Relation via derivative of two angle-based probabilities for n points on a circle.Suppose we sample n points uniformly (and independently) on the unit circle in the sense that the probability that a point lies within some circular arc is proportional to its length. For this question, we'll choose the convention to measure angles from 0 to 1 (rather than from 0 to 2 π when we use radians) to make the formulae come out nicer.Then if we are given a fixed circular arc subtending an angle 0 ≤ α ≤ 1 2 (the bounds chosen due to technicalities from reflex angles), we know that the probability that all n points lie in this arc is α n . It is also a known result/elementary exercise that the probability that n such points lie within some arc which subtends an angle of α is n α n − 1 .Is there some probabilistic reason that these formulae are related by differentiation with respect to α?

Question

Relation via derivative of two angle-based probabilities for n points on a circle.Suppose we sample n points uniformly (and independently) on the unit circle in the sense that the probability that a point lies within some circular arc is proportional to its length. For this question, we&#039;ll choose the convention to measure angles from 0 to 1 (rather than from 0 to   2  π when we use radians) to make the formulae come out nicer.Then if we are given a fixed circular arc subtending an angle   0  ≤  α  ≤      1    2   (the bounds chosen due to technicalities from reflex angles), we know that the probability that all n points lie in this arc is       α    n  . It is also a known result/elementary exercise that the probability that n such points lie within some arc which subtends an angle of   α is   n      α          n      −      1      .Is there some probabilistic reason that these formulae are related by differentiation with respect to   α?

Jason Petersen · Accepted Answer

Step 1I’m not sure whether this is what you’d call a probabilistic reason for this relationship, but I think it throws at least some light on why these two results should be the same.The derivative of the probability that all points lie in a fixed arc is the rate of change of that probability as you expand the arc. The probability increases due to the possiblity that one of the n points lies in the newly added segment and the   n  −  1 others were already in the arc. There are n ways to choose the point, the probability that it lies in the added segment of length       d    α is just       d    α (since the density is 1), and the probability that the other   n  −  1 points were already in the arc is       α          n      −      1      , so the probability increase is       d    p  =  n      α          n      −      1            d    α.Step 2This is pretty much the same construction that leads to the result that the probability for all points to lie in some arc of length   α is   n      α          n      −      1      : There are n ways to choose the point that is farthest from its nearest neighbour in clockwise direction, and the probability that the other   n  −  1 points are all within   α of it in counterclockwise direction is       α          n      −      1      .

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