OP ATTRIBUTES IN STATISTICS. 2H3 



9. Case of Equality of Contraries. 



Before proceeding to the determination of the numbers of independent frequencies 

 in this case, we shall first prove the following three theorems : 



Theorem I. If equality of contraries subsist for frequencies of any given order, 



then it subsists for all lower orders. 

 Theorem II. If equality of contraries subsist for any even order of frequencies, say 



2m, then it need not in general subsist for order 2m + 1. If, however, it be 



assumed to subsist for order 2m + 1> then the frequencies of this order can be 



expressed in terms of those of the lower order 2m. 

 Theorem III. If equality of contraries subsist for any odd order of frequencies, say 



2ni I, thru it must subsist ;ilsu in t'iv<|urn<-irs {' tin- m-xt lii-'lin- ..nlrr ij;/i. 

 But frequencies of this higher order cannot be expressed in terms of those 

 of order 2m 1. 



The first theorem may be very simply proved. 



10. Suppose we are given, for example, that equality of contraries subsists for 

 frequencies of the fifth order ; then we have 



(ABODE) = (a0ySe) 



(ABCD/) = (aySE) 

 or adding 



(ABCD) = (aygyS) 



and so on. The expansions of contrary frequencies are in fact necessarily contrary 

 themselves. 



11. Next for Theorem II. To take the simplest case, let us suppose equality of 

 contraries given for the second order frequencies. Take any third order group and 

 expand it in terms of its contrary and second order frequencies. This may be done 

 most elementarily step by step. Thus 



(aBC) = (BC) - (ABC) 

 (aC) = (aC) - (BC) + (ABC) 

 /. (a/3y) = (aft) - (aC) + (BC) - (ABC). 



Evidently no equality of contraries amongst second order groups will give us 

 (*/fty) = (ABC). But if we assume this relation to hold we must have 



2 (ABC) = (aft] - (aC) + (BC) 1 



= (AB) + (BC) - (aC) J 



an equation which expresses the third order frequencies in terms of the second. 

 Similarly if equality of contraries is to subsist amongst fifth order groups when it 

 subsists amongst those of the fourth order, we must have 



2 (ABCDE) = (ABCD) + (BCDE) + (ABS) - (ABC) - (aCDE). . . (3). 

 As the method of expansion used is evidently quite general, this proves Theorem II. 



