No uniform distribution on infinite sets

[tom.stoer]

What exactly prevents us from ruling out a uniform distribution on infinite sets? To be more precise, why are distributions and limits like

\int_{-\infty}^{+\infty}dx\,\lim_{\sigma\to\infty}f_{\mu,\sigma}(x) = 1
\int_{-\infty}^{+\infty}dx\,\lim_{\Lambda\to\infty}\frac{1}{\Lambda} \chi_{[a,a+L]} = 1

not allowed or not reasonable in probability theory? What prevents us from interpreting this as uniform distributions on infinite intervals?

(f is a normal distribution with mean μ and standard deviation sigma; χ is the characteristic function on some interval)

 

[caveman1917]

What exactly do the limits represent?

Assuming that it's $g = \lim_{\sigma \to \infty} f_{\mu, \sigma}$, so you're defining a probability function $g$ as the limit of a sequence of probability functions indexed by $\sigma$, can you show that $g$ is a probability function?

 

[caveman1917]

Or more directly to the point, can you show that for all strictly positive $\epsilon$ the constant function to $\epsilon$ is a strict upper bound to $g$? If so, what must $g$ then be?

 

[tom.stoer]

Assuming that it's $g = \lim_{\sigma \to \infty} f_{\mu, \sigma}$, so you're defining a probability function $g$ as the limit of a sequence of probability functions indexed by $\sigma$, can you show that $g$ is a probability function?

That's the idea. I don't think that one work directly with g, but with any $f_{\mu\,\sigma}$ for every positive $\sigma$.

 

[tom.stoer]

can you show that for all strictly positive $\epsilon$ the constant function to $\epsilon$ is a strict upper bound to $g$? If so, what must $g$ then be?

What is epsilon?

 

[caveman1917]

That's the idea. I don't think that one work directly with g, but with any $f_{\mu\,\sigma}$ for every positive $\sigma$.

The idea doesn't work because the integrals evaluate to zero, not one. You have presumably changed the order of integration and taking that limit, but that doesn't necessarily give the same answer.

 

[caveman1917]

What is epsilon?

Any strictly positive real number. I'll be more precise, define $g = \lim_{\sigma \to \infty} f_{\mu, \sigma}$ for any given $\mu$ and where we consider the limit defined if it evaluates to the same function for all possible sequences of strictly positive real numbers $\sigma$ who go to infinity.

Show that $\forall \epsilon > 0, \forall x \in \mathbb{R}: 0 \leq g(x) < \epsilon$

 

[tom.stoer]

The idea doesn't work because the integrals evaluate to zero, not one. You have presumably changed the order of integration and taking that limit, but that doesn't necessarily give the same answer.

The functions are defined such that the integral is equal to one $\forall\sigma > 0$ and $\forall\Lambda < \infty$. In these cases its OK to change the order of integration and taking the limit.

The idea is now to interpret the limiting case directly as probability distribution. Yes, you are right, in order to do that one must first integrate and then take the limit (which is trivial). The question is whether this possible in some sense; or why it isn't possible.

 

[Stephen Tashi]

What exactly prevents us from ruling out a uniform distribution on infinite sets?

caveman1917 has shown the way to answer that.

An interesting digression is to ask "What mathematical tricks can be used to imitiate the effect of a uniform distribution on an infinite set?"

For example, when people say things like "The probability that a randomly chosen integer is odd is 1/2" they appear to have in mind some process that gives each integer an equal probability of being chosen. (I've read the statement that Erdos and others developed ways to use probability in number theory, but I've never seen an example of how this is done.)

If we have a process that produces some answer A_n as a function of a probability distribution f_n then the limit of A_n as n approaches infinity may exist even though the limit of f_n as n approaches infinity is not a probability distribution.

In the case of "picking an integer at random", let f_n be the uniform distribution on the integers in [-n,n] and let A_n be the probability that the chosen integer is odd. We can say the limit of A_n as n approaches infinity = 1/2 even though the limit of f_n as n approaches infinity is not a probability distribution.

As another example, suppose we take a f_n to be a uniform distribution on the real numbers in the interval [-n,n] and use it as a prior distribution in computing a Bayesian "credible interval". The limit A_n of as n approaches infinity of A_n may be a well defined interval - even though the function given by the limit of f_n as n approaches infinity is not a probability distribution.

 

[caveman1917]

The functions are defined such that the integral is equal to one $\forall\sigma > 0$ and $\forall\Lambda < \infty$. In these cases its OK to change the order of integration and taking the limit.

No it is not. Just because something is true for a sequence of functions doesn't mean it is true for the limiting function, if it exists. Take the sequence $2^{-n}$, all these numbers are strictly positive, does that mean its limit is also strictly positive?

 

[caveman1917]

In the case of "picking an integer at random", let f_n be the uniform distribution on the integers in [-n,n] and let A_n be the probability that the chosen integer is odd. We can say the limit of A_n as n approaches infinity = 1/2 even though the limit of f_n as n approaches infinity is not a probability distribution.

That method relies on the ordering of the integers. If i permute the set of integers such that for every n the set [-n,n] (taken as indices in the permuted set) contains 3 times as many odd as even numbers, then the result is 3/4 and not 1/2

 

[Stephen Tashi]

That method relies on the ordering of the integers.

Well naturally, since [-n, n] is assumed to define an interval.

 

[tom.stoer]

@caveman: thanks, I see that there are elementary mathematical problems

Anyway, my idea goes into the direction of Stephen's "mathematical tricks to imitiate the effect of a uniform distribution on an infinite set." Intuitively that should work.

 

[caveman1917]

Well naturally, since [-n, n] is assumed to define an interval.

Sure, the question is how much the result can be considered meaningful. You are defining a sequence $S_n \subseteq \mathbb{Z}$ such that $\lim_{i \to \infty} S_i = \mathbb{Z}$, and defining a function $p: \mathcal{P}(\mathbb{Z}) \to [0,1]: S \mapsto p(S) = \frac{n_{odd}}{N}$, and define $p(\mathbb{Z}) = \lim_{i \to \infty}p(S_i)$.

The issue is that $p(\mathbb{Z})$ depends on the specific sequence of $S_n$ you choose, the limit depends on how you approach the full set $\mathbb{Z}$. I just don't see the argument as being stronger than for instance saying the limit of the sequence $1,2,1,2,1,2,...$ equals $1$ because you can choose a subsequence such that that is true, even though you could just as well choose a subsequence such that the answer is $2$, or one where the answer doesn't exist.

 

[caveman1917]

@caveman: thanks, I see that there are elementary mathematical problems

Anyway, my idea goes into the direction of Stephen's "mathematical tricks to imitiate the effect of a uniform distribution on an infinite set." Intuitively that should work.

If you are free to choose your result then there's no need to appeal to conveniently chosen sequences of subsets of $\mathbb{Z}$, you can just as well declare 1/2 to be the correct answer and be done with it :)

If this is supposed to work then it should work for all sequences approaching $\mathbb{Z}$.

 

[Stephen Tashi]

Sure, the question is how much the result can be considered meaningful..

I agree. From an applied point of view, the idea that "The probability is 1/2 that a randomly chosen integer is odd" only makes sense if a person is picking randomly chosen integers in a particular manner.

Defining something like a "normal distribution" is discussed using the natural order of the real numbers, so the idea of a defining a family of distributions by referring to natural order isn't so scandalous.

Being curious how probability theory is applied to number theory, I looked at the PDF Probabilistic Number Theory by Dr. Jorn Steuding [ http://www.google.com/url?sa=t&rct=...=mqT4vo2CZ6j-HDdWeKasVQ&bvm=bv.75097201,d.aWw ].

Glancing at Theorem 2.1 and the subsequent definition of a "density" ( which is not a "probability density"), there seems to be an emphasis on "natural density", which is a definition that depends on the natural order of the integers. I wonder if "natural density" is emphasized just to give the reader an easy introduction to the material or whether it plays the key role in applying probability (actually "probability-like" theory) to number theory.

 

[caveman1917]

Defining something like a "normal distribution" is discussed using the natural order of the real numbers, so the idea of a defining a family of distributions by referring to natural order isn't so scandalous.

True, but there the order is implicitly induced by having a normal distribution depend on a distinguished element $\mu$ and a metric (or at least a notion of distance from each element to the distinguished element).

I suppose the same could be said about $\mathbb{N}$. If number theoretical conjectures, absent proof, are researched in terms of checking increasingly large numbers, then it makes sense to pose questions in terms of "the odds of success if I look at the next larger batch of numbers".

I just don't see it with the probability of having an odd randomly picked integer. If I propose to do the same but replacing odd integers with red balls and even integers with blue balls then it seems the same answer should be given, yet there is no natural order.

 

[caveman1917]

I'm also not sure in how much this can be considered imitating the effect of a uniform distribution to infinite sets. For instance one could define a uniform distribution over a finite set as the unique distribution that is invariant under permutations of the set, something that is distinctly not imitated.

 

[WWGD]

Sure, the question is how much the result can be considered meaningful. You are defining a sequence $S_n \subseteq \mathbb{Z}$ such that $\lim_{i \to \infty} S_i = \mathbb{Z}$, and defining a function $p: \mathcal{P}(\mathbb{Z}) \to [0,1]: S \mapsto p(S) = \frac{n_{odd}}{N}$, and define $p(\mathbb{Z}) = \lim_{i \to \infty}p(S_i)$.

The issue is that $p(\mathbb{Z})$ depends on the specific sequence of $S_n$ you choose, the limit depends on how you approach the full set $\mathbb{Z}$. I just don't see the argument as being stronger than for instance saying the limit of the sequence $1,2,1,2,1,2,...$ equals $1$ because you can choose a subsequence such that that is true, even though you could just as well choose a subsequence such that the answer is $2$, or one where the answer doesn't exist.

Isn't there a way to restrict to the subcollection of $P (\mathbb {Z})$ of all sequences of consecutive terms, i.e., to the sequences :
{n, n+1, n+2,...} ?

 

[caveman1917]

Isn't there a way to restrict to the subcollection of $P (\mathbb {Z})$ of all sequences of consecutive terms, i.e., to the sequences :
{n, n+1, n+2,...} ?

Certainly, you can restrict to any subcollection that you want. The point is that the answer depends on which one you choose.

 

[WWGD]

Certainly, you can restrict to any subcollection that you want. The point is that the answer depends on which one you choose.

But if you restrict to this subcollection, then P(odd)=1/2 , independently of the choice of sequence, isn't it? P(odd) in {n, n+1,...} will be 1/2 indepently of the choice, insn't it? Maybe I am missing your point?

 

[caveman1917]

But if you restrict to this subcollection, then P(odd)=1/2

And if you choose another sequence then it can be something else.

, independently of the choice of sequence, isn't it?

How can it be independent of the choice if it depends on the choice?

P(odd) in {n, n+1,...} will be 1/2 indepently of the choice, insn't it?

If you need choosing subsets of the form n,n+1,... then in what way is the answer independent of the choice? Strictly speaking your way of doing it won't work at all, you need to consider finite subsets, otherwise it's just begging the question.

Maybe I am missing your point?

Presumably.

 

[WWGD]

And if you choose another sequence then it can be something else.
How can it be independent of the choice if it depends on the choice?
If you need choosing subsets of the form n,n+1,... then in what way is the answer independent of the choice? Strictly speaking your way of doing it won't work at all, you need to consider finite subsets, otherwise it's just begging the question.
Presumably.

Notice I did not state that the limit depends on the choice of sequence; we may be talking about different things.

I am referring to working only with infinite subsets {n, n+1, n+2,...} . Then P(Odd)=

Lim_i --> oo P({n, n+1, n+2,..,n+i})= Lim _N--> oo n_Odd/ N = Lim_N-->oo N/(2N+1)= 1/2.

If n is odd, then the sequence n_Odd/N is {1/1, 1/2, 2/3, 2/4, 3/5, 3/6,..., n/(2n-1), n/2n,..}

For n even, the sequence is given by { 0/1, 1/2, 2/3, 2/4,...} both converge to 1/2.

 

[caveman1917]

I am referring to working only with infinite subsets {n, n+1, n+2,...} . Then P(Odd)=

Lim_i --> oo P({n, n+1, n+2,..,n+i})

Those are not infinite sets, that is the limit of a sequence of functions of finite subsets of $\mathbb{N}$. The answer depends on the chosen sequence of subsets, as you can easily check. Also, you have to set $n = 0$ in your example otherwise it doesn't limit to $\mathbb{N}$.

If you require another sequence, take the first odd number, then the 3 first even ones and keep switching taking 1 odd and 3 even. The answer is now that the probability of odd is 1/4.