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C6L1V@xxxxxxx wrote:
> Ralf Goertz wrote:
> > Hi,
> >
> > I've been working on this problem for quite a while now and although it
> > didn't seem to be too difficult, I got stuck.
>
> The book "An Intro. to Prob. Theory and its Applications", Volume 2, by
> W. Feller, Wiley (1971) deals with problems like this one in Chapter I.
> Feller says to view the problem by fixing one of the points, then
> opening up the circle at that point, so that the other (n-1) points are
> at the order statistics X(1), X(2),...,X(n-1), giving n intervals X(1),
> X(2)-X(1), ...X(n-1)-X(n-2), 2*pi-X(n-1). Problems 22 and 23 on page 42
> give a solution to your problem, with formulas given. It seems that the
> final result for your problem is
>
> P{all gaps > y} = (2*pi)^(-(n-1))*(2*pi - n*y)^(n-1) *1{y <= 2*pi/n}
>
> (where 1{} is the indicator function).
>
> For n = 2 this gives P{gaps > y} = 1 - y/pi for 0 <= y <= pi. For n = 3
> the result is P{gaps > y} = (2*pi)^(-2) * (2*pi - 3*y)^2 for 0 <= y <=
> 2*pi/3. Results for higher n are equally easy. While Feller does not
> actually derive the results, his material in Chapter I and his problems
> develop everything needed, via intermediate results and hints, etc.
This seems to be a different problem from the one stated. According to
my scribblings it's much easier to solve (because the integration
bounds don't fragment into a zillion different cases), but how do you
get from here to the answer to the original problem?
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