Hi! There's a detail I can't get from the demonstration seen in the last lecture, about the sphere cap and the fact that, even with a small probability of making a mistake (true risk), we have a high adversarial risk:
At first, we say that the datapoints (the Xs) are uniformly distributed on the surface of the sphere. I assume that this means that the probability of finding a datapoint is the same everywhere on the surface of the sphere, right?
Well, then we say that the mass (maybe I miss something but I have no idea of what this mass is) is concentrated on the equator, so basically we are saying that the probability of finding a point on the surface of the sphere is not the same everywhere anymore. What is going on? What am I missing?
I have just replied to a very similar question. You should have a look.
You are correct with your first point but you have to think twice of what you are exactly saying with it: "The probability of finding a datapoint is the same everywhere." What does it mean?
If you consider a certain region on the sphere with an area s then the probability of having a uniform point on this area will be s over the area of the sphere. And this probability depends only on the area of the region and not where it is on the sphere.
But what we are proving is that the area of a thin stripe around the equator is very large (almost the whole surface of the sphere) and the reminding area of the two large spherical caps is very small. So as you say, it is very likely to pick a point around the equator where the mass is concentrated and very unlikely to do so in the spherical caps.
That is just that when you visualize it in low dimension you imagine the surface of the stripe around the equator to be small with respect to the caps but in high dimension this is not true anymore. You cant visualize it but you can still imagine that for an high dimensional sphere the surface around the equator is of the size of the largest country in the world and the rest is of the size of the smallest country in the world.
Cap demonstration - some questions
Hi! There's a detail I can't get from the demonstration seen in the last lecture, about the sphere cap and the fact that, even with a small probability of making a mistake (true risk), we have a high adversarial risk:
At first, we say that the datapoints (the Xs) are uniformly distributed on the surface of the sphere. I assume that this means that the probability of finding a datapoint is the same everywhere on the surface of the sphere, right?
Well, then we say that the mass (maybe I miss something but I have no idea of what this mass is) is concentrated on the equator, so basically we are saying that the probability of finding a point on the surface of the sphere is not the same everywhere anymore. What is going on? What am I missing?
Thanks in advance :)
Hi Giorgio,
I have just replied to a very similar question. You should have a look.
You are correct with your first point but you have to think twice of what you are exactly saying with it: "The probability of finding a datapoint is the same everywhere." What does it mean?
If you consider a certain region on the sphere with an area s then the probability of having a uniform point on this area will be s over the area of the sphere. And this probability depends only on the area of the region and not where it is on the sphere.
But what we are proving is that the area of a thin stripe around the equator is very large (almost the whole surface of the sphere) and the reminding area of the two large spherical caps is very small. So as you say, it is very likely to pick a point around the equator where the mass is concentrated and very unlikely to do so in the spherical caps.
That is just that when you visualize it in low dimension you imagine the surface of the stripe around the equator to be small with respect to the caps but in high dimension this is not true anymore. You cant visualize it but you can still imagine that for an high dimensional sphere the surface around the equator is of the size of the largest country in the world and the rest is of the size of the smallest country in the world.
Best,
Nicolas
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