MR. K. PEARSON ON THE MATHEMATICAL THEORY OF EVOLUTION. . 347 
We shall now proceed to find the first four moments of the system of rectangles 
round GN. Jf the inertia of each rectangle might be considered as concentrated along 
its mid vertical, we should have for the s™ moment round NG, writing d= c(1-+ gq), 
ap, = S{y,c X (re — d)°}. 
The resulting values are 
Hy = npge 
Hs = mpg (p — 9g) 
Hy = npg {1 + 3 (7m — 2) pq} ct, 
whence, remembering that p + ¢ = 1, we find that p and q are roots of 
(Bpg" = Hy) Me + Ms” 



9 
home ag - = 0 
2 4 (Bply” — ply) Ma + Bg” ‘ 
Wes 2 p98 2 iS VS 12 (Bp? = by) Ma + 8ps"} 
(Bug — Ha) Ba + Ms? P2 
Thus, when pz, pz, and p, have been calculated for the frequency curve, the 
elements of the point-binomial are known. ‘These results were given by me in a 
letter to ‘ Nature, October 26, 1893. 
They give quite a fair solution so long as n is large and ¢ small, 2.e., so long as the 
asymmetry and the “excess” (‘ Phil. Trans.,’ vol. 185, A, p. 93), measured respec- 
tively by ps; and py — 3p.” (which vanish for the normal curve) are not considerable.* 
In many cases, however, they are considerable, and the following solution is perfectly 
general. 
* If y, denote the largest term in (p+ q)" and y, the ‘th term beyond it, then an application of 
Srieiine’s theorem—if n be large—shows that 
EE Nt= f t \-—t-—qn—-k 
vin =(1— =) (+2) : 
‘ | qu 
Take 
t X 
log wu = (tf — pn — 3) tog (1 Fp) 
t 
log ¢ = (—t— qn = }) og (1 + ra 
and expand the right hand side in powers of #, we find 
1 Za il me if 1 es 3 
ee ‘(1 ug om) ~ Qn { IL zs}  6p?n? (1 me =) ~ 12p8n3 (1 = al Ue: 



Hence, remembering that p + ¢g = 1, we have 
AUD Oy ( cs he200) reid) ae =) 
2pqn 2npq 2upq ) 29? 


log w = — 
te 3 (1 = 4pq + 2p*q") 
= 12piqin’ (1 — 3pq — Qnpq + ete. 
Now, making use of the values given in § 2 for jy, wz, and m,, and writing t X c= a, and y;= y, 
we find 

4 XB 
