474 
Proceedings of the Royal Society of Edinburgh. [Sess. 
and (2) where the dominant includes the hybrid. It is obvious that the 
correlation between parent and offspring will be much greater in the former 
case than in the latter. This argument will be made clearer if the element- 
ary Mendelian formula is examined. In the first place, consider a population 
consisting of two pure races. Let them be denoted by (a, a) and (b, b) 
respectively and let (a, b) be the hybrid between them. Then the whole 
population may be expressed by a parentage of both sexes each repre- 
sented by 
x 2 (a, a) + 2xy (a, b) + y 2 (b, b), 
where x 2 , 2 xy and y 2 denote the numbers respectively of each type. If 
mating is random and fertility equal, we have offspring in the following 
proportions : — 
x 2 (a, a) mating with x 2 (a, a) gives x 4 (a, a) 
55 
„ 2 xy (a, b) „ 
x 3 y (a, a ) + x 3 y (a, b) 
55 
„ y 2 (b, b) „ 
x 2 y 2 (a, b) 
%xy (a, b) 
55 
,, x 2 (a, a) ,, 
x 3 y (a, a )-\-x 3 y (a, b) 
55 
„ 2 xy (a, b) „ 
x 2 y 2 (a, a) + 2x 2 y 2 (a, b) + y 2 x 2 (b, b) 
5 5 
>, y 2 (b, b) „ 
xy 3 (a, b) + xy 3 (b, b) 
y 2 (b, b) 
55 
,, x 2 (a, a) ,, 
xhy 2 (a, b) 
5 5 
„ 2 xy (a, b) „ 
xy 3 (a, b) + xy 3 (b, b) 
55 
„ y 2 (E b) „ 
2/ 4 (b, b) 
Adding together and arranging the terms, we have the population of 
offspring given by 
x 2 (x + y) 2 (a, a), 2 xy(x + y) 2 (a, b), y\x + y) 2 (b, b), 
or the numbers of the offspring are in the same proportions as those of the 
parents ; that is, the population is stable. Stability, then, depends on the 
number of the hybrid being equal to twice the geometric mean of the 
number of the pure races. It is also easily shown that even though these 
proportions are not originally present they at once appear. 
4. When these figures are arranged so as to show the correlation from 
parent to child the following table is formed : — 
Number of Parents of each Type. 
Number of 
Offspring of 
each Type. 
(a, a). 
(a, b). 
(b, b). 
(a, a) . 
x 4 + x 3 y 
x 3 y + x 2 y 2 
(a, b) . 
x 3 y + x 2 y 2 
x s y + 2x 2 y 2 + xy 3 
x 2 y 2 + xy 3 
(b,b). . 
x 2 y 2 -f xy 3 
xy 3 + ?/ 4 
