), we choose one of the points as the pole of a set of spherical coordinates.
We do this only so that we can use the "latitude" in that coordinate system to measure the angle between the two points.
The chance that the angle will be between $phi_1$ and $phi_2$
(where these are both angles between $0$ and $180$ degrees)
is then proportional to how much of the sphere lies between the lines of latitude at angles $phi_1$ and $phi_2$ from the pole.
A general formula for this probability in high dimensions is rather ugly, I think. It is related to the question
what is the surface area of a cap on a hypersphere?
The probability density, however, is dealt with in the answer to
Distribution of an angle between a random and fixed unit-length $n$-vectors.
If the angle $phi$ is measure in radians, the density over the range
$0 leq phileq pi$ is
$$
f(phi)=frac{sin^{n-2}phi}{int_0^pisin^{n-2}theta,mathrm dtheta}.
$$
Now let's take our $10000$-dimensional hypersphere again and let's see what the probability is that the angle is within $1/50$ radian (a little more than one degree) from a right angle. This is
$$
P(fracpi2 - frac1{50} leq phi leq fracpi2 + frac1{50})
= frac{int_{pi/2 - 1/{50}}^{pi/2 + 1/{50}}sin^{9998}phi,mathrm dphi}{int_0^pisin^{9998}theta,mathrm dtheta}
approx 0.95449.
$$
So there is a better than $95%$ chance that the angle will be between
$88.85$ and $91.15$ degrees.
answered Jan 2 at 18:17
David K
52.7k340115
So, all locations are equally probable, but there are more locations near the equator, increasing the probability of finding the second point there. Does it also mean distance between any 2 random points is roughly equal in the higher-dimensional sphere?
– Pagadala
Jan 2 at 18:21
That's the idea.
– David K
Jan 2 at 18:22
I have accepted the answer, because it does best effort to explain the argument. But I still think the argument makes no sense. Selection of two random points involves 2(N-1) pieces of coordinate information in the N-1 dimensional surface of the sphere. Instead, first select a 2D angle randomly as the span between the points, and then select N more random angles to position the span and orient it. I don't think the points will have any affinity to equator or perpendicularity, in this case.
– Pagadala
2 days ago
I gave a thorough refutation of the argument at prettypebble.com/notes/250.html
– Pagadala
2 days ago
"The idea is that, the points inside the volume can be sequentially labeled ... ." No. No, no, no, no, no. It is easy to refute such a "proof," because such a "proof" is not mathematical at all. One of the fundamental facts about any continuous distribution is that it is not possible to sequentially label the atomic outcomes. And even if you could, there is no uniform distribution over all the integers. I was going to say shred the book where you read this "proof," but on second thought I think it is more likely that the proof is completely different and you misread it.
– David K
2 days ago
|
show 6 more comments
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I'm assuming a uniform continuous distribution, assuming the probability distributions of the two points are independent, and assuming you mean "close to orthogonal" rather than strictly orthogonal.
It is true that "all locations on the sphere are equally probable candidates for both points." Assuming independent distributions,
all locations are equally probable for the second point even given a known position for the first point.
The point is that the part of the sphere close to the "equator" that you randomly selected when you placed the first point
is much larger than the part that is close to that point or to the antipodal point.
If you go to a high enough dimension, most of the sphere will be "close" to the "equator," for whatever your definition of "close" is.
The remarks above, of course, amount merely to a restatement of the claim in slightly different terms.
Better would be an actual proof.
Suppose we have a hypersphere of $n$ dimensions.
To simplify the calculations, let the radius of the sphere be $1.$
The total volume inside this sphere is therefore
$$
V_n(1) = frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}.
$$
We assume a uniform distribution over the volume inside the sphere. That is, when we pick a random point according to this distribution, the probability that the point lands inside a particular region of hyperspace inside the sphere is proportional to the volume of that region.
Now arbitrarily pick a number $q$ with $0 < q leq 1$ and consider the probability that the distance $r$ between the center of the hypersphere and the random point will be less than $q.$
This probability is the volume inside the hypersphere of radius $q$ divided by the total volume inside the hypersphere, namely,
$$
P(r leq q) = frac{V_n(q)}{V_n(1)}
= frac{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)} q^n}
{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}}
= q^n.
$$
Now suppose $n$ is very large, for example, $n = 10000,$ and let's see what is the probability that the random point lands within a distance $q = 0.999$ from the center of the hypersphere. This is
$$
P(r leq 0.999) = 0.999^{10000} approx 0.000045173.
$$
So the random point will land in this region less than $0.005%$ of the time.
The other $99.995%$ of the time, the point lands in the thin hyperspherical shell within just $0.001$ units from the surface.
This doesn't happen because the point was somehow attracted to the surface of the hypersphere or because we skewed the distribution to something non-uniform to make it close to the surface. This is a distribution that is strictly uniform by volume, with no other influences. But it almost always lands very close to the surface simply because that's where almost all the volume is.
High dimensions are weird!
Now let's randomly choose two points and measure the angle between them.
In order to measure the angle, after the points have already been chosen
(so that we can no longer unduly influence the random placement of the points), we choose one of the points as the pole of a set of spherical coordinates.
We do this only so that we can use the "latitude" in that coordinate system to measure the angle between the two points.
The chance that the angle will be between $phi_1$ and $phi_2$
(where these are both angles between $0$ and $180$ degrees)
is then proportional to how much of the sphere lies between the lines of latitude at angles $phi_1$ and $phi_2$ from the pole.
A general formula for this probability in high dimensions is rather ugly, I think. It is related to the question
what is the surface area of a cap on a hypersphere?
The probability density, however, is dealt with in the answer to
Distribution of an angle between a random and fixed unit-length $n$-vectors.
If the angle $phi$ is measure in radians, the density over the range
$0 leq phileq pi$ is
$$
f(phi)=frac{sin^{n-2}phi}{int_0^pisin^{n-2}theta,mathrm dtheta}.
$$
Now let's take our $10000$-dimensional hypersphere again and let's see what the probability is that the angle is within $1/50$ radian (a little more than one degree) from a right angle. This is
$$
P(fracpi2 - frac1{50} leq phi leq fracpi2 + frac1{50})
= frac{int_{pi/2 - 1/{50}}^{pi/2 + 1/{50}}sin^{9998}phi,mathrm dphi}{int_0^pisin^{9998}theta,mathrm dtheta}
approx 0.95449.
$$
So there is a better than $95%$ chance that the angle will be between
$88.85$ and $91.15$ degrees.
answered Jan 2 at 18:17
David K
52.7k340115
So, all locations are equally probable, but there are more locations near the equator, increasing the probability of finding the second point there. Does it also mean distance between any 2 random points is roughly equal in the higher-dimensional sphere?
– Pagadala
Jan 2 at 18:21
That's the idea.
– David K
Jan 2 at 18:22
I have accepted the answer, because it does best effort to explain the argument. But I still think the argument makes no sense. Selection of two random points involves 2(N-1) pieces of coordinate information in the N-1 dimensional surface of the sphere. Instead, first select a 2D angle randomly as the span between the points, and then select N more random angles to position the span and orient it. I don't think the points will have any affinity to equator or perpendicularity, in this case.
– Pagadala
2 days ago
I gave a thorough refutation of the argument at prettypebble.com/notes/250.html
– Pagadala
2 days ago
"The idea is that, the points inside the volume can be sequentially labeled ... ." No. No, no, no, no, no. It is easy to refute such a "proof," because such a "proof" is not mathematical at all. One of the fundamental facts about any continuous distribution is that it is not possible to sequentially label the atomic outcomes. And even if you could, there is no uniform distribution over all the integers. I was going to say shred the book where you read this "proof," but on second thought I think it is more likely that the proof is completely different and you misread it.
– David K
2 days ago
|
show 6 more comments
I'm assuming a uniform continuous distribution, assuming the probability distributions of the two points are independent, and assuming you mean "close to orthogonal" rather than strictly orthogonal.
It is true that "all locations on the sphere are equally probable candidates for both points." Assuming independent distributions,
all locations are equally probable for the second point even given a known position for the first point.
The point is that the part of the sphere close to the "equator" that you randomly selected when you placed the first point
is much larger than the part that is close to that point or to the antipodal point.
If you go to a high enough dimension, most of the sphere will be "close" to the "equator," for whatever your definition of "close" is.
The remarks above, of course, amount merely to a restatement of the claim in slightly different terms.
Better would be an actual proof.
Suppose we have a hypersphere of $n$ dimensions.
To simplify the calculations, let the radius of the sphere be $1.$
The total volume inside this sphere is therefore
$$
V_n(1) = frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}.
$$
We assume a uniform distribution over the volume inside the sphere. That is, when we pick a random point according to this distribution, the probability that the point lands inside a particular region of hyperspace inside the sphere is proportional to the volume of that region.
Now arbitrarily pick a number $q$ with $0 < q leq 1$ and consider the probability that the distance $r$ between the center of the hypersphere and the random point will be less than $q.$
This probability is the volume inside the hypersphere of radius $q$ divided by the total volume inside the hypersphere, namely,
$$
P(r leq q) = frac{V_n(q)}{V_n(1)}
= frac{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)} q^n}
{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}}
= q^n.
$$
Now suppose $n$ is very large, for example, $n = 10000,$ and let's see what is the probability that the random point lands within a distance $q = 0.999$ from the center of the hypersphere. This is
$$
P(r leq 0.999) = 0.999^{10000} approx 0.000045173.
$$
So the random point will land in this region less than $0.005%$ of the time.
The other $99.995%$ of the time, the point lands in the thin hyperspherical shell within just $0.001$ units from the surface.
This doesn't happen because the point was somehow attracted to the surface of the hypersphere or because we skewed the distribution to something non-uniform to make it close to the surface. This is a distribution that is strictly uniform by volume, with no other influences. But it almost always lands very close to the surface simply because that's where almost all the volume is.
High dimensions are weird!
Now let's randomly choose two points and measure the angle between them.
In order to measure the angle, after the points have already been chosen
(so that we can no longer unduly influence the random placement of the points), we choose one of the points as the pole of a set of spherical coordinates.
We do this only so that we can use the "latitude" in that coordinate system to measure the angle between the two points.
The chance that the angle will be between $phi_1$ and $phi_2$
(where these are both angles between $0$ and $180$ degrees)
is then proportional to how much of the sphere lies between the lines of latitude at angles $phi_1$ and $phi_2$ from the pole.
A general formula for this probability in high dimensions is rather ugly, I think. It is related to the question
what is the surface area of a cap on a hypersphere?
The probability density, however, is dealt with in the answer to
Distribution of an angle between a random and fixed unit-length $n$-vectors.
If the angle $phi$ is measure in radians, the density over the range
$0 leq phileq pi$ is
$$
f(phi)=frac{sin^{n-2}phi}{int_0^pisin^{n-2}theta,mathrm dtheta}.
$$
Now let's take our $10000$-dimensional hypersphere again and let's see what the probability is that the angle is within $1/50$ radian (a little more than one degree) from a right angle. This is
$$
P(fracpi2 - frac1{50} leq phi leq fracpi2 + frac1{50})
= frac{int_{pi/2 - 1/{50}}^{pi/2 + 1/{50}}sin^{9998}phi,mathrm dphi}{int_0^pisin^{9998}theta,mathrm dtheta}
approx 0.95449.
$$
So there is a better than $95%$ chance that the angle will be between
$88.85$ and $91.15$ degrees.
answered Jan 2 at 18:17
David K
52.7k340115
So, all locations are equally probable, but there are more locations near the equator, increasing the probability of finding the second point there. Does it also mean distance between any 2 random points is roughly equal in the higher-dimensional sphere?
– Pagadala
Jan 2 at 18:21
That's the idea.
– David K
Jan 2 at 18:22
I have accepted the answer, because it does best effort to explain the argument. But I still think the argument makes no sense. Selection of two random points involves 2(N-1) pieces of coordinate information in the N-1 dimensional surface of the sphere. Instead, first select a 2D angle randomly as the span between the points, and then select N more random angles to position the span and orient it. I don't think the points will have any affinity to equator or perpendicularity, in this case.
– Pagadala
2 days ago
I gave a thorough refutation of the argument at prettypebble.com/notes/250.html
– Pagadala
2 days ago
"The idea is that, the points inside the volume can be sequentially labeled ... ." No. No, no, no, no, no. It is easy to refute such a "proof," because such a "proof" is not mathematical at all. One of the fundamental facts about any continuous distribution is that it is not possible to sequentially label the atomic outcomes. And even if you could, there is no uniform distribution over all the integers. I was going to say shred the book where you read this "proof," but on second thought I think it is more likely that the proof is completely different and you misread it.
– David K
2 days ago
|
show 6 more comments
I'm assuming a uniform continuous distribution, assuming the probability distributions of the two points are independent, and assuming you mean "close to orthogonal" rather than strictly orthogonal.
It is true that "all locations on the sphere are equally probable candidates for both points." Assuming independent distributions,
all locations are equally probable for the second point even given a known position for the first point.
The point is that the part of the sphere close to the "equator" that you randomly selected when you placed the first point
is much larger than the part that is close to that point or to the antipodal point.
If you go to a high enough dimension, most of the sphere will be "close" to the "equator," for whatever your definition of "close" is.
The remarks above, of course, amount merely to a restatement of the claim in slightly different terms.
Better would be an actual proof.
Suppose we have a hypersphere of $n$ dimensions.
To simplify the calculations, let the radius of the sphere be $1.$
The total volume inside this sphere is therefore
$$
V_n(1) = frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}.
$$
We assume a uniform distribution over the volume inside the sphere. That is, when we pick a random point according to this distribution, the probability that the point lands inside a particular region of hyperspace inside the sphere is proportional to the volume of that region.
Now arbitrarily pick a number $q$ with $0 < q leq 1$ and consider the probability that the distance $r$ between the center of the hypersphere and the random point will be less than $q.$
This probability is the volume inside the hypersphere of radius $q$ divided by the total volume inside the hypersphere, namely,
$$
P(r leq q) = frac{V_n(q)}{V_n(1)}
= frac{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)} q^n}
{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}}
= q^n.
$$
Now suppose $n$ is very large, for example, $n = 10000,$ and let's see what is the probability that the random point lands within a distance $q = 0.999$ from the center of the hypersphere. This is
$$
P(r leq 0.999) = 0.999^{10000} approx 0.000045173.
$$
So the random point will land in this region less than $0.005%$ of the time.
The other $99.995%$ of the time, the point lands in the thin hyperspherical shell within just $0.001$ units from the surface.
This doesn't happen because the point was somehow attracted to the surface of the hypersphere or because we skewed the distribution to something non-uniform to make it close to the surface. This is a distribution that is strictly uniform by volume, with no other influences. But it almost always lands very close to the surface simply because that's where almost all the volume is.
High dimensions are weird!
Now let's randomly choose two points and measure the angle between them.
In order to measure the angle, after the points have already been chosen
(so that we can no longer unduly influence the random placement of the points), we choose one of the points as the pole of a set of spherical coordinates.
We do this only so that we can use the "latitude" in that coordinate system to measure the angle between the two points.
The chance that the angle will be between $phi_1$ and $phi_2$
(where these are both angles between $0$ and $180$ degrees)
is then proportional to how much of the sphere lies between the lines of latitude at angles $phi_1$ and $phi_2$ from the pole.
A general formula for this probability in high dimensions is rather ugly, I think. It is related to the question
what is the surface area of a cap on a hypersphere?
The probability density, however, is dealt with in the answer to
Distribution of an angle between a random and fixed unit-length $n$-vectors.
If the angle $phi$ is measure in radians, the density over the range
$0 leq phileq pi$ is
$$
f(phi)=frac{sin^{n-2}phi}{int_0^pisin^{n-2}theta,mathrm dtheta}.
$$
Now let's take our $10000$-dimensional hypersphere again and let's see what the probability is that the angle is within $1/50$ radian (a little more than one degree) from a right angle. This is
$$
P(fracpi2 - frac1{50} leq phi leq fracpi2 + frac1{50})
= frac{int_{pi/2 - 1/{50}}^{pi/2 + 1/{50}}sin^{9998}phi,mathrm dphi}{int_0^pisin^{9998}theta,mathrm dtheta}
approx 0.95449.
$$
So there is a better than $95%$ chance that the angle will be between
$88.85$ and $91.15$ degrees.
answered Jan 2 at 18:17
David K
52.7k340115
I'm assuming a uniform continuous distribution, assuming the probability distributions of the two points are independent, and assuming you mean "close to orthogonal" rather than strictly orthogonal.
It is true that "all locations on the sphere are equally probable candidates for both points." Assuming independent distributions,
all locations are equally probable for the second point even given a known position for the first point.
The point is that the part of the sphere close to the "equator" that you randomly selected when you placed the first point
is much larger than the part that is close to that point or to the antipodal point.
If you go to a high enough dimension, most of the sphere will be "close" to the "equator," for whatever your definition of "close" is.
The remarks above, of course, amount merely to a restatement of the claim in slightly different terms.
Better would be an actual proof.
Suppose we have a hypersphere of $n$ dimensions.
To simplify the calculations, let the radius of the sphere be $1.$
The total volume inside this sphere is therefore
$$
V_n(1) = frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}.
$$
We assume a uniform distribution over the volume inside the sphere. That is, when we pick a random point according to this distribution, the probability that the point lands inside a particular region of hyperspace inside the sphere is proportional to the volume of that region.
Now arbitrarily pick a number $q$ with $0 < q leq 1$ and consider the probability that the distance $r$ between the center of the hypersphere and the random point will be less than $q.$
This probability is the volume inside the hypersphere of radius $q$ divided by the total volume inside the hypersphere, namely,
$$
P(r leq q) = frac{V_n(q)}{V_n(1)}
= frac{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)} q^n}
{frac{pi^{n/2}}{Gammaleft(frac n2 + 1right)}}
= q^n.
$$
Now suppose $n$ is very large, for example, $n = 10000,$ and let's see what is the probability that the random point lands within a distance $q = 0.999$ from the center of the hypersphere. This is
$$
P(r leq 0.999) = 0.999^{10000} approx 0.000045173.
$$
So the random point will land in this region less than $0.005%$ of the time.
The other $99.995%$ of the time, the point lands in the thin hyperspherical shell within just $0.001$ units from the surface.
This doesn't happen because the point was somehow attracted to the surface of the hypersphere or because we skewed the distribution to something non-uniform to make it close to the surface. This is a distribution that is strictly uniform by volume, with no other influences. But it almost always lands very close to the surface simply because that's where almost all the volume is.
High dimensions are weird!
Now let's randomly choose two points and measure the angle between them.
In order to measure the angle, after the points have already been chosen
(so that we can no longer unduly influence the random placement of the points), we choose one of the points as the pole of a set of spherical coordinates.
We do this only so that we can use the "latitude" in that coordinate system to measure the angle between the two points.
The chance that the angle will be between $phi_1$ and $phi_2$
(where these are both angles between $0$ and $180$ degrees)
is then proportional to how much of the sphere lies between the lines of latitude at angles $phi_1$ and $phi_2$ from the pole.
A general formula for this probability in high dimensions is rather ugly, I think. It is related to the question
what is the surface area of a cap on a hypersphere?
The probability density, however, is dealt with in the answer to
Distribution of an angle between a random and fixed unit-length $n$-vectors.
If the angle $phi$ is measure in radians, the density over the range
$0 leq phileq pi$ is
$$
f(phi)=frac{sin^{n-2}phi}{int_0^pisin^{n-2}theta,mathrm dtheta}.
$$
Now let's take our $10000$-dimensional hypersphere again and let's see what the probability is that the angle is within $1/50$ radian (a little more than one degree) from a right angle. This is
$$
P(fracpi2 - frac1{50} leq phi leq fracpi2 + frac1{50})
= frac{int_{pi/2 - 1/{50}}^{pi/2 + 1/{50}}sin^{9998}phi,mathrm dphi}{int_0^pisin^{9998}theta,mathrm dtheta}
approx 0.95449.
$$
So there is a better than $95%$ chance that the angle will be between
$88.85$ and $91.15$ degrees.
answered Jan 2 at 18:17
David K
52.7k340115
edited 2 days ago
answered Jan 2 at 18:17
David K
52.7k340115
answered Jan 2 at 18:17
David K
52.7k340115
answered Jan 2 at 18:17
David K
52.7k340115
52.7k340115
So, all locations are equally probable, but there are more locations near the equator, increasing the probability of finding the second point there. Does it also mean distance between any 2 random points is roughly equal in the higher-dimensional sphere?
– Pagadala
Jan 2 at 18:21
That's the idea.
– David K
Jan 2 at 18:22
I have accepted the answer, because it does best effort to explain the argument. But I still think the argument makes no sense. Selection of two random points involves 2(N-1) pieces of coordinate information in the N-1 dimensional surface of the sphere. Instead, first select a 2D angle randomly as the span between the points, and then select N more random angles to position the span and orient it. I don't think the points will have any affinity to equator or perpendicularity, in this case.
– Pagadala
2 days ago
I gave a thorough refutation of the argument at prettypebble.com/notes/250.html
– Pagadala
2 days ago
"The idea is that, the points inside the volume can be sequentially labeled ... ." No. No, no, no, no, no. It is easy to refute such a "proof," because such a "proof" is not mathematical at all. One of the fundamental facts about any continuous distribution is that it is not possible to sequentially label the atomic outcomes. And even if you could, there is no uniform distribution over all the integers. I was going to say shred the book where you read this "proof," but on second thought I think it is more likely that the proof is completely different and you misread it.
– David K
2 days ago
|
show 6 more comments
So, all locations are equally probable, but there are more locations near the equator, increasing the probability of finding the second point there. Does it also mean distance between any 2 random points is roughly equal in the higher-dimensional sphere?
– Pagadala
Jan 2 at 18:21
That's the idea.
– David K
Jan 2 at 18:22
I have accepted the answer, because it does best effort to explain the argument. But I still think the argument makes no sense. Selection of two random points involves 2(N-1) pieces of coordinate information in the N-1 dimensional surface of the sphere. Instead, first select a 2D angle randomly as the span between the points, and then select N more random angles to position the span and orient it. I don't think the points will have any affinity to equator or perpendicularity, in this case.
– Pagadala
2 days ago
I gave a thorough refutation of the argument at prettypebble.com/notes/250.html
– Pagadala
2 days ago
"The idea is that, the points inside the volume can be sequentially labeled ... ." No. No, no, no, no, no. It is easy to refute such a "proof," because such a "proof" is not mathematical at all. One of the fundamental facts about any continuous distribution is that it is not possible to sequentially label the atomic outcomes. And even if you could, there is no uniform distribution over all the integers. I was going to say shred the book where you read this "proof," but on second thought I think it is more likely that the proof is completely different and you misread it.
– David K
2 days ago
So, all locations are equally probable, but there are more locations near the equator, increasing the probability of finding the second point there. Does it also mean distance between any 2 random points is roughly equal in the higher-dimensional sphere?
– Pagadala
Jan 2 at 18:21
So, all locations are equally probable, but there are more locations near the equator, increasing the probability of finding the second point there. Does it also mean distance between any 2 random points is roughly equal in the higher-dimensional sphere?
– Pagadala
Jan 2 at 18:21
That's the idea.
– David K
Jan 2 at 18:22
That's the idea.
– David K
Jan 2 at 18:22
I have accepted the answer, because it does best effort to explain the argument. But I still think the argument makes no sense. Selection of two random points involves 2(N-1) pieces of coordinate information in the N-1 dimensional surface of the sphere. Instead, first select a 2D angle randomly as the span between the points, and then select N more random angles to position the span and orient it. I don't think the points will have any affinity to equator or perpendicularity, in this case.
– Pagadala
2 days ago
I have accepted the answer, because it does best effort to explain the argument. But I still think the argument makes no sense. Selection of two random points involves 2(N-1) pieces of coordinate information in the N-1 dimensional surface of the sphere. Instead, first select a 2D angle randomly as the span between the points, and then select N more random angles to position the span and orient it. I don't think the points will have any affinity to equator or perpendicularity, in this case.
– Pagadala
2 days ago
I gave a thorough refutation of the argument at prettypebble.com/notes/250.html
– Pagadala
2 days ago
I gave a thorough refutation of the argument at prettypebble.com/notes/250.html
– Pagadala
2 days ago
"The idea is that, the points inside the volume can be sequentially labeled ... ." No. No, no, no, no, no. It is easy to refute such a "proof," because such a "proof" is not mathematical at all. One of the fundamental facts about any continuous distribution is that it is not possible to sequentially label the atomic outcomes. And even if you could, there is no uniform distribution over all the integers. I was going to say shred the book where you read this "proof," but on second thought I think it is more likely that the proof is completely different and you misread it.
– David K
2 days ago
"The idea is that, the points inside the volume can be sequentially labeled ... ." No. No, no, no, no, no. It is easy to refute such a "proof," because such a "proof" is not mathematical at all. One of the fundamental facts about any continuous distribution is that it is not possible to sequentially label the atomic outcomes. And even if you could, there is no uniform distribution over all the integers. I was going to say shred the book where you read this "proof," but on second thought I think it is more likely that the proof is completely different and you misread it.
– David K
2 days ago
|
show 6 more comments
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Pick the first point; rotate the hypersphere so this point is on the north pole, see that other points are more likely to be close to orthogonal to this North Pole than at a very narrow angle or a very wide angle; rotate back to the original orientation and the second point remains more likely to be close to orthogonal. Or just run a simulation
– Henry
Jan 2 at 17:53
I think about it this way: . Let X and Y be two Gaussian random vectors (normalized to lie on the sphere - note this yields to points uniformly distributed on the sphere). The expected value of $Xcdot Y$ is 0, since each component in the sum is the product of two independent Gaussians with mean 0. Intuitively, if the first point is fixed, then each coordinate of the second point is equally likely to agree/disagree with the respective coordinate of the second point. Hence, Our best guess for the inner product is 0.
– mm8511
Jan 2 at 17:53
1
The north pole is a reference point. It is a trick to ease calculation. It works because rotations do not change the angles between pairs of points. You can do the same thing in the plane. Imagine you want to calculate the distance between two points. Distance is preserved under translation so you can move the plane so that one of the points is at the origin.
– John Douma
Jan 2 at 18:00
How about this? Instead of picking two points, pick the distance between them randomly and the orient that distance randomly. The end points of that distance will be the random two points. Will they still be orthogonal?
– Pagadala
Jan 2 at 18:01
In fact, you can learn the dot product's distribution here.
– J.G.
Jan 2 at 18:04