ON INDEPENDENT SETS IN PURELY ATOMIC PROBABILITY SPACES WITH GEOMETRIC DISTRIBUTION. 1. Introduction. 1 r r. r k for every set E A, E \ {0},

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1 ON INDEPENDENT SETS IN PURELY ATOMIC PROBABILITY SPACES WITH GEOMETRIC DISTRIBUTION E. J. IONASCU and A. A. STANCU Abstact. We ae inteested in constucting concete independent events in puely atomic pobability spaces with geometic distibution. Among othe facts we pove that thee ae uncountable many sequences of independent events. 1. Intoduction Let us assume a fixed atio is given, (0, 1). In the pape we will wok with the discete pobability space N 0 = {0, 1, 2, 3,...} and the usual geometic pobability on A (all subsets of N 0 ) defined by P (E) := 1 k E\{0} k fo evey set E A, E \ {0}, 0 if E = {0} o E =. We ae inteested in studying the class of independent sets in this pobability space. We ae going to follow [2] and define: A, B subsets of N 0 ae called independent if P (A B) = P (A)P (B). Received Septembe 18, 2008; evised Febuay 22, Mathematics Subject Classification. Pimay 60A10. Key wods and phases. independence; puely atomic pobability; geometic distibution.

2 With this definition fo evey E N 0, N 0 and E ae independent. Sets and E ae also independent. These ae clea tivial examples. Thee o moe subsets of N 0, A 1,..., A n ae called mutually independent o simply independent if fo evey choice of k (n k 2) such sets, say A i1,...,a ik, we have (1) k P k = P ( ) A ij. j=1 A ij j=1 So, fo n (n 2) independent sets one needs to have 2 n n 1 elations as in (1) to be satisfied. An infinite family of subsets is called independent if each finite collection of these subsets is independent. Events ae called tivial if thei pobability is 0 o 1. If n N, then Ω(n) usually denotes the numbe of pimes dividing n counting thei multiplicities (see [8]). In [1] and [6], independent families of events have been studied fo finite pobability spaces with unifom distibution. Eisenbeg and Ghosh [6] show that the numbe of nontivial independent events in such spaces cannot be moe than Ω(m) whee m is the cadinality of the space. This esult should be seen in view of the known fact (see [7, Poblem 50, Section 4.1]) that if A 1, A 2,..., A n ae independent non-tivial events of a sample space X, then X 2 n. One can obseve that in geneal Ω(m) is consideably smalle than log 2 m. It is woth mentioning that accoding to [5] the fist pape dealing with this poblem in unifom finite pobability spaces is [9]. In thei pape, Shiflett and Shultz [9] aise the question of the existence of spaces with no non-tivial independent pais, called dependent pobability spaces. A space containing non-tivial independent events is called independent. Fo a unifom distibuted pobability space X, as a esult of the wok in [6] and [1], X is dependent if X is a pime numbe and independent if X is composite. Fo denumeable sets X one can see the constuction given in [5] o look at [10, Example 1.1]. Fo ou spaces, the Example 1.1 does not apply to and in fact, we will constuct explicitly lots of independent sets.

3 Fo evey n N, one can conside the following space of geometic pobability distibution, denoted hee by G n := ([n], P([n]), P ) whee P (k) = q k with k [n] := {1, 2, 3,..., n} and of couse, q is the positive solution of the equation n q k = 1. k=1 This space is independent fo evey n 4 with n composites. Indeed, if n = st with s, t N s, t 2, one can check that the sets A := {1, 2, 3,..., s}, B := {1, s + 1, 2s , (t 1)s + 1} epesent non-tivial independent events. To match the unifom distibution situation it would be inteesting if G n was a dependent space fo evey pime n. The class of independent sets is impotant in pobability theoy fo vaious easons. Philosophically speaking, the concept of independence is at the heat of the axiomatic system of moden pobability theoy intoduced by A. N. Kolmogoov in Moe ecently, it was shown in [3] that two pobability measues on the same space which have the same independent (pais of) events must be equal if at least one of them is atomless. This was in fact a esult of A. P. Yuachkivsky fom 1989 as the same authos of [3] pointed out in the addendum to thei pape that appeaed in [4]. On the othe hand, Szekely and Moi [10] showed that if the pobability space is atomic then thee may be no independent sets o one may have a sequence of such sets. The following esult that appeaed in [10] is a sufficient condition fo the existence of a sequence of independent events in the pobability space. Theoem 1. If the ange of a puely atomic pobability measue contains an inteval of the fom [0, ɛ) fo some ɛ > 0, then thee ae infinitely many independent sets in the undelying pobability space.

4 Let us obseve that if = 1/2, the pobability space (N 0, A, P 1/2 ) does satisfy the hypothesis of the above theoem with ɛ = 1 because evey numbe in [0, 1] has a epesentation in base 2. On the othe hand, if let us say = 1/3, then the ange of P 1/3 is the usual Canto set which has Lebesgue measue zeo, so Theoem 1 does not apply to (N 0, A, P 1/3 ). Howeve, we will show that thee ae uncountably many pais of sets that ae independent in (N 0, A, P ) fo evey 0 < < 1 (these sets do not depend on ). 2. Independent pais of events fo denumeable spaces The fist esult we would like to include is in fact a chaacteization, unde some estictions of, of all pais of independent events (A, B), in which one of them, say B, is fixed and of a cetain fom. This will show in paticula that thee ae uncountably many such pais. In ode to state this theoem we need to stat with a peliminay ingedient. Lemma 1. Fo m 1, conside the function given by f(x) = (2x 1)(1 + x m ) x m fo all x [0, 1]. The function f is stictly inceasing and it has unique zeo in [0, 1] denoted by t m. Moeove, fo all m we have t m > 1/2, the sequence {t m } is deceasing and lim m t m = 1 2. Having t m defined as above we can state ou fist theoem.

5 (2) Theoem 2. Fo evey natual numbe n 2, we define the events E := {0, n 1} and B :={1, 2,..., n 1, 2n 1, 2n,..., 3n 3, }{{}}{{} 4n 3, 4n 2,..., 5n 5 }{{},... }. Also, fo an abitay nonempty subset T B we set A := E + T with the usual definition of addition of two sets in a semigoup. Then A and B ae independent events in (N 0, A, P ). Convesely, if < t m (whee m = n 1 and t m as in Lemma 1), B is given as in (2) and A foms an independent pai with B, then A must be of the above fom, i.e. A = E + T fo some T B. Poof of Lemma 1. The function f has deivative f (x) = 2(1+x m ) 2m(1 x)x m 1, x (0, 1]. Fo m 2, using the Geometic-Aithmetic Mean inequality we have m (m 1)(1 x) + x + x x }{{} ( ) (m 1)(1 x)x m 1 m 1 m m 1 m = m and so m(1 x)x m 1 ( m 1 m )m 1 1 which implies m(1 x)x m 1 1. This last inequality is tue fo m = 1, too. This implies that f (x) = 2(1 + x m ) 2m(1 x)x m 1 2x m > 0 fo all x (0, 1]. Theefoe the function f is stictly inceasing and because f(1/2) = 1 2 m < 0 and f(1) = 1 > 0, by the Intemediate Values Theoem thee must be a unique solution x = t m,

6 ( of the equation f(x) = 0 in the inteval (1/2, 1). Because f(t m 1 ) = 1 t m 1 1+t m 1 m 1 ) t m 1 m 1 > 0 we see that t m < t m 1 fo all m 2. Since (2t m 1)(1 + t m m) = t m m we can let m go to infinity in this equality and obtain t m 1/2. Using Maple, we got some numeical values fo the sequence t m : t 1 = , t , t , t and, fo instance, t Poof of Theoem 2. Fist let us check that E 1 = E + 1 = {1, n} and B ae independent. Since E 1 B = {1}, P ({1}) = 1 = 1 and P (E 1 ) = 1 ( + n ) = (1 )(1 + ), we have to 1 show that P (B) = 1+. We have ( P (B) = 1 m ) j 2mi = m m = m j=1 i=0 which is what we needed. Now, suppose b B and conside E b = E + b = {b, b + n 1}. We notice that by the definition of B, the intesection B E b is {b}. Hence, P (B E b ) = 1 b = (1 ) c (with c = b 1) and P (B)P (E b ) = Hence, B and E b ae independent fo evey b B. 1 1 ( b 1 + m + b+m) = (1 ) c. Next we would like to obseve that if (F 1, B) and (F 2, B) ae independent pais of events and F 1 F 2 =, then F 1 F 2 and B ae independent events as well.

7 Indeed, by the given assumption we can wite P (B (F 1 F 2 )) = P ((B F 1 ) (B F 2 )) = P (B F 1 ) + P (B F 2 ) = P (B)P (F 1 ) + P (B)P (F 2 ) = P (B)(P (F 1 ) + P (F 2 )) = P (B)P (F 1 F 2 ). In fact, the above statement can be genealized to a sequence of sets F k which ae paiwise disjoint, due to the fact that P is a genuine finite measue and so it is continuous (fom below and above). Then if T B is nonempty, A = E + T = b B E b is a countable union and since E b E b = fo all b, b B (b b ) the above obsevation can be applied to {E b } b T. So, we get that B and A ae independent. Fo the convese, we need the following lemma. Lemma 2. If L N 0 \ B and the smallest element of L is s = (2i 1)m + j, whee i, j N, j m, then Poof of Lemma 2. Indeed, we have P (L) s 1 2im 1 + m. P (L) 1 [( s + s im ) + ( (2i+1)m )] = s 1 2im + 2im P (Ω \ B) = s 1 2im + 2im (1 ) m = s 1 2im 1 + m.

8 So, let us assume that < t m, B is as in (2) and A is independent of B. We let T be the intesection of A and B and put α := P (T )/P (B). Also, we define A := T +{0, }, L = A\A and L = A \ A. We have clealy L, L Ω \ B. By the fist pat of ou theoem P (A ) = α. Because A and B ae independent P (A) must be equal to α as well. Hence P (A) = P (A ) which attacts (3) k = k k = 2 k. k L k L k L L k L Fom (3), it is clea that L = if an only if L = and so if L is empty then A = A which is what we need in ode to conclude ou poof. By way of contadiction, suppose L (o equivalently L ). We can assume without loss of geneality that L contains the smallest numbe of L L, say s, which is witten as in Lemma 2. Thus fom equality (3) we have P (L L ) 2P (L ) and then by Lemma 2 we get s 1 Theefoe fo evey n and 1 j m, = 2im 1 + m 2(1 )s im+1 s 1 + m n j 1 + m 1 + m 1 + m f() = (2 1)(1 + m ) m 0. n j 1 + m. By Lemma 1 we see that t m which is a contadiction. It emains that L and L must be empty and so A = A.

9 In the pevious theoem, since T is an abitay subset of an infinite set we obtain an uncountable family of pais of independent sets. Remak 1. If = 1 φ whee φ stands fo the classical notation of the golden atio (i.e. φ = ), n = 2, B = {1, 3, 5, 7,...} as in (2), and A = {1, 4, 6}, then one can check that P (B) = 1 1+, P (A B) = 1, P (A) = (1 )( ). So the equality P (A B) = P (A)P (B) is equivalent to 1 + = which is the same as = 0. One can easily see that this last equation is satisfied by = 1 φ. Hence A and B ae independent so clealy A is not a tanslation of {0, 1} with a subset of B. Theefoe the convese pat in Theoem 1 cannot be extended to numbes t m such as = 1 φ. In fact, we believe that the constants t m ae shap in the sense that fo all > t m the convese pat is false, but an agument fo showing this is beyond the scope of this pape. Remak 2. Anothe family of independent events which seems to have no connection with those constucted so fa is given by A = {1, 2, 3, 4,..., n 1, n} and B = {n, 2n, 3n,...}, with n N. A natual question aises as a esult of this wealth of independent events: can one chaacteize all pais (A, B) which ae independent egadless the value of the paamete? 3. Thee independent events The next theoem deals with the situation in which two sets the same as in the constuction of Theoem 2 and B given by (2), fom a tiple of independent sets. Let us obseve that if A 1, A 2, and B ae mutually independent, then by Theoem 2 (at least if (0, t m )), A 1 and A 2 must be given by A i = T i + E with T i B, i = 1, 2. Theefoe A 1 A 2 = (T 1 T 2 ) + E.

10 Also, we note that P (A i ) = P (T i )(1+ ), i = 1, 2, and P (A 1 A 2 ) = P (T 1 T 2 )(1+ ). This means that the equality P (A 1 A 2 ) = P (A 1 )P (A 2 ) is equivalent to (4) P (T 1 T 2 ) = P (T 1 )P (T 2 )(1 + ). On the othe hand the condition P (A 1 A 2 B) = P (A 1 )P (A 2 )P (B) educes to P (T 1 T 2 ) = P (T 1 )P (T 2 )(1 + ) 2 P (B), which is the same as (4). So, thee sets A 1, A 2 and B ae independent if and only if (4) is satisfied. Let us notice that the condition (4) may be intepeted as a conditional pobability independence elation: (5) P (T 1 T 2 B) = P (T 1 B)P (T 2 B). At this point the constuction we have in Theoem 2 can be epeated. As a esult, egadless of what is, we obtain an uncountable family of thee events which ae mutually independent in (N 0, A, P ). Theoem 3. Fo a fixed n 3, we conside B as in (2), and pick b {2,..., n 1} such that 2(b 1) divides m = n 1 (m = 2(b 1)k). Fo F := {0, b 1}, we let (6) B 1 := {1, 2,..., b 1, 2b 1, 2b,..., 3b 3, 4b 3, 4b 2,..., 5b 5, }{{}}{{}}{{} b 1 b 1 b 1..., (2k 2)(b 1) + 1,..., (2k 1)(b 1) }, }{{} b 1 B 1 := B 1 (B 1 + 2m) (B 1 + 4m) (B 1 + 6m)...

11 and T be a subset of B 1. Then T 1 := F + T and B 1 ae independent sets elative to the induced pobability measue on B. Moeove, A 1 := T 1 + {0, n 1}, A 2 := B 1 + {0, n 1} and B fom a tiple of mutually independent sets in (N 0, A, P ) fo all. Poof. The second pat of the theoem follows fom the consideations we made befoe the theoem and fom the fist pat. To show the fist pat we need to check (4) fo T 1 and T 2 = B 1. Let us emembe that B = {1, 2,..., n 1, 2n 1, 2n,..., 3n 3, }{{}}{{} 4n 3, 4n 2,..., 5n 5 }{{},... 1 }, and P (B) = 1 + m. We obseve that B 1 {1, 2,..., n 1} and so B 1 B. Let us fist take into consideation the case T = {1}. Since T 1 = {1, b} we get T 1 T 2 = {1}, P (T 1 ) = (1 )(1 + b 1 ), and = So, it emains to calculate P (B 1): P (B 1) = 1 P (B 1 ) = P (B 1)(1 + 2m + 4m + 6m +...) = P (B 1) 1 2m. ( b 1) ( 1 + 2(b 1) + 4(b 1) (k 1)(b 1)) = (1 b 1 ) 1 2k(b 1) 1 2(b 1) = 1 m 1 + b 1 P (B 1 ) = 1 (1 + b 1 )(1 + m ).

12 This shows that (4) is satisfied. In the geneal case, i.e. an abitay subset T of B 1, we poceed as in the poof of Theoem Uncountable sequences of independent events In [10], Szekely and Moi give an example of an infinite sequence of independent sets in (N 0,A,P 1/2 ). Given an infinite sequence of independent sets {A n } n we may assume that P (A k ) 1 2 and so by [10, Poposition 1.1] we must have P (A k ) <. k=1 Let us obseve that Theoem 2 can be applied to a diffeent space now that can be constucted within B given by (2) in tems of classes: N0 = {ˆ0, ˆ1, ˆ2,...} whee ˆ0 =, ˆ1 := {1, 2,..., n 1}, ˆ2 := {2n 1, 2n,..., 3n 3}, ˆ3 := {4n 3, 4n 2,..., 5n 5},..., and the pobability on this space is the conditional pobability as subsets of B. Hence fo k N, one can check that P (ˆk) = 1 2m 2m 2km with m = n 1. It shows that this space is isomophic to (N 0, A, P s ) with s = 2m. One can check the following poposition by induction. Poposition 1. Let n N, n 2. If A 1,...,A n ae independent in N 0, then A 1 +T, A 2 +T,..., A n + T and B ae indepenedent in (N 0, A, P ). This constuction can be then iteated indefinitely giving ise of a sequence B, B 1, B 2,..., which is going to be independent and its constuction is in tems of a sequence (n, n 1, n 2,...) with

13 n k 2. As a esult, we have a uncountable way of constucting sequences of independent sets. This constuction coincides with the one in [10] if n k = 2 fo all k N. 1. Bayshnikov Y. M. and Eisenbeg B., Independent events and independent expeiments, Poc. Ame. Math. Soc., 118(2) (1993), Billingsley P., Pobability and Measue, 3d ed. J. Wiley & Sons, New Yok, Chen Z., Rubin H. and Vitale R. A., Independence and detemination of pobabilities, Poc. Ame. Math. Soc. 125(12) (1997), , Addendum to Independence and detemination of pobabilities, Poc. Ame. Math. Soc. 129(9), (2001), Edwads W., Shiflett R. and Shultz H., Dependent pobability spaces, Collega Mathematics Jounal 39(3) (2008), Eisenbeg B. and Ghosh B. K., Independent events in a discete unifom pobability space, Ame. Statist. 41 (1987), Ginstead C. M. and Snell J. L., Intoduction to Pobability, AMS, 1997, Niven I., Zuckeman H. S. and Montgomey H. L., An intoduction to the theoy of numbes, 5th ed. J. Wiley & Sons, New Yok, Shiflett R. C. and Shultz H. S., An appoach to independent sets, Mathematical Spectum 12 (1997/80), Skekely G. J. and Moi T. F., Independence and atoms, Poc. Ame. Math. Soc. 130(1) (2001), E. J. Ionascu, Depatment of Mathematics, Columbus State Univesity, 4225 Univesity Avenue, Columbus, GA 31907, USA, ionascu eugen@colstate.edu A. A. Stancu, Depatment of Mathematics, Columbus State Univesity, 4225 Univesity Avenue, Columbus, GA 31907, USA, stancu alin1@colstate.edu

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