Hi,
In my experiments I choose (A,B) independent and uniform on (0,1), but
I reject the cases where A > B (and respectively the cases where the
size constraints are not satisfied). This seemed to me the most
appropriate way to generate intervals uniformly distributed in (0,1),
although I am not sure of its correctness.
When no size constraints are imposed, the experimental evaluation of the
intersection probability (generate a large number of interval pairs and
verify how many of them overlap) by choosing (A,B) independent and
uniform on (0,1) and rejecting the cases A > B seems to be equivalent to
the third alternative below, namely choose (X,Y) independent and uniform
on (0,1) and set A=min{X,Y} and B=max{X,Y}.
I would appreciate if someone could tell me how I could generate points
uniformly distributed in the set {(X,Y) | 0 <= X <= 1, 0 <= Y <= 1, X <=
Y} and respectively in the set {(X,Y) | 0 <= X <= 1, 0 <= Y <= 1, X <=
Y, a <= Y - X <= b, a, b constants, 0 <= a <= b <= 1}. Of course, all
the four alternatives bellow, suggested by David, plus my proposition
(based on rejection) could candidate for an answer, but which is the
correct one?
Thank you very much,
Cristian Saita
David Jones wrote:
> "Cristian-Augustin Saita" <[EMAIL PROTECTED]> wrote in
> message as86v0$9kh$[EMAIL PROTECTED]">news:as86v0$9kh$[EMAIL PROTECTED]...
>
>>I'm interested in the probability that two intervals selected "at
>>random" from among those that satisfy the constraints will
>
> intersect.
>
>>"At random" here means that the intervals satisfying the constraints
>>follow an uniform distribution.
>>
>>
>
>
> Well maybe, but what do you think this should mean? The following four
> ways of specifying an interval (A,B) yield different answers:
>
> (1) choose B uniformly on (0,1), then A uniformly on (0,A).
>
> (2) choose A uniformly on (0,1), then B uniformly on (A,1).
>
> (3) choose (X,Y) independent and uniform on (0,1), then set
> A=min(X,Y), B=max(X,Y).
>
> (4) choose the length L to be uniform on (0,1), then choose A to be
> uniform on (0,1-L), and set B=A+L.
>
> David Jones
>
>
.
.
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