Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss
complement of Uniform Distribution?
15 views
Skip to first unread message
Norman B. Grover
Feb 15, 2013, 4:51:11 AM2/15/13
to
I'm not sure the subject makes much sense so I will give the actual
example.
Consider N empty indistinguishable urns placed in a row and m
indistinguishable balls. I place one of the balls in one of the urns chosen
at random. The probability that a particular urn remain empty is of course
1-1/N. I now place another ball in one of the urns and, if all the urns
(including the occupied urn)still have the same chance of being chosen, the
probability that a particular urn remain empty is (1-1/N)^2. After placing
all m balls at random, the probability that a particular urn remain empty is
(1-1/N)^m or 1-mu, where mu is the probability that a particular urn contain
at least one ball.
The position of the m balls forms a Discrete Uniform Distribution over
the interval 1 to N. That is the easy part. What I am interested is the
'complement', the distribution of the number of empty urns p(k) between the
occupied urns: how many occupied urns are preceded by exactly k empty urns,
where k can range from 0 (two contiguous occupied urns) to N-1 (all m balls
in same urn). For simplicity, I consider the urns in a circle,so that urn #N
is followed by urn #1.
At first I tried the Geometric Distribution: p(k)= mu*(1-mu)^k, but that
cannot be correct since the probability of two consecutive urns being empty
is not (1-mu)*(1-mu) because these probabilities are not independent; that
is, the intersection of these two events (call them A and B), p(A)*p(B|A),
is not p(A)*p(B) since the occurrence of A decreases the probability of B.
I could also compute the total number of empty urns and so the average
value of k, but I have no idea how to proceed from there to get its
distribution.
Any help or suggestion or reference would be greatly appreciated.
--
Norman B. Grover
Jerusalem, Israel
example.
Consider N empty indistinguishable urns placed in a row and m
indistinguishable balls. I place one of the balls in one of the urns chosen
at random. The probability that a particular urn remain empty is of course
1-1/N. I now place another ball in one of the urns and, if all the urns
(including the occupied urn)still have the same chance of being chosen, the
probability that a particular urn remain empty is (1-1/N)^2. After placing
all m balls at random, the probability that a particular urn remain empty is
(1-1/N)^m or 1-mu, where mu is the probability that a particular urn contain
at least one ball.
The position of the m balls forms a Discrete Uniform Distribution over
the interval 1 to N. That is the easy part. What I am interested is the
'complement', the distribution of the number of empty urns p(k) between the
occupied urns: how many occupied urns are preceded by exactly k empty urns,
where k can range from 0 (two contiguous occupied urns) to N-1 (all m balls
in same urn). For simplicity, I consider the urns in a circle,so that urn #N
is followed by urn #1.
At first I tried the Geometric Distribution: p(k)= mu*(1-mu)^k, but that
cannot be correct since the probability of two consecutive urns being empty
is not (1-mu)*(1-mu) because these probabilities are not independent; that
is, the intersection of these two events (call them A and B), p(A)*p(B|A),
is not p(A)*p(B) since the occurrence of A decreases the probability of B.
I could also compute the total number of empty urns and so the average
value of k, but I have no idea how to proceed from there to get its
distribution.
Any help or suggestion or reference would be greatly appreciated.
--
Norman B. Grover
Jerusalem, Israel
Norman B. Grover
Feb 15, 2013, 8:52:42 AM2/15/13
to
In article <MPG.2b8831851...@news.albasani.net>,
nor...@md.huji.ac.il says...
number of empty urns,it may be possible to use the Negative Hypergeometric
Distribution with a=1.
nor...@md.huji.ac.il says...
> I'm not sure the subject makes much sense so I will give the actual
> example.
> Consider N empty indistinguishable urns placed in a row and m
> indistinguishable balls. I place one of the balls in one of the urns chosen
> at random. The probability that a particular urn remain empty is of course
> 1-1/N. I now place another ball in one of the urns and, if all the urns
> (including the occupied urn)still have the same chance of being chosen, the
> probability that a particular urn remain empty is (1-1/N)^2. After placing
> all m balls at random, the probability that a particular urn remain empty is
> (1-1/N)^m or 1-mu, where mu is the probability that a particular urn contain
> at least one ball.
> The position of the m balls forms a Discrete Uniform Distribution over
> the interval 1 to N. That is the easy part. What I am interested in is the
> example.
> Consider N empty indistinguishable urns placed in a row and m
> indistinguishable balls. I place one of the balls in one of the urns chosen
> at random. The probability that a particular urn remain empty is of course
> 1-1/N. I now place another ball in one of the urns and, if all the urns
> (including the occupied urn)still have the same chance of being chosen, the
> probability that a particular urn remain empty is (1-1/N)^2. After placing
> all m balls at random, the probability that a particular urn remain empty is
> (1-1/N)^m or 1-mu, where mu is the probability that a particular urn contain
> at least one ball.
> The position of the m balls forms a Discrete Uniform Distribution over
> 'complement', the distribution of the number of empty urns p(k) between the
> occupied urns: how many occupied urns are preceded by exactly k empty urns,
> where k can range from 0 (two contiguous occupied urns) to N-1 (all m balls
> in same urn). For simplicity, I consider the urns in a circle,so that urn #N
> is followed by urn #1.
> At first I tried the Geometric Distribution: p(k)= mu*(1-mu)^k, but that
> cannot be correct since the probability of two consecutive urns being empty
> is not (1-mu)*(1-mu) because these probabilities are not independent; that
> is, the intersection of these two events (call them A and B), p(A)*p(B|A),
> is not p(A)*p(B) since the occurrence of A decreases the probability of B.
> I could also compute the total number of empty urns and so the average
> value of k, but I have no idea how to proceed from there to get its
> distribution.
> Any help or suggestion or reference would be greatly appreciated.
>
It just occurred to me that if I really am able to compute the total
> occupied urns: how many occupied urns are preceded by exactly k empty urns,
> where k can range from 0 (two contiguous occupied urns) to N-1 (all m balls
> in same urn). For simplicity, I consider the urns in a circle,so that urn #N
> is followed by urn #1.
> At first I tried the Geometric Distribution: p(k)= mu*(1-mu)^k, but that
> cannot be correct since the probability of two consecutive urns being empty
> is not (1-mu)*(1-mu) because these probabilities are not independent; that
> is, the intersection of these two events (call them A and B), p(A)*p(B|A),
> is not p(A)*p(B) since the occurrence of A decreases the probability of B.
> I could also compute the total number of empty urns and so the average
> value of k, but I have no idea how to proceed from there to get its
> distribution.
> Any help or suggestion or reference would be greatly appreciated.
>
number of empty urns,it may be possible to use the Negative Hypergeometric
Distribution with a=1.
Norman B. Grover
Feb 15, 2013, 11:46:39 AM2/15/13
to
In article <MPG.2b886a204...@news.albasani.net>,
nor...@md.huji.ac.il says...
be applied to the present problem.
nor...@md.huji.ac.il says...
> It just occurred to me that if I really am able to compute the total
> number of empty urns,it may be possible to use the Negative Hypergeometric
> Distribution with a=1.
My enthusiasm was premature; I doubt that the Negative Hypergeometric can
> number of empty urns,it may be possible to use the Negative Hypergeometric
> Distribution with a=1.
be applied to the present problem.
0 new messages