Can nonstandard analysis give a uniform probability distribution over the integers?

Question

There exists no uniform probability distribution over the non-negative integers. This is because we would need to have $p(i)=q$ for all $i$, for some real number $0\le q\le 1$. But normalisation requires $$\sum_{i=1}^\infty q = 1,\tag{*} $$ and there exists no real number $q$ with this property, since the sum diverges for $q\ne 0$.

However, there are various formulations of non-standard analysis that extend the real numbers to include infinitesimals. These include the hyperreal numbers, the surreal numbers, and I believe others.

Does there exist such a system in which equation $(*)$ can be satisfied by an infinitesimal $q$? If so, does it allow uniform probability distributions to be defined on infinite supports, such as the integers, reals etc.?

Answer 1

Nonstandard analysis can give you a uniform distribution on the set $$\{-n, -n+1, -n+2,\ldots, n-2, n-1, n\}, $$ where $n$ is an infinitely large integer.

It cannot give you a uniform distribution on the set of finite integers, because that is an external set.

Internal and external sets are defined in a way that satisfies the "transfer principle," which states that every sentence in a certain language that is true of the real numbers and of functions and sets of real numbers, remains true if all quantifiers over infinite sets $A$ of real number, such as $\forall x\in A$ or $\exists x\in A,$ are replaced by quantifiers over the nonstandard counterparts of those sets, and quantifiers over sets of reals, saying something is true of all such sets or some such sets, are replaced by quantifiers over all internal sets saying for all internal sets or for some internal sets, and quantifiers over all functions likewise are over all internal functions.

Thus for every one-to-one function from $\{1,2,3,\ldots,n\}$ to $\{1,2,3,\ldots,n,n+1,n+2,n+3\},$ there are exactly three elements not in its image, and that is true of standard positive integers $n;$ hence true of nonstandard positive integers $n$ if one specifies that the function $f$ must be internal. And every nonempty set of reals that has an upper bound has a least upper bound, so every nonempty internal set of nonstandard reals that has an upper bound has a least upper bound. It follows that the set of finite numbers is external. Likewise the set of finite integers, being a set of integers with an integer upper bound, is external, since every internal set of integers that has an integer upper bound has a largest member.

Answer 2

As Michael Hardy observed, it is possible to define a uniform probability distribution over sets of the form $\{0,1,\ldots,n−2,n−1,n\}$ or $\{−n,−n+1,−n+2,\ldots,n−2,n−1,n\}$, where $n$ is an infinitely large natural number. These probability distributions provide good nonstandard models for a fair lottery over $\mathbb{N}$ and over $\mathbb{Z}$, respectively.

Similarly, it is possible to define a uniform probability distribution on the set $\left\{-n, -n+\frac{1}{n}, -n+\frac{2}{n}, \ldots,n-\frac{2}{n}, n-\frac{1}{n}, n\right\}$: if $n$ is an infinitely large natural number, this is a nonstandard model of a fair lottery over $\mathbb{Q}$.

If you want to define a similar probability distribution on $\mathbb{R}$, you need to develop some nonstandard measure theory. You might want to take a look at the papers "On the nonstandard representation of measures" by Henson (http://dx.doi.org/10.1090/S0002-9947-1972-0315082-2), "Non-Archimedean Probability" by Benci, Horsten & Wenmackers (https://philpapers.org/rec/BENNP-2), "Elementary numerosity and measures" by Benci, Bottazzi & Di Nasso (http://logicandanalysis.org/index.php/jla/article/view/212/93), and "Some applications of numerosities in measure theory", also by Benci, Bottazzi & Di Nasso (http://www.ems-ph.org/journals/show_abstract.php?issn=1120-6330&vol=26&iss=1&rank=3), where these and similar examples are thoroughly discussed.

Answer 3

No, this cannot be done, as Countable Additivity fails for the hyper-reals. This can be easily shown.

Suppose that e>0 is a hyper-real number such that the countably infinite sum:

e + e + ... = 1

Note that e must therefore be an infinitesimal (since any appreciable e would give us a non-finite sum). Now, if we were assume this Countable sum is well-defined, then we should be able to multiply both sides by 2:

2 * (e + e + ...) = 2

Distributing the 2, we get:

2e + 2e + ... = 2

Since 2e = (e + e), this can be expanded as:

(e + e) + (e + e) + ... = 2

By associativity, we can remove the parentheses:

e + e + e + ... = 2

But from first our assumption, the lefthand side of the equation equal 1, yielding the contradiction:

1 = 2

The reason this fails is that a Countably Infinite sum of positive infinitesimals is undefined in *R. Finite Addivity is defined, as well as Hyper-finite Additivity (the sum of hyper-reals, enumerated from 1 to any unlimited hyper-natural number h). But a Countably Infinite sum falls through the cracks.