Good afternoon,
I have a question about the spherical cap example to explain how we can get a very low standard risk with an adversarial risk close to 1.
I'm struggling with the fact that the all mass (is it the probability mass ?) tends to sit near the equator. Is it due to our assumption: p being very small, and E covering nearly half of the sphere, with the probability of picking a point at random in E being therefore very small ?
I can't see why this should be symmetric, and how the distribution of x on the sphere is still uniform. I cannot visualize it.
Thank you for your help !
Indeed it is a very counterintuitive fact and it is not easy (even impossible) to visualize.
So what are we doing here: we consider an uniform point of the sphere, what does it mean? You pick a point and it can be everywhere with the same probability. For example since the surface of the South and North hemispheres are equal and their union forms the whole sphere you have that the probability of ending on the South is the same than the one of ending in the North and is 1/2.
Now we consider a spherical cap we parametrize by a real \(\beta\). If \(\beta=0\) then the cap is the complete North hemisphere and if \(\beta=0\) the cap is only the North Pole. Clear?
We will go first in one direction: I give you \(\beta\), that defines a certain spherical cap, this spherical cap will have a certain surface and therefore you will be able to compute the probability \(p\) that an uniform point on the sphere ends on your spherical cap.
Then we want to go in the other way around: I give you a certain probability \(p\) and you want to know what is the associated \(\beta\). What we show in the course is that:
$$ \beta = \frac{fct(p)}{\sqrt{D}} $$
Therefore in very large dimension you see that the value of \(\beta\) is very small and depends only slightly on the value of \(p\). And therefore the size of your cap is very large (close to the hemisphere)...
Therefore you see that the probability an uniform point on the sphere sits on a stripe of width \(2\beta\) around the equator would be by symmetry:
$$ 1- 2p$$
i.e. the probability your point is on the sphere minus the probability of being on the South cap minus the probability being on the North cap.
And you see that even for very small value of \(\beta\), this probability will be very close to one.
If it still not clear do not hesitate to come to the Q&A session of Thursday.
Thanks ! It's getting clearer.
I've a small question left. I don't really see how the ending distribution is still uniform, since the probability of picking a point at random on the sphere is not uniform anymore...
The probability of picking a point at random on the sphere is still uniform!
You should check the other question on the forum about the spherical cap. It will explain you why this is the case.
Spherical cap case
Good afternoon,
I have a question about the spherical cap example to explain how we can get a very low standard risk with an adversarial risk close to 1.
I'm struggling with the fact that the all mass (is it the probability mass ?) tends to sit near the equator. Is it due to our assumption: p being very small, and E covering nearly half of the sphere, with the probability of picking a point at random in E being therefore very small ?
I can't see why this should be symmetric, and how the distribution of x on the sphere is still uniform. I cannot visualize it.
Thank you for your help !
1
Hello,
Indeed it is a very counterintuitive fact and it is not easy (even impossible) to visualize.
So what are we doing here: we consider an uniform point of the sphere, what does it mean? You pick a point and it can be everywhere with the same probability. For example since the surface of the South and North hemispheres are equal and their union forms the whole sphere you have that the probability of ending on the South is the same than the one of ending in the North and is 1/2.
Now we consider a spherical cap we parametrize by a real \(\beta\). If \(\beta=0\) then the cap is the complete North hemisphere and if \(\beta=0\) the cap is only the North Pole. Clear?
We will go first in one direction: I give you \(\beta\), that defines a certain spherical cap, this spherical cap will have a certain surface and therefore you will be able to compute the probability \(p\) that an uniform point on the sphere ends on your spherical cap.
Then we want to go in the other way around: I give you a certain probability \(p\) and you want to know what is the associated \(\beta\). What we show in the course is that:
$$ \beta = \frac{fct(p)}{\sqrt{D}} $$
Therefore in very large dimension you see that the value of \(\beta\) is very small and depends only slightly on the value of \(p\). And therefore the size of your cap is very large (close to the hemisphere)...
Therefore you see that the probability an uniform point on the sphere sits on a stripe of width \(2\beta\) around the equator would be by symmetry:
$$ 1- 2p$$
i.e. the probability your point is on the sphere minus the probability of being on the South cap minus the probability being on the North cap.
And you see that even for very small value of \(\beta\), this probability will be very close to one.
If it still not clear do not hesitate to come to the Q&A session of Thursday.
Cheers,
Nicolas
1
Thanks ! It's getting clearer.
I've a small question left. I don't really see how the ending distribution is still uniform, since the probability of picking a point at random on the sphere is not uniform anymore...
The probability of picking a point at random on the sphere is still uniform!
You should check the other question on the forum about the spherical cap. It will explain you why this is the case.
Best,
Nicolas
1
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