Nonextensive Statistical Mechanics

3.3 Correlations, Occupancy of Phase-Space, and Extensivity of S q 71subsystem. For example, p (A 1)+A 2 +...+A Ni 2 i 3 ...i N≡ ∑ W 1i 1 =1 p A 1+A 2 +...+A Ni 1 i 2 ...i N, p (A 1)+(A 2 )+...+A Ni 3 i 4 ...i N≡ ∑ W 1∑ W2i 1 =1 i 2 =1 p A 1+A 2 +...+A Ni 1 i 2 ...i N, and so on. The binary case that we introduced abovecorresponds of course to the particular case W r = 2(∀r). Let us go now back to it.Let us assume the simple case in which all N binary subsystems are equal. Tables3.2 and 3.3 then become Tables 3.5 and 3.6 respectively.The general case of N equal subsystems has the joint probabilities {r Nn } (n =0, 1, 2,...,N), which satisfyN∑n=0N!(N − n)! n! r Nn = 1 (N = 1, 2, 3,...) . (3.123)N!The probability r Nn equals all the(N−n)! n! joint probabilities {p A 1+A 2 +...+A Ni 1 i 2 ...i N} thatare associated with (N − n) subsystems in state 1 and n subsystems in state 2, inwhatever order. 12The instance of the N subsystems being equal admits a representation which ismuch simpler than the hypercubic one used up to now. They admit a “triangular”representation: see Table 3.4.A particular case of this probabilistic triangle is indicated in Table 3.7. The set ofall the left members of the pairs constitute the so-called Pascal triangle, where eachelement equals the sum of its “North-West” and “North-East” neighbors. The setof all the right members of the pairs constitute the so-called Leibnitz triangle [844],where each element equals the sum of its “South-West” and “South-East” neighbors.In other words, Leibnitz triangle satisfies the ruler N,n + r N,n+1 = r N−1,n (∀n, ∀N) . (3.124)Table 3.4 Merging of the Pascal triangle (left member of each pair) with the probabilities {r Nn }(right member of each pair) associated with N equal subsystems(N = 0) (1, 1)(N = 1) (1, r 10 )(1, r 11 )(N = 2) (1, r 20 )(2, r 21 )(1, r 22 )(N = 3) (1, r 30 )(3, r 31 )(3, r 32 )(1, r 33 )(N = 4) (1, r 40 )(4, r 41 )(6, r 42 )(4, r 43 )(1, r 44 )12 If we consider the outcomes 1 and 2 in a specific order, we can think of them as being a timeseries. In such a case, for say N = 3, the probabilities p 112 , p 121 ,andp 211 , might not coincidedue to memory effects. If they did, that would be a case in which we have no memory of theirorder of appearance. Within this interpretation, the case we are addressing above would correspondto having memory of how many 1s and 2s we have, but not having memory of their order. Ifwe have no memory at all, that would correspond to equal probabilities, i.e., r NnNormalization of these probabilities is in this case preserved through ∑ Nn=0= 1/2 N , ∀n.N!(N−n)! n! = 2N .