TRANSLATION INVARIANCE AND FINITE ADDITIVITY IN A PROBABILITY MEASURE ON THE NATURAL NUMBERS

Inspired by the “two envelopes exchange paradox,” a finitely additive probability measure m on the natural numbers is introduced. The measure is uniform in the sense that m({i})=m({j}) for all i,j ∈N. The measure is shown to be translation invariant and has such desirable properties as m({i∈N | i≡ 0(mod2)})= 1/2. For any r ∈ [0,1], a set A is constructed such thatm(A)= r ; however,m is not defined on the power set of N. Finally, a resolution to the two envelopes exchange paradox is presented in terms of m.


Introduction.
In recent work on the so-called "two envelopes exchange paradox" (cf.Broome [1] and Nalebuff [4]), it has been proposed that the commonly used assumption of countable additivity for probability distributions be dropped [5].The two envelopes problem is as follows: a random (positive) amount of money is put in an envelope O.A coin is flipped and if the coin comes up heads, twice the amount of money in envelope O is put in a second envelope (call it T ) and if the coin comes up tails, half the amount of money in envelope O is put in envelope T .
The paradox arises by reasoning that if we choose one envelope (no matter which one) then there is a 50% chance that the other envelope contains one-half the amount we hold, and there is a 50% chance that the other envelope contains twice the amount we hold.That is, the other envelope has an expected value of 1.25 times the amount in the envelope we hold.This expected value is greater, regardless of whether we hold envelope O or T .Therefore, if we hold envelope O it appears to be to our advantage to swap envelope O for envelope T (we might even be willing to pay a certain amount of money to swap).Also, if we should hold envelope T we also have a desire to swap since we can argue that the expected value in O is 1.25 times the amount in T .Of course, the "paradox" is resolved if a probability distribution is given by which the amount of money to be put in envelope O is determined.The expected amount in O is then the expected value of this distribution and the expected amount in T is 1.25 times the expected amount in O [3].
Rawling [5] has suggested addressing the paradox by exploring what happens when the amount in envelope O is based on a natural number n (he chose to put $2 n in envelope O) which is chosen according to a uniform probability distribution on the natural numbers N.With such a distribution, countable additivity of the probability measure must be abandoned.Motivated by this approach, we explore the implications of postulating a uniform probability distribution on N, which satisfies the properties of finite additivity and translation invariance.For each real number r such that 0 ≤ r ≤ 1, we construct a subset of N, with measure r .Finally, we propose a method for calculating expected values for the two envelopes problem which resolves the paradox.

The definition and results
. We center our discussion around the following definition. (2.1) For S ⊂ N define the (probability) measure of S as provided this limit exists.
The probability measure of S, P (S), is occasionally called the asymptotic density of the set S [2].This definition is also briefly discussed in [6].Notice that if Now, we show that P satisfies several desired conditions for the probability measure of the two envelopes problem.Theorem 2.2.Probability measure P satisfies: P (∅) = 0, P ({n}) = P ({m}) = 0 for all m, n ∈ N, P (N) = 1, and Proof.This result follows trivially from the definition of P .Theorem 2.2 includes the idea that P should be determined by a uniform distribution.Although finite sets have measure zero, the converse of this result does not hold.Consider, for example, the set Then since c A (n) ≤ log 10 (n), we have P (A) = 0.
Theorem 2.3.The probability measure P is finitely additive.That is, if P (S 1 ), P (S 2 ),...,P(S k ) are defined, S i ∩ S j = ∅ for i ≠ j, and (2.4) Therefore, (2.5) Notice that, in general, P is not countably additive.This follows from the facts that N = ∞ n=1 {n}, P ({n}) = 0 for each n ∈ N, and P (N) = 1.We define for S ⊂ N and x ∈ N, (2.6) The set S + x is commonly called a translation for S. We have the following theorem.
Theorem 2.4.The probability measure P is translation invariant.That is, P (S + x) = P (S) for all S ⊂ N and for x ∈ N.

Sets of given measures.
In this section, for any r ∈ [0, 1], we construct a set A ⊂ N such that P (A) = r .First, we define for i ≤ j in N.
Theorem 3.1.Let p, q ∈ N with p ≤ q.Then there exists A ⊂ N with P (A) = p/q.
Proof.Notice that for all p, q ∈ N, p ≤ q we have Therefore, P (A p/q ) = lim n→∞ (c A p/q (n)/n) = 1/q by the Sandwich theorem.By Theorem 2.3,

.3)
We now use the decimal expansion of irrational numbers to construct sets of irrational measure.Theorem 3.2.For any irrational r ∈ [0, 1], there exists a set A with P (A) = r .
Proof.Let the decimal expansion of r be 0. where Similarly, there exists N ∈ N with With We would now like to return to the two envelopes problem and draw some conclusions from the properties we have developed.We associate a value of n with set {n}.If we calculate the expected value in envelope O using infinite sums, then we get It is not surprising that we get this absurdity when taking an infinite sum, since we have calculated probabilities without having the property of countable additivity.An alternative approach is to calculate a "cumulative expected value" and then take a limit.That is, we can argue that the expected amount in envelope O is In this way, we calculate a limit of finite sums and never directly deal with an infinite sum.Notice that this gives an infinite expected value for the contents of both envelopes O and T , and the paradox is resolved.We therefore propose that, in the setting of the two envelopes problem, probabilities and expected values be computed as above.

4 .
Discussion.Unfortunately, it is fairly easy to construct subsets of N which are not measurable under our definition.We simply alternate the inclusion and exclusion of larger and larger numbers of natural numbers.For example, defineA i = {10 2i−2 +1, 10 2i−2 + 2,...,10 2i−1 } for i ∈ N, and define A = ∞ i=1 A i .Then for n = 10 k where k is odd, c A (n) = 9 × (k+1)/2 i=110 2i−2 and for n = 10 k where k is even, c A (n) = 9 × k/2 i=1 10 2i−2 .If we restrict n to values in the set {n | n = 10 k where k is odd}, then lim n→∞ (c A (n)/n) = 10/11.If we restrict n to values in the set {n | n = 10 k where k is even}, then lim n→∞ (c A (n)/n) = 1/11.Therefore P (A) is not defined.