On Independence and Determination of Probability Measures

J Theor Probab (2015) 28:968 975 DOI 10.1007/s10959-013-0513-0 On Independence and Determination of Probability Measures Iddo Ben-Ari Received: 18 March 2013 / Revised: 27 June 2013 / Published online: 13 September 2013 Springer Science+Business Media New York 2013 Abstract We show that two probability measures defined on the same measure space having the same pairs of independent events are either purely atomic or equal. In the former case, either measures are trivial (taking values in {0, 1}) and singular, or equivalent. We also characterize nonatomic measures having the same events independent of a fixed (nontrivial) event and provide sufficient condition on pairs on independent intervals to determine a probability measure on the real line. Keywords Independence Nonatomic Mathematics Subject Classification (2010) 60A99 1 Introduction and Previous Results The paper has two goals. The first is to publicize the fact that under appropriate conditions, independence of events uniquely determines a probability measure. We think this is theoretically fundamental because most uniqueness results for probability measure are measure-theoretic (e.g., monotone class) or analytic (e.g., characteristic functions, Stein s equation), and the independence approach discussed here is to us somewhat more elementary and genuinely probabilistic. One still has to show that the independence approach is useful in applications, and perhaps the readers of this work will... The second goal is to expand and complement the existing results. This work was partially supported by NSA Grant H98230-12-1-0225. I. Ben-Ari (B) Department of Mathematics, University of Connecticut, 196 Auditorium Rd, Storrs, CT 06269-3009, USA e-mail: iddo.ben-ari@uconn.edu

J Theor Probab (2015) 28:968 975 969 Throughout the discussion, we will fix a measurable space (Ω, F). In the sequel, whenever we refer to a subset of Ω, we tacitly assume it is an event, that is, an element of F, and all measures are probability measures on (Ω, F). LetP be a probability measure. We say that events A and B are P-independent and write A P B if P(A B) = P(A)P(B). We identify P as a relation on F, that is, P ={(A, B) F F : P(A B) = P(A)P(B)}. An event A is P-null if P(A) = 0, and trivial if A or A c is P-null (equivalently, A is trivial if A P A c ). The measure P is trivial if every event is P-trivial. An event is a P-atom if P(A) >0, and for every B A, B is P-null or A B is P-null (equivalently, P(A) >0, and for every B A, B P (A B)). An event is P-nonatomic if it is not P-null and does not contain P-atoms. The measure P is purely atomic if every event which is not P-null contains a P-atom (equivalently, P is purely atomic if none of the events are P-nonatomic). The measure P is nonatomic if it has no P-atoms. We may omit reference to P when there is no risk of ambiguity. Finally, the probability measures P and Q are singular if there exists L with P(L) = 1 and Q(L) = 0 and are equivalent if their null events coincide. We begin with a review of the results that motivated our work. Theorem A ([3,9]) Suppose that P and Q are two probability measures at least one of which is nonatomic. If they have identical independent events, then they coincide. The theorem is a corollary to the main result of [9] (in Russian), predating [3]. Yet, the derivation in the latter paper was done independently. We comment that [3] presents two proofs. The second relies on Lyapunov s convexity theorem for vector-valued measures [1,4] and is short and elegant. That proof also yields the following corollary. Corollary B ([3]) Suppose that P is nonatomic, and for some α (0, 1), Then, P = Q. {B : P(B) = α} {B : Q(B) = α}. In [3], it was chosen to state the result for α = 2 1 with equality between the sets, although the proof gives the more general statement in the corollary. An alternative proof is given in the Appendix. In [7], the authors relaxed the assumption in Theorem A that one of the measures is nonatomic. We quote the result below (with minor changes), but we need to fix some additional notation. Recall the following key elementary fact. For every P,Ω could be uniquely decomposed, up to P-null events, into a countable (possibly empty or finite) union of disjoint atoms, and an event containing no atoms. We refer to the latter event as the atomless part of P, and to its complement as the atomic part of P. We also say that P and Q have the same conditionally independent events if for any L with P(L)Q(L) >0, P( L) = Q( L). This of course implies P = Q as seen by taking L = Ω. Theorem C ([7]) Suppose that P and Q are not singular. If P and Q have the same collection of conditionally independent events, then the atomless part of P and Q coincide, and the atomic parts of P and Q vanish outside of the union of a countably

970 J Theor Probab (2015) 28:968 975 many pairwise disjoint events, which are atoms of P as well as of Q. The condition on P and Q to be not singular can be omitted when the atomless part of P (or Q) does not vanish. We close this section mentioning several related publications. The problem of triviality or nontriviality of the independence relation was addressed by several authors. In [6,8], the authors provide sufficient conditions for having infinitely many pairs of nontrivial independent events. In [5], the authors provide a sufficient conditions for nonexistence of pairs of nontrivial independent events. We also mention [2] which discusses independence of events under uniform distribution on finite sample spaces. 2 Main Results Before stating our results, we recall the following well-known result which is our main tool. Lemma 1 Let P be nonatomic. Then, the range of P is [0, 1]. Note that this is a corollary to Lyapunov s convexity theorem, but can be derived independently. In order to keep this note self-contained, we include a proof in the Appendix. Observe that if P is a probability measure and C is not P-null and P- nonatomic, then P( C) is a nonatomic probability measure and it follows from Lemma 1 applied to this measure that the restriction of P to subsets of C has range [0, P(C)]. Our main results are Theorems 1 and 2 below. Theorem 1 characterizes the relation between two probability measures having identical pairs of independent events. This extends and combines Theorem A and Theorem C. Theorem 2 characterizes the relation between two probability measures P and Q with a weaker assumption on the pairs of independent events: There exists some P nontrivial event such that the set of events that are P-independent of it is contained in the set of events that are Q-independent of it. 2.1 On Probabilities with Identical Independent Events Theorem 1 Suppose that P = Q. Then, either 1. P and Q are singular and trivial; or 2. P is not purely atomic, and then P = Q; or 3. P is purely atomic, and then Q is purely atomic and every P-atom is a Q-atom. Independence determines whether P is trivial or not. Indeed, P is trivial if and only if every event is P-independent of itself. When P is trivial, then only 1. or 3. can hold, but since all events are trivial, dependence does not determine whether it is 1. or 3. that holds (of course, under assumption of triviality, 3. is equality of P and Q). When P is not trivial, A is a P-atom if and only if A is not P-independent of itself, and every B A is P-independent of A B. This determines whether 2. or 3. holds. The proof of the theorem will be obtained through a sequence of lemmas. We begin with a simple dichotomy.

J Theor Probab (2015) 28:968 975 971 Lemma 2 Suppose that P = Q. Then, either 1. P and Q are singular and trivial; or 2. Q and P are equivalent. Note that the first case complements Theorem C and gives an answer to what can happen when P and Q are singular: They are also trivial measures, having an atom of full measure. In particular, in this case, P = Q = F F. Proof Observe that under any probability measure, an event A is trivial if and only if it is independent of its complement. Since P = Q, we conclude that the set of P-trivial events is equal to the set of Q-trivial events. Thus, either Q and P are equivalent, or, for some A, P(A) = 1, Q(A) = 0. In the latter case, let B A. Since B is Q-trivial, it is Q-independent of its complement, and as a result, B is P-independent of its complement. Thus, B is P-trivial. Thus, A is a P-atom. Lemma 3 Suppose that C is P-nonatomic, and D C is not P-null. Then, there exist A, B D with A P B and P(A)P(B) >0. Proof Indeed, let A D satisfy P(A) = P(D)/2, and choose B 1 A with P(B) = P(D) 2 /4, and B 2 D A 1 with P(B 2 ) = P(D)/2 P(D) 2 /4 < P(D A 1 ) = P(D)/2. Letting B = B 1 B 2 completes the claim. Lemma 4 Suppose that C is P-nonatomic and that P and Q coincide on subsets of C. Then P(C) = Q(C). Proof Since C is nonatomic, C is not P-independent of itself. Therefore, C is not Q- independent of itself. Thus, Q(C) <1. Now, if A is P-null, then A P A; therefore, Q(A) {0, 1}, but since Q(A) Q(C) <1, Q(A) = 0. Conversely, if A C is Q- null, then it is P-independent of itself, and since P(C) <1, P(C) = 0. Summarizing, Q(C) (0, 1), and the measures Q( C) and P( C) are equivalent. Let f = dq dp. To complete the proof, we need to show that on C, { f = 1}, P-a.s. It is enough to show that on C, { f 1} P-a.s, because the same argument, with roles of P and Q interchanged, shows that on C, { 1 f 1} Q-a.s., hence P-a.s. To prove the claim, we argue by contradiction and assume that there exists t 0 > 1 satisfying P( f t 0 )>0. Let ɛ := 1+t 0 2, and for n Z +,letc n := {t 0 ɛ n f < t 0 ɛ n+1 } C. Then, by assumption, P(C n0 )>0forsomen 0 Z +. By Lemma 3, wehavea, B C n0 which are not P-null and satisfy A P B. This gives the following: Q(A B) = f dp t 0 ɛ n0+1 P(A B) = t 0 ɛ n0+1 P(A)P(B) A B = t 0 ɛ n 0+1 A 1 f dq < ɛ t 0 Q(A)Q(B) <Q(A)Q(B). B 1 f dq t 0ɛ n0+1 (t 0 ɛ n 0) 2 Q(A)Q(B)

972 J Theor Probab (2015) 28:968 975 Lemma 5 Suppose that P is not purely atomic. If P = Q, then P = Q. Proof Lemma 4 guarantees that P and Q coincide on all P-nonatomic events. Since every event is a countable union (possibly empty) of P-atoms or a countable union of P-atoms and a P-nonatomic event, it is sufficient to prove that Q(L) = P(L) for all P-atoms L. Fix such L and let R L c be P-nonatomic. Choose R 0 R with P(R 0 ) (0, P(R)). LetA = L R 0 and observe that P(A) (0, 1) since 0 < P(R 0 )<P(R) 1. Since both R 0 and R R 0 are P-nonatomic, we can find D 0 R 0 satisfying 1 P(A) P(A) P(D 0 ) = P(R R 0 ).LetB = D 0 (R R 0 ). We claim that A P B. Indeed, A B = D 0, while P(A)P(B) = P(A)(P(D 0 ) + 1 P(A) P(D 0 )) = P(D 0 ). P(A) Therefore, A Q B. Since A = L R 0, the independence statements with respect to P and Q could be rewritten as (P(L) + P(R 0 ))P(B) = P(D 0 ) and (Q(L) + Q(R 0 ))Q(B) = Q(D 0 ). However, since R 0, B, D 0 R, and R is P-nonatomic, it follows from Lemma 4 that Q(R 0 ) = P(R 0 ), Q(B) = P(B) and Q(D 0 ) = P(D 0 ), and all are strictly positive. Thus, Q(L) = Q(D 0) Q(B) Q(R 0) = P(D 0) P(B) P(R 0) = P(L). Proof of Theorem 1 From Lemma 2, we conclude that either 1. holds or that P and Q are equivalent. In the latter case, 2. follows from Lemma 5. To prove 3., assume that 1. and 2. do not hold. Then, P and Q are equivalent, and P is purely atomic. Let A be a P-atom. Then, for B A, B P (A B). Consequently, B Q (A B). It follows that Q(B) = 0orQ(A B) = 0. This shows that A is Q-null or a Q-atom. Since P and Q share the same null events, and since A is a P-atom, P(A) >0, which implies Q(A) >0; thus, A is a Q-atom. Finally, since Ω is a countable union of disjoint P-atoms, it is also a countable union of disjoint Q-atoms, proving that Q is purely atomic. 2.2 On Independence with Respect to a Fixed Event Theorem 2 Let P be nonatomic, and let L be an event. Then, {A : A P L} {A : A Q L} F, (1) if and only if Q(L) (0, 1) and Q = Q(L)P( L) + Q(L c )P( L c ). The equivalent condition to (1) could be expressed in terms of the Radon-Nykodim derivative as Q P and dq dp = Q(L) P(L) 1 L + Q(Lc ) P(L c ) 1 L c. Corollary 1 Let P be nonatomic and suppose that L satisfies (1). Then, P = Qif P(L) = Q(L). In particular, P(L) = Q(L) if there exist A, B L which are not P-null and satisfy A P BaswellasA Q B

J Theor Probab (2015) 28:968 975 973 Proof of Corollary 1 If Q(L) = P(L), then equality between P and Q follows immediately from the conclusion of Theorem 2. Assume now that there exist A, B L which are not P-null and satisfy A P B and A Q B.Letρ = Q(L) P(L). Then, ρ>0 because Q(L) > 0. Observe that Q(A B) = ρ P(A B), and Q(A)Q(B) = ρ 2 P(A)P(B). Since P(A B) = P(A)P(B) >0 and A Q B,wehaveρ = ρ 2. Since ρ>0, it follows that ρ = 1. Proof of Theorem 2 Observe that a A P L if and only if P(A L) = P(A L c ). Assume that Q = ρ P( L) + (1 ρ)p( L c ) and let A satisfy A P L. Clearly, Q(L) = ρ. Furthermore, Q(A L) = ρ P(A L) and Q(A L c ) = (1 ρ)p(a L c ) = (1 ρ)p(a L). Adding both equalities, we obtain Q(A) = P(A L). Thus, Q(A L) = ρ P(A L) = Q(L)Q(A), that is, A Q L. Observe that L is not Q-trivial because the set of events which are Q- independent of L is a proper subset of F. Next, we show that if B 1, B 2 L c satisfy P(B 1 ) = P(B 2 ), then Q(B 1 ) = Q(B 2 ). Since P is nonatomic, there exists L 0 L with P(L 0 ) = P(B 1) P(L c ) P(L) = P(B 2) P(L c ) P(L). LetA 1 = B 1 L 0 and let A 2 = B 2 L 0. Observe that P(A 1 L) = P(L 0) P(L) = P(B 1) P(L c ) = P(A 1 L c ), with a similar statement for A 2. Therefore, A 1 P L and A 2 P L, and from our assumption, we conclude that A 1 Q L and A 2 Q L. But rewriting these statements, we obtain (Q(B 1 ) + Q(L 0 ))Q(L) = Q(L 0 ), and (Q(B 2 ) + Q(L 0 ))Q(L) = Q(L 0 ). Since Q(L) >0, it follows that Q(B 1 ) = Q(B 2 ). This in particular implies that if B 1 L c is P-null, then since P(B 1 ) = P( ) = 0, we also have Q(B 1 ) = 0. Thus, Q( L c ) is absolutely continuous with respect to P( L c ).Let f denote the Radon-Nykodim derivative. We want to show that f is constant P-a.s. Otherwise, there exist disjoint B 1, B 2 L c, with P(B 1 )P(B 2 )>0 and sup x B1 f < inf x B2 f (x). But this implies Q(B 1 ) = B 1 fdp < B 2 fdp = Q(B 2 ), a contradiction. Therefore, f is some constant β, P-a.s. This implies Q(L c ) = β P(L c ), hence β = Q(L c )/P(L c ). 3 Additional Results In this section, we present two additional simple results on characterizations of measures through independence of events. The first provides a mechanism for finding events with rational probabilities from independence. The second discusses equality of probability measures on the real line through independence of certain pairs of intervals. Proposition 1 Let A be an event and let k N. Suppose that there exists P-nontrivial B and a partition of B, B 0,...,B k 1 with B 0 = B A, such that for every j = 0,...,k 1, ((A B) B j ) P B. Then, P(A) = 1 k. Applying the proposition for k = 2, we obtain that P(A) = 2 1 provided A is not P-trivial, and for some P-nontrivial B, wehavea P B and (A B) P B, where A B is the symmetric difference ( A B) (B A). Proof We have P(B j ) = (P(A B) + P(B j ))P(B). That, is P(B j )(1 P(B)) = P(A B)P(B). Since P(B) (0, 1), it follows that P(B j ) is independent of j, but

974 J Theor Probab (2015) 28:968 975 then P(B) = kp(b 0 )>0and the independence between A = (A B) B 0 and B implies P(B 0 ) = P(A B) = P(A)P(B 0 )k = P(A)P(B). In analogy with monotone class results, one may ask whether independence of events in a small (and usually easily identifiable) subset of F determines P. The following result provides a positive answer for distributions on R, under a mild regularity condition. Proposition 2 Let X be a random variable with distribution function F. If, for some differentiability point b of F, F (b) >0, then F is uniquely determined by the pairs of intervals of the form (a, c) and (b, d) where a < b < c < d, which are independent under the Borel measure induced by X. The proof is essentially a restatement of the fact that the probability of an interval with one end point equal to b is proportional to its length as the length tends to 0, and so conditional probabilities involving intervals around b reduce to ratios of lengths of the intervals. Also, as the proof shows, the condition in the theorem could be relaxed to the following: There exists a dense A with the property that for every d A and d > b, there exist sequences a n < b < c n satisfying lim n (c n a n ) = 0 and (a n, c n ) (b, d), and similarly, that for every a A with a < b, there exist sequences b < c n < d n with lim n d n b = 0 such that (a, c n ) (b, d n ). Proof of Proposition 2 Let P denote the Borel probability measure induced by X. That is, P(A) = P(X A).Fixd > b. Observe that for a < b < c, (a, c) (b, d) if and only if P((b, d)) = P((b,c)) P((a,c)). However, P((b, c)) = F(c) F(b) = F (b)(c b)+ o(c b), and similarly, P((a, c)) = P((b, c))+ P((a, b)) = (F(c) F(b))+(F(b) F(a)) = F (b)(c b)+o(b c) (F (b)(a b)+o(a b)) = F (b)(c a)+o(c a). As a result for any ɛ>0, the ratios { P((b,c)) P((a,c)) : a < b < c, c a <ɛ} are dense in [0, 1]. In particular, there exist sequences d n d, a n b and c n b such that (a n, c n ) (b, d n ). In particular, P((b, d n )) = F (b)(c n b)(1 + o(1)) F (b)(c n a n )(1 + o(1)) = c n b c n a n (1 + o(1)). By letting d n d, P((b, d n )) P((b, d]) = F(d) F(b). Therefore, these differences are uniquely determined by the pairs (a n, c n ) (b, d n ).Byletting d, F(b) is determined as well. Therefore, F is determined for all x b. The remainder of the argument is similar. First, observe that for any ɛ>0, the ratios : c > d > b, d b <ɛ} are dense in [0, 1], because the ratio is asymptotically equivalent to d b c b as d, c b. As a result, given any a < b, then there exist sequences a n a, d n > c n with d n b, such that { P((b,c) P((b,d)) P((b, c n )) P((b, d n )) P([b, c n)) = P((a n, b)), that is, (a, c n ) (b, d n ). Since all terms on the right-hand side are already determined, and the right-hand side converges to F(b ) F(a) = F(b) F(a), it follows that F(a) is determined as well.

J Theor Probab (2015) 28:968 975 975 Acknowledgments I would like to thank R. A. Vitale for introducing me to the topic as well as for making useful comments and suggestions. I would also like to thank and A. P. Yurachkovskiĭ and A. L. Rukhin for sending me copies of [9] and[7], and an anonymous referee for comments and suggestions that greatly improved the manuscript. Appendix Proof of Lemma 1 First, we observe that any event A with P(A) >0 has an event with arbitrarily small but strictly positive measure. Indeed, let such A be given. Then, since P is nonatomic, there exists A 1 A with P(A 1 ), P(A A 1 )>0. Without loss of generality, P(A 1 ) P(A)/2. Continue inductively. Given A n A with 0 < P(A n ) P(A)/2 n, there exists A n+1 A n, with 0 < P(A n+1 )<P(A n )/2. We turn to the main claim. Fix α (0, 1), and let A 0 be such that P(A 0 ) (0,α). Continue inductively, assuming P(A n ) (0,α).LetI n = {B A c n : P(B) α P(A n )}. Choose B n+1 I n that satisfies P(B n ) (1 n 1 ) sup B I n P(B), and let A n+1 = A n B n+1. Note that the events (B n : n N) are disjoint. Therefore, P(B n ) 0. Consequently, sup B In P(B) 0. Let A = n N A n. We continue the proof arguing by contradiction, assuming that P(A )<α. Then, by the first paragraph, there exists B A c satisfying 0 < P(B )<α P(A ). Since A c Ac n for all n N, it follows that B I n for all n. But this contradicts the fact that sup B In P(B) 0. Proof of Corollary B Let ɛ = min(α, 1 α). It is enough to show that if P(A) = P(B) ɛ, then Q(A) = Q(B). Without loss of generality, A B = ; otherwise, we will consider the pair A (A B) and B (A B) instead of A and B. Since A B c and P(B c ) α, by Lemma 1, there exists an event L B c, with P(L) = α and A L. But since P((L A) B) = P(L) P(A) + P(B) = P(L) = α,itfollows that Q((L A) B) = α. However, Q((L A) B) = Q(L) Q(A) + Q(B) = α Q(A) + Q(B). Thus, Q(A) = Q(B). References 1. Artstein, Z.: Yet another proof of the Lyapunov convexity theorem. Proc. Am. Math. Soc. 108(1),89 91 (1990) 2. Baryshnikov, Y.M., Eisenberg, B.: Independent events and independent experiments. Proc. Am. Math. Soc. 118(2), 615 617 (1993) 3. Chen, Z., Rubin, H., Vitale, R.A.: Independence and determination of probabilities. Proc. Am. Math. Soc. 125(12), 3721 3723 (1997) 4. Diestel, J., Jerry U., Jr., J.: Progress in vector measures 1977 1983. Measure theory and its applications (Sherbrooke, Que., 1982). Lecture Notes in Mathematics, vol. 1033. Springer, Berlin, pp. 144 192 (1983) 5. Edwards, W.F., Shiflett, R.C., Shultz, H.S.: Dependent probability spaces. College Math. J. 39(3), 221 226 (2008) 6. Ionascu, E.J., Stancu, A.A.: On independent sets in purely atomic probability spaces with geometric distribution. Acta Math. Univ. Comenian. (N.S.) 79(1), 31 38 (2010) 7. Plachky, D., Rukhin, A.L.: Characterization of probability distributions by independence structures. Math. Methods Stat. 9(1), 90 105 (2000) 8. Székely, G.J., Móri, T.F.: Independence and atoms. Proc. Am. Math. Soc. 130(1), 213 216 (2002) (electronic) 9. Yurachkovskiĭ, A.P.: An axiomatic independence relation as an equivalent of probability measure. Akad. Nauk Ukrain. SSR Inst. Mat. Preprint, no. 6, 35 (1990)