In article <asdbe7$qg7$[EMAIL PROTECTED]>,
Cristian-Augustin Saita <[EMAIL PROTECTED]> wrote:
>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.
It's certainly "correct", _if_ this is your interpretation of a "random
interval". It is equivalent to David Jones's #3, which is more efficient
since it doesn't reject anything.
In answer to your initial question, then: if you select two random
intervals by your method or his #3 using X1, X2, X3, X4 chosen
independently from any continuous distribution (getting the first interval
from X1 and X2), your intervals will intersect with probability
exactly 2/3 (consider the 24 equally-likely possible orderings of
X1, X2, X3, X4, which come down to 6 if X1 < X2 and X3 < X4).
>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?
{(X,Y) | 0 <= X <= 1, 0 <= Y <= 1, X <= Y} is David's #3, as I already
said.
For {(X,Y) | 0 <= X , Y <= 1, a <= Y - X <= b} (the other inequalities
you listed are redundant in this case), one way is to choose
X uniform in [0, 1-a], then Y uniform in [X+a,X+b], and reject if
Y > 1.
Robert Israel [EMAIL PROTECTED]
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
>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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