Map angles to probabilities: Say that we have n angles which together sum into 2pi...

mathematiker

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Hello. I am struggling with a very specific mathematics problem, which I think is highly non-trivial. Can somebody please help me?

Say that we have n angles which together sum into 2pi. These angles can all be different from each other.
Now we pick one angle at random with probability, such that smaller angles are more likely to be picked than larger angles.

Here are some examples how this would work in the limiting cases:
- if all angles are the same, probability of picking a particular one is 1/n
- if one angle is 0 while all others are non-zero, probability of picking the one that is zero is going to be 1 (or very close to 1)
- if two angles are zero while others are non-zero, probability of picking a zero angle is going to be close to 0.5

Is it possible to find a mapping that would map a particular distribution of angles (that together sum into 2pi) to probabilities
that would satisfy the above limiting cases?

Thanks!
 
Say that we have n angles which together sum into 2pi. These angles can all be different from each other.
Now we pick one angle at random with probability, such that smaller angles are more likely to be picked than larger angles.

Here are some examples how this would work in the limiting cases:
- if all angles are the same, probability of picking a particular one is 1/n
- if one angle is 0 while all others are non-zero, probability of picking the one that is zero is going to be 1 (or very close to 1)
- if two angles are zero while others are non-zero, probability of picking a zero angle is going to be close to 0.5

Is it possible to find a mapping that would map a particular distribution of angles (that together sum into 2pi) to probabilities
that would satisfy the above limiting cases?

So, you just want some probability distribution that fits these requirements, without having to have any actual justification? Is this supposed to model some actual situation in which smaller angles are more likely to be chosen?

The first thing that comes to mind is that, if 0 were not allowed, you could just take the reciprocal of each angle, add them up, and divide each reciprocal by the sum to get its probability. That is, for angles x1, x2, x3, P(xi) would be (1/x1) / (1/x1+1/x2+1/x3), and so on.

If you need to allow an angle to actually be zero, you could adjust by some small amount, like this:

P(xi) = (1/(x1+.0001)) / (1/(x1+.0001)+1/(x2+.0001)+1/(x3+.0001)).

Does that fit your needs?

If there is a context, that might suggest a better idea.
 
So, you just want some probability distribution that fits these requirements, without having to have any actual justification? Is this supposed to model some actual situation in which smaller angles are more likely to be chosen?

The first thing that comes to mind is that, if 0 were not allowed, you could just take the reciprocal of each angle, add them up, and divide each reciprocal by the sum to get its probability. That is, for angles x1, x2, x3, P(xi) would be (1/x1) / (1/x1+1/x2+1/x3), and so on.

If you need to allow an angle to actually be zero, you could adjust by some small amount, like this:

P(xi) = (1/(x1+.0001)) / (1/(x1+.0001)+1/(x2+.0001)+1/(x3+.0001)).

Does that fit your needs?

If there is a context, that might suggest a better idea.

Yes, I think this is what I have been looking for. Thanks!
 
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