Author 
Message 
G.E. Ivey science forum Guru
Joined: 29 Apr 2005
Posts: 308

Posted: Mon Jul 10, 2006 12:05 pm Post subject:
Re: random selections of integers from the infinite set of all integers



Quote:  By "perfectly random manner" are you still holding
out hope for a
"hypothetical" uniformly random selection process
over all integers?
I'd be interested to see how that would work
because I don't think it
can be done  not even hypothetically.
I don't know how you would do it, either, but then, I
would worry about that
after you actually presented me with a real infinite
ditribution from which
to select elements at random....

Excuse me? You were the one who repeatedly posed the problem about "selecting numbers at random from the set of all itegers", in spite of having been told repeatedly that there is NO way to do that there is no way to assign a probability distribution to the set of all integers. If you believe it can be done, then it is YOUR responsibility to display such a probability distribution, not those who have said it is impossible
Quote:  Perhaps it would simply be the entire infinite set of
integers, in some
totally random order, with no pattern possible. Of
course, don't ask me how
you would get that, either.....
:=)

Pete B
matt271829news@yahoo.co.uk> wrote in message
news:1152214244.827384.204850@m73g2000cwd.googlegroups
.com...
Pete B wrote:
Ok I misunderstood what your objection was. I was
aware of the points
you
present, I was really looking for a speculation as
to whether the
outcomes
were possible or not possible.
On that basis, let me modify this, what I am
really interested in is
primarily this. Given the selection hypothesis I
outlined, where one
selects integers, in a hypothetically perfect
random manner, from the
infinite set of integers,
By "perfectly random manner" are you still holding
out hope for a
"hypothetical" uniformly random selection process
over all integers?
I'd be interested to see how that would work
because I don't think it
can be done  not even hypothetically.
doing so for all eternity, do you think that, at
some point, every integer would have been selected
at least one time? Or
do
you think that even when you select integers at
random for all eternity,
IOW
forever, in an infinite neverending selection
process, there is a
nonzero
but otherwise undefinable possibility that there
would be an integer that
would never be selected?
Just looking for a yes/no for both hypothetical
outcomes, I really do not
care about actual probabilities. One can imagine
that oit is possible to
have infinite sets, and one can imagine that it is
possible to make truly
random selections from those sets. So what I seek
is really just an
intelligent speculation as to the outcome.

Pete B
"Randy Poe" <poespamtrap@yahoo.com> wrote in
message
news:1152212295.020790.94860@j8g2000cwa.googlegroups.c
om...
Topposting repaired.
Pete B wrote:
"Randy Poe" <poespamtrap@yahoo.com> wrote in
message
news:1152202617.267205.167000@j8g2000cwa.googlegroups.
com...
Pete B wrote:
Suppose one spends an eternity making
perfectly random selections
of
integers from the infinite set of all
integers.
Unfortunately, your scenario has a problem
with the first line.
There is no way to have "perfectly random
selection from
the set of all integers".
Well, of course not. Similarly, there is no
way to do anything with
infinite sets because they are simply mind
constructs,
You misunderstood. There's no way to "mind
construct" a
uniform distribution over all the integers.
There's no way to
define such a probability function.
not real collections.
There is no way to do anything for eternity,
either.
I'm talking about what is possible to construct
selfconsistently
using the normal axioms of mathematics and
probability. A
uniform distribution over the integers is not.
But I am posing this
is a hypothetical question, one which assumes
that you can do such
imaginary
operations. Cantor did the same, I see no
reason why I should not be
able
to.
Cantor constructed things which can be
selfconsistently
defined on a set of axioms. A nonuniform
distribution over
the integers can be so defined, and I did so in
my other reply.
A uniform distribution can't.
 Randy



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Pubkeybreaker science forum Guru
Joined: 24 Mar 2005
Posts: 333

Posted: Mon Jul 10, 2006 4:44 pm Post subject:
Re: random selections of integers from the infinite set of all integers



Randy Poe wrote:
Quote:  Pete B wrote:
Suppose one spends an eternity making perfectly random selections of
integers from the infinite set of all integers.
Unfortunately, your scenario has a problem with the first line.
There is no way to have "perfectly random selection from
the set of all integers".
 Randy

huh? Of course there is. Your claim is false.
What *is* true is that it is impossible to make a *uniformly*
random selection. It is possible to make perfectly random
draws from a nonuniform distribution. For example
consider the pdf f(n) = 1/2 * 1/2^abs(n) n != 0, and f(0) = 0 

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Pubkeybreaker science forum Guru
Joined: 24 Mar 2005
Posts: 333

Posted: Mon Jul 10, 2006 4:47 pm Post subject:
Re: random selections of integers from the infinite set of all integers



G.E. Ivey wrote:
Quote:  Excuse me? You were the one who repeatedly posed the problem about
"selecting numbers at random from the set of all itegers", in spite of 
having
been told repeatedly that there is NO way to do that there is no way
to assign
a probability distribution to the set of all integers. If you believe
it can be done,
then it is YOUR responsibility to display such a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible. It *IS* impossible to
have a *UNIFORM* density function, but a nonuniform density
function (where the probability of selecting n decays (say)
exponentially
with abs(n)) is quite possible. 

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Tom111 science forum Guru Wannabe
Joined: 21 Jan 2006
Posts: 173

Posted: Mon Jul 10, 2006 6:22 pm Post subject:
Re: random selections of integers from the infinite set of all integers



Quote: 
G.E. Ivey wrote:
Excuse me? You were the one who repeatedly
posed the problem about
"selecting numbers at random from the set of all
itegers", in spite of
having
been told repeatedly that there is NO way to do that
there is no way
to assign
a probability distribution to the set of all
integers. If you believe
it can be done,
then it is YOUR responsibility to display such a
a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible. It *IS*
impossible to
have a *UNIFORM* density function, but a
nonuniform density
function (where the probability of selecting n
decays (say)
exponentially
with abs(n)) is quite possible.
It *IS* impossible to have a *UNIFORM* density function

I think its fair to say that using singlevalued algebra, a uniform density probability function over the integers doesn't exist (since it has a mean which is not defined in the algebra).
using multivalued algebra, it would exist. Its variance and standard deviation would be infinity, its typical values in the set would be + infinity, and its mean would be the set of all numbers, or 0/0 (unconstrained but defined).
Similarly, using singlevalued algebra, the question cannot be answered (as the distribution can't be defined).
With multivalued algebra we can answer these questions (questions 1  4 seem to give the result [0,1], and 5 could well be {yes, no}). 

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matt271829news@yahoo.co. science forum Guru
Joined: 11 Sep 2005
Posts: 846

Posted: Mon Jul 10, 2006 8:40 pm Post subject:
Re: random selections of integers from the infinite set of all integers



Tom wrote:
Quote: 
G.E. Ivey wrote:
Excuse me? You were the one who repeatedly
posed the problem about
"selecting numbers at random from the set of all
itegers", in spite of
having
been told repeatedly that there is NO way to do that
there is no way
to assign
a probability distribution to the set of all
integers. If you believe
it can be done,
then it is YOUR responsibility to display such a
a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible. It *IS*
impossible to
have a *UNIFORM* density function, but a
nonuniform density
function (where the probability of selecting n
decays (say)
exponentially
with abs(n)) is quite possible.
It *IS* impossible to have a *UNIFORM* density function
I think its fair to say that using singlevalued algebra, a uniform density probability function over the integers doesn't exist (since it has a mean which is not defined in the algebra).
using multivalued algebra, it would exist. Its variance and standard deviation would be infinity, its typical values in the set would be + infinity, and its mean would be the set of all numbers, or 0/0 (unconstrained but defined).

I'm not very clear what you mean by "multivalued algebra" here. How
would this probability density function be defined? For arbitrary
integer n, what is Pr(n)?
Quote: 
Similarly, using singlevalued algebra, the question cannot be answered (as the distribution can't be defined).
With multivalued algebra we can answer these questions (questions 1  4 seem to give the result [0,1], and 5 could well be {yes, no}). 


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Tom111 science forum Guru Wannabe
Joined: 21 Jan 2006
Posts: 173

Posted: Tue Jul 11, 2006 12:48 pm Post subject:
Re: random selections of integers from the infinite set of all integers



Quote: 
Tom wrote:
G.E. Ivey wrote:
Excuse me? You were the one who repeatedly
posed the problem about
"selecting numbers at random from the set of all
itegers", in spite of
having
been told repeatedly that there is NO way to do
that
there is no way
to assign
a probability distribution to the set of all
integers. If you believe
it can be done,
then it is YOUR responsibility to display such a
a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible. It
*IS*
impossible to
have a *UNIFORM* density function, but a
nonuniform density
function (where the probability of selecting n
decays (say)
exponentially
with abs(n)) is quite possible.
It *IS* impossible to have a *UNIFORM* density
function
I think its fair to say that using singlevalued
algebra, a uniform density probability function over
the integers doesn't exist (since it has a mean which
is not defined in the algebra).
using multivalued algebra, it would exist. Its
variance and standard deviation would be infinity,
its typical values in the set would be + infinity,
and its mean would be the set of all numbers, or 0/0
(unconstrained but defined).
I'm not very clear what you mean by "multivalued
algebra" here. How
would this probability density function be defined?
For arbitrary
integer n, what is Pr(n)?

Pr(n) = 0 for arbitrary n
Quote:  I'm not very clear what you mean by "multivalued
algebra" here.
No I haven't been... I mean an algebra (quantities and operations on them) which is a superset of real number algebra. The algebra accepts multivalues as quantities in their own right. For example +sqrt(4) = {2,2}, so the right hand side of this equation is still a quantity which can be used. 
similarly 0/0 = { all reals }.
I still haven't found any decent texts on this, but I'm interested in finding them. My only references I can give are from a previous thread:
http://mathforum.org/kb/message.jspa?messageID=4829865&tstart=0
and this book:
R. C. Young, "The algebra of manyvalued quantities", _Math. Annalen_
104 (1931) 260290. 

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matt271829news@yahoo.co. science forum Guru
Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jul 11, 2006 8:23 pm Post subject:
Re: random selections of integers from the infinite set of all integers



Tom wrote:
Quote: 
Tom wrote:
G.E. Ivey wrote:
Excuse me? You were the one who repeatedly
posed the problem about
"selecting numbers at random from the set of all
itegers", in spite of
having
been told repeatedly that there is NO way to do
that
there is no way
to assign
a probability distribution to the set of all
integers. If you believe
it can be done,
then it is YOUR responsibility to display such a
a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible. It
*IS*
impossible to
have a *UNIFORM* density function, but a
nonuniform density
function (where the probability of selecting n
decays (say)
exponentially
with abs(n)) is quite possible.
It *IS* impossible to have a *UNIFORM* density
function
I think its fair to say that using singlevalued
algebra, a uniform density probability function over
the integers doesn't exist (since it has a mean which
is not defined in the algebra).
using multivalued algebra, it would exist. Its
variance and standard deviation would be infinity,
its typical values in the set would be + infinity,
and its mean would be the set of all numbers, or 0/0
(unconstrained but defined).
I'm not very clear what you mean by "multivalued
algebra" here. How
would this probability density function be defined?
For arbitrary
integer n, what is Pr(n)?
Pr(n) = 0 for arbitrary n
I'm not very clear what you mean by "multivalued
algebra" here.
No I haven't been... I mean an algebra (quantities and operations on them) which is a superset of real number algebra. The algebra accepts multivalues as quantities in their own right. For example +sqrt(4) = {2,2}, so the right hand side of this equation is still a quantity which can be used.
similarly 0/0 = { all reals }.
I still haven't found any decent texts on this, but I'm interested in finding them. My only references I can give are from a previous thread:
http://mathforum.org/kb/message.jspa?messageID=4829865&tstart=0
and this book:
R. C. Young, "The algebra of manyvalued quantities", _Math. Annalen_
104 (1931) 260290.

Right, I remember that thread actually. But I don't see the relevance
of multivalued algebra to the problem of constructing a uniform random
distribution over all integers, or how it might make such a thing in
some sense possible. 

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Tom111 science forum Guru Wannabe
Joined: 21 Jan 2006
Posts: 173

Posted: Wed Jul 12, 2006 9:47 am Post subject:
Re: random selections of integers from the infinite set of all integers



Quote: 
Tom wrote:
Tom wrote:
G.E. Ivey wrote:
Excuse me? You were the one who
repeatedly
posed the problem about
"selecting numbers at random from the set of
all
itegers", in spite of
having
been told repeatedly that there is NO way to
do
that
there is no way
to assign
a probability distribution to the set of all
integers. If you believe
it can be done,
then it is YOUR responsibility to display
such a
a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible. It
*IS*
impossible to
have a *UNIFORM* density function, but a
nonuniform density
function (where the probability of selecting
n
decays (say)
exponentially
with abs(n)) is quite possible.
It *IS* impossible to have a *UNIFORM*
density
function
I think its fair to say that using
singlevalued
algebra, a uniform density probability function
over
the integers doesn't exist (since it has a mean
which
is not defined in the algebra).
using multivalued algebra, it would exist. Its
variance and standard deviation would be
infinity,
its typical values in the set would be +
infinity,
and its mean would be the set of all numbers, or
0/0
(unconstrained but defined).
I'm not very clear what you mean by "multivalued
algebra" here. How
would this probability density function be
defined?
For arbitrary
integer n, what is Pr(n)?
Pr(n) = 0 for arbitrary n
I'm not very clear what you mean by "multivalued
algebra" here.
No I haven't been... I mean an algebra (quantities
and operations on them) which is a superset of real
number algebra. The algebra accepts multivalues as
quantities in their own right. For example +sqrt(4)
= {2,2}, so the right hand side of this equation is
still a quantity which can be used.
similarly 0/0 = { all reals }.
I still haven't found any decent texts on this, but
I'm interested in finding them. My only references I
can give are from a previous thread:
http://mathforum.org/kb/message.jspa?messageID=4829865
&tstart=0
and this book:
R. C. Young, "The algebra of manyvalued
quantities", _Math. Annalen_
104 (1931) 260290.
Right, I remember that thread actually. But I don't
see the relevance
of multivalued algebra to the problem of
constructing a uniform random
distribution over all integers, or how it might make
such a thing in
some sense possible.

Well it simply means that some results are defined instead of undefined.
for example if we're using natural number algebra, there is no solution to x+2 = 1. x doesn't exist in this algebra.
if we're using real number algebra, there is no sqrt(1), it doesn't exist and is usually exressed as undefined (or error on a calculator).
if we're using singlevalued algebra then 0/0 is undefined / doesn't exist.
if we're using multivalued algebra then 0/0 is the set of all numbers, +sqrt(4) is {2,2} etc.
so whereas a uniform random distribution over all the integers in singlevalued algebra has undefined mean (and so its interpreted that this distribution can't exist), in multivalued algebra the mean is the set of all numbers.
Also the variance is infinite and the probability at any number is 0.
Its probably not a very helpful to use these values directly, but all attempts to make use of this infinite distribution should be replaced with attempts to use the convergence of a finite uniform distribution as its size approaches infinity. This is true of all infinities... they are short hand for 'what you converge on as you approach infinity'. If you converge on different numbers when you approach in different ways then the value at infinity is multivalued (like the mean of this uniform distribution).
Anyway, I should say that this is my interpretation, and not necessarily a classical view of things.. so I'm speaking somewhat out of the box I think.. but hopefully it makes sense.
Tom. 

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matt271829news@yahoo.co. science forum Guru
Joined: 11 Sep 2005
Posts: 846

Posted: Wed Jul 12, 2006 11:13 am Post subject:
Re: random selections of integers from the infinite set of all integers



Tom wrote:
Quote: 
Tom wrote:
Tom wrote:
G.E. Ivey wrote:
Excuse me? You were the one who
repeatedly
posed the problem about
"selecting numbers at random from the set of
all
itegers", in spite of
having
been told repeatedly that there is NO way to
do
that
there is no way
to assign
a probability distribution to the set of all
integers. If you believe
it can be done,
then it is YOUR responsibility to display
such a
a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible. It
*IS*
impossible to
have a *UNIFORM* density function, but a
nonuniform density
function (where the probability of selecting
n
decays (say)
exponentially
with abs(n)) is quite possible.
It *IS* impossible to have a *UNIFORM*
density
function
I think its fair to say that using
singlevalued
algebra, a uniform density probability function
over
the integers doesn't exist (since it has a mean
which
is not defined in the algebra).
using multivalued algebra, it would exist. Its
variance and standard deviation would be
infinity,
its typical values in the set would be +
infinity,
and its mean would be the set of all numbers, or
0/0
(unconstrained but defined).
I'm not very clear what you mean by "multivalued
algebra" here. How
would this probability density function be
defined?
For arbitrary
integer n, what is Pr(n)?
Pr(n) = 0 for arbitrary n
I'm not very clear what you mean by "multivalued
algebra" here.
No I haven't been... I mean an algebra (quantities
and operations on them) which is a superset of real
number algebra. The algebra accepts multivalues as
quantities in their own right. For example +sqrt(4)
= {2,2}, so the right hand side of this equation is
still a quantity which can be used.
similarly 0/0 = { all reals }.
I still haven't found any decent texts on this, but
I'm interested in finding them. My only references I
can give are from a previous thread:
http://mathforum.org/kb/message.jspa?messageID=4829865
&tstart=0
and this book:
R. C. Young, "The algebra of manyvalued
quantities", _Math. Annalen_
104 (1931) 260290.
Right, I remember that thread actually. But I don't
see the relevance
of multivalued algebra to the problem of
constructing a uniform random
distribution over all integers, or how it might make
such a thing in
some sense possible.
Well it simply means that some results are defined instead of undefined.
for example if we're using natural number algebra, there is no solution to x+2 = 1. x doesn't exist in this algebra.
if we're using real number algebra, there is no sqrt(1), it doesn't exist and is usually exressed as undefined (or error on a calculator).
if we're using singlevalued algebra then 0/0 is undefined / doesn't exist.
if we're using multivalued algebra then 0/0 is the set of all numbers, +sqrt(4) is {2,2} etc.
so whereas a uniform random distribution over all the integers in singlevalued algebra has undefined mean (and so its interpreted that this distribution can't exist),

There are perfectly fine probability distributions that definitely
exist and yet do not have a defined mean or variance. One simple
example is the discrete distribution P(x) = 1/x, where x = 2, 4, 8, 16,
.... So I think your "so" doesn't follow.
Quote:  in multivalued algebra the mean is the set of all numbers.
Also the variance is infinite and the probability at any number is 0.
Its probably not a very helpful to use these values directly, but all attempts to make use of this infinite distribution should be replaced with attempts to use the convergence of a finite uniform distribution as its size approaches infinity. This is true of all infinities... they are short hand for 'what you converge on as you approach infinity'. If you converge on different numbers when you approach in different ways then the value at infinity is multivalued (like the mean of this uniform distribution).
Anyway, I should say that this is my interpretation, and not necessarily a classical view of things.. so I'm speaking somewhat out of the box I think.. but hopefully it makes sense.

Not really, not to me anyway. You say "the probability at any number is
0", but I confess I don't see how the distribution P(n) = 0, for n
integer, is in any sense a uniform distribution over all integers.
But good luck anyway! 

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Jonathan Hoyle science forum Guru Wannabe
Joined: 04 Sep 2005
Posts: 260

Posted: Wed Jul 12, 2006 11:39 am Post subject:
Re: random selections of integers from the infinite set of all integers



Quote:  Not really, not to me anyway. You say "the probability at any number is
0", but I confess I don't see how the distribution P(n) = 0, for n
integer, is in any sense a uniform distribution over all integers.
But good luck anyway!

That P(n) = 0 for all integer n is not the problem. After all, in
U(0,1), the probability of picking any specific real number is also 0.
The problem lies with the fact that any probability distribution
function must sum to 1 over the entire space. That is to say: although
the probability of picking any particular real in U(0,1) is 0, the
probaility of picking *something* in the range of (0,1) is 1, and for
any subset X in (0,1), we have 0 <= Prob(X) <= 1. Unfortunately, since
the integers are countable, assigning P(n) = 0 for all n means that
Prob(N) = 0 as well, and thus it is not a pdf.
It is a theorem of mathematics that you cannot assign a pdf to a
countably infinite set such that each element of the set is
equiprobable. 

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Tom111 science forum Guru Wannabe
Joined: 21 Jan 2006
Posts: 173

Posted: Wed Jul 12, 2006 11:43 am Post subject:
Re: random selections of integers from the infinite set of all integers



Quote:  Tom wrote:
Tom wrote:
Tom wrote:
G.E. Ivey wrote:
Excuse me? You were the one who
repeatedly
posed the problem about
"selecting numbers at random from the set
of
all
itegers", in spite of
having
been told repeatedly that there is NO way
to
do
that
there is no way
to assign
a probability distribution to the set of
all
integers. If you believe
it can be done,
then it is YOUR responsibility to display
such a
a probability
distribution, not
those who have said it is impossible
As I said before, it is NOT impossible.
It
*IS*
impossible to
have a *UNIFORM* density function, but
a
nonuniform density
function (where the probability of
selecting
n
decays (say)
exponentially
with abs(n)) is quite possible.
It *IS* impossible to have a *UNIFORM*
density
function
I think its fair to say that using
singlevalued
algebra, a uniform density probability
function
over
the integers doesn't exist (since it has a
mean
which
is not defined in the algebra).
using multivalued algebra, it would exist.
Its
variance and standard deviation would be
infinity,
its typical values in the set would be +
infinity,
and its mean would be the set of all numbers,
or
0/0
(unconstrained but defined).
I'm not very clear what you mean by
"multivalued
algebra" here. How
would this probability density function be
defined?
For arbitrary
integer n, what is Pr(n)?
Pr(n) = 0 for arbitrary n
I'm not very clear what you mean by
"multivalued
algebra" here.
No I haven't been... I mean an algebra
(quantities
and operations on them) which is a superset of
real
number algebra. The algebra accepts multivalues
as
quantities in their own right. For example
+sqrt(4)
= {2,2}, so the right hand side of this equation
is
still a quantity which can be used.
similarly 0/0 = { all reals }.
I still haven't found any decent texts on this,
but
I'm interested in finding them. My only
references I
can give are from a previous thread:
http://mathforum.org/kb/message.jspa?messageID=4829865
&tstart=0
and this book:
R. C. Young, "The algebra of manyvalued
quantities", _Math. Annalen_
104 (1931) 260290.
Right, I remember that thread actually. But I
don't
see the relevance
of multivalued algebra to the problem of
constructing a uniform random
distribution over all integers, or how it might
make
such a thing in
some sense possible.
Well it simply means that some results are defined
instead of undefined.
for example if we're using natural number algebra,
there is no solution to x+2 = 1. x doesn't exist in
this algebra.
if we're using real number algebra, there is no
sqrt(1), it doesn't exist and is usually exressed as
undefined (or error on a calculator).
if we're using singlevalued algebra then 0/0 is
undefined / doesn't exist.
if we're using multivalued algebra then 0/0 is the
set of all numbers, +sqrt(4) is {2,2} etc.
so whereas a uniform random distribution over all
the integers in singlevalued algebra has undefined
mean (and so its interpreted that this distribution
can't exist),
There are perfectly fine probability distributions
that definitely
exist and yet do not have a defined mean or variance.
One simple
example is the discrete distribution P(x) = 1/x,
where x = 2, 4, 8, 16,
... So I think your "so" doesn't follow.
in multivalued algebra the mean is the set of all
numbers.
Also the variance is infinite and the probability
at any number is 0.
Its probably not a very helpful to use these values
directly, but all attempts to make use of this
infinite distribution should be replaced with
attempts to use the convergence of a finite uniform
distribution as its size approaches infinity. This is
true of all infinities... they are short hand for
'what you converge on as you approach infinity'. If
you converge on different numbers when you approach
in different ways then the value at infinity is
multivalued (like the mean of this uniform
distribution).
Anyway, I should say that this is my
interpretation, and not necessarily a classical view
of things.. so I'm speaking somewhat out of the box I
think.. but hopefully it makes sense.
Not really, not to me anyway. You say "the
probability at any number is
0", but I confess I don't see how the distribution
P(n) = 0, for n
integer, is in any sense a uniform distribution over
all integers.
But good luck anyway!

Well thanks, we'll agree to disagree..
perhaps its quite a philosophical point of contention anyway. 

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Math1723 science forum beginner
Joined: 17 Dec 2005
Posts: 13

Posted: Wed Jul 12, 2006 11:52 am Post subject:
Re: random selections of integers from the infinite set of all integers



Jonathan Hoyle wrote:
Quote:  It is a theorem of mathematics that you cannot assign a pdf to a
countably infinite set such that each element of the set is
equiprobable.

Not without dropping Countable Addivity.
You can in theory do it by dropping Countable Addivity, but then you
are not left with much. For any finite subset X, we know Prob(X) = 0,
but for infinite X, P(X) is not defined. I suppose you can create a
nonprinciple ultrafilter on N so that "large" sets X have P(X) = 1 and
all other sets havce P(X) = 0, but this would be ad hoc and is a major
departure from standard Probability Theory. It is also not within the
spirit of the original problem. 

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Tom111 science forum Guru Wannabe
Joined: 21 Jan 2006
Posts: 173

Posted: Wed Jul 12, 2006 1:50 pm Post subject:
Re: random selections of integers from the infinite set of all integers



Quote:  Jonathan Hoyle wrote:
It is a theorem of mathematics that you cannot
assign a pdf to a
countably infinite set such that each element of
the set is
equiprobable.
Not without dropping Countable Addivity.
You can in theory do it by dropping Countable
Addivity, but then you
are not left with much. For any finite subset X, we
know Prob(X) = 0,
but for infinite X, P(X) is not defined. I suppose
you can create a
nonprinciple ultrafilter on N so that "large" sets X
have P(X) = 1 and
all other sets havce P(X) = 0, but this would be ad
hoc and is a major
departure from standard Probability Theory. It is
also not within the
spirit of the original problem.

Infinities are outside real numbers, so when we talk of a property at infinity, this is short hand for 'the convergence of the property as we approach infinity'.
Now lets say (hypothetically) that we could pick from a uniform distribution over all integers.
For each value we pick (x), lets inverse it (1/x) and plot a distribution of these values.
the exact (infinitely precise) distribution you get is a twofold convergence... firstly we see what the distribution converges to for more and more sample points on a finite uniform distribution. Secondly we see what this converges to as the finite distribution gets larger (approaches a distribution over all the integers).
Plotting these graphically we see that as the number of samples>infinity and the width of the uniform distribution>infinity, we approach a discrete (spike) distribition of P(0) = 1.
The point of this rather involved example is that we've used a uniform distribution over all the integers in a calculation to get a useful result (in this case the distribution of the inverse of uniformly chosen integers).
The general point being that you can use this infinite uniform distributions in calculations to get definite answers to questions. Which is what the original poster was after.
Thought of the day: no infinites exist, but if you can measure what is converged towards as you approach them, then you can pretend they exist. 

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