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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 matt271829-news@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 never-ending selection process, there is a non-zero 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" wrote in message news:1152212295.020790.94860@j8g2000cwa.googlegroups.c om... Top-posting repaired. Pete B wrote: "Randy Poe" 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 self-consistently 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 self-consistently defined on a set of axioms. A non-uniform distribution over the integers can be so defined, and I did so in my other reply. A uniform distribution can't. - Randy
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 non-uniform distribution. For example
consider the pdf f(n) = 1/2 * 1/2^abs(n) n != 0, and f(0) = 0
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 non-uniform density
function (where the probability of selecting n decays (say)
exponentially
with abs(n)) is quite possible.
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 non-uniform 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 single-valued 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 multi-valued 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 single-valued algebra, the question cannot be answered (as the distribution can't be defined).
With multi-valued algebra we can answer these questions (questions 1 - 4 seem to give the result [0,1], and 5 could well be {yes, no}).
matt271829-news@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 non-uniform 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 single-valued 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 multi-valued 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 "multi-valued algebra" here. How
would this probability density function be defined? For arbitrary
integer n, what is Pr(n)?

 Quote: Similarly, using single-valued algebra, the question cannot be answered (as the distribution can't be defined). With multi-valued algebra we can answer these questions (questions 1 - 4 seem to give the result [0,1], and 5 could well be {yes, no}).
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 non-uniform 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 single-valued 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 multi-valued 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 "multi-valued 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 "multi-valued 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 multi-values 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 many-valued quantities", _Math. Annalen_
104 (1931) 260-290.
matt271829-news@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 non-uniform 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 single-valued 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 multi-valued 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 "multi-valued 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 "multi-valued 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 multi-values 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 many-valued quantities", _Math. Annalen_ 104 (1931) 260-290.

Right, I remember that thread actually. But I don't see the relevance
of multi-valued 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.
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 non-uniform 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 single-valued 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 multi-valued 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 "multi-valued 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 "multi-valued 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 multi-values 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 many-valued quantities", _Math. Annalen_ 104 (1931) 260-290. Right, I remember that thread actually. But I don't see the relevance of multi-valued 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 single-valued algebra then 0/0 is undefined / doesn't exist.
if we're using multi-valued 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 single-valued algebra has undefined mean (and so its interpreted that this distribution can't exist), in multi-valued 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 multi-valued (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.
matt271829-news@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 non-uniform 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 single-valued 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 multi-valued 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 "multi-valued 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 "multi-valued 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 multi-values 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 many-valued quantities", _Math. Annalen_ 104 (1931) 260-290. Right, I remember that thread actually. But I don't see the relevance of multi-valued 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 single-valued algebra then 0/0 is undefined / doesn't exist. if we're using multi-valued 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 single-valued 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 multi-valued 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 multi-valued (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!
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.
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 non-uniform 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 single-valued 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 multi-valued 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 "multi-valued 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 "multi-valued 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 multi-values 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 many-valued quantities", _Math. Annalen_ 104 (1931) 260-290. Right, I remember that thread actually. But I don't see the relevance of multi-valued 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 single-valued algebra then 0/0 is undefined / doesn't exist. if we're using multi-valued 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 single-valued 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 multi-valued 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 multi-valued (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.
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.

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
non-principle 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.
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 non-principle 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 two-fold 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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