from (A.16) that almost surely, for some constant c ′′ > 0. Consider now a random permutation ( πy$_{1}$ ) , . . . , ( πy$_{N}$ ) of the y -values y$_{1}$, . . . , y$_{N}$ . Let S π m × m be the test statistic that is computed from the data ( x$_{1}$, πy$_{1}$ ) , . . . , ( x$_{N}$, πy$_{N}$ ). By Markov’s inequality, where ⃗ x = ( x$_{1}$, . . . , x$_{N}$ ) and ⃗ y = ( y$_{1}$, . . . , y$_{N}$ ). The approximation in (A.18) becomes more accurate the larger N is, since each of the contingency tables is approximately χ 2 with ( m − 1) 2 degrees of freedom. The right hand side of equation (A.18) goes to zero as N → ∞ , as long as lim$_{N}$$_{→∞}$ m √ N = 0. Thus, We now have all the necessary results to complete the proof. Specifically, where the inequality follows from (A.19), since S tab 1 − α is below c ′′ ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$N for N large enough, and the equality follows from (A.17), thus proving item 1 of Theorem 1 for the ADP summation statistic. 53 lim N $_{→∞}$P r ( S$_{m}$$_{×}$$_{m}$ ≤ S tab 1 − $_{α}$) ≤ P r$^{(}$S π m × m ≤ c ′′ $^{(}$N − 1 m − 1$^{)(}$ N − 1 m − 1$^{)}$ N$^{)}$ = 0 , (A.20) lim N →∞ ,m/ √ N → 0 P r$^{(}$S π m × m ≥ c ′′ $^{(}$N − 1 m − 1$^{)(}$ N − 1 m − 1$^{)}$ N | ⃗ x, ⃗ y$^{)}$ = 0 . (A.19) P r$^{(}$S π m × m ≥ c ′′ $^{(}$N − 1 m − 1$^{)(}$ N − 1 m − 1$^{)}$ N | ⃗ x, ⃗ y$^{)}$ ≤ E ( S π m × m | ⃗ x, ⃗ y ) c ′′ ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$N ≈ ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$( m − 1) 2 c ′′ ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$N , (A.18) lim N →∞ S$_{m}$$_{×}$$_{m}$ ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$N ≥ c $^{′′}$, (A.17)