Difference Problem 
393 
(3) Another method of reducing the integral in (iv) without quadratures is, perhaps, of 
interest. I have found it convenient in other cases, where the integral limits are, or can be 
safely extended to, ±00 . Suppose we require to find : 
C+ao 
/= Udx. 
■f 
Let m be the value of .x for which U reaches the maximum value U,^ and let M = log (7 ; thus 
{du/dx),„ = 0, unless U^=oo . Then we find : 
4- terms in and higher powersj (v). 
Now since ?7 is a maximum, d^Ujd.i^ and generally d'^ii/dx^ will be negative. The limits of 
I where x = ni + ^ will also be ±00 , and the integral of U can thus be expressed in terms of the 
well-known area and moments of the probability curve. In the first place, if l/o-^= -d'^ujdx^ 
if i be an integer. 
Further : 
/ 
r+cc 
/d^\ 
g - ^r/<r2 (2i - 1 ) (2i - 3) 3 . W2,r a^' + 1. 
Hence writing a,j= ! we find : 
.(vi). 
The successive terms often converge with such rapidity that two or three of them are quite 
sufficient for practical purposes. 
To apply this to our special case, we note 
f/=a™-J'(l -a)P, 
■M==log Z7=(?i-^)loga + /)log(l -a), 
du 1 dU _ /n—p p \da 
dx U dx \ n l — a/dx' 
Hence if ^7 be a maximum, we have du/dx=0, and 
a = {n—p)ln, l~a=p/n (vii). 
Thus m is to be found from 
n — p f™ , 
n J —'^ 
Biometrika i 42 
