Karl Pearson 287 
It is clear that by simple expansion of the trinomial expression, we can always 
find Is in terms oi qg^'. 
We have accordingly to study the expansion of 
'\ n = s m = s -u ( g\ ) 
(l-?-2)»„=o ,«=o r ^ {s-a-m)\m\u\ j 
and if this value be substituted in the integral expression for we find 
1 u=s m=s-u ( g I 
(V). 
The lower values can be equally readily found by the expansion of 
in powers of r by aid of the binomial. 
The first few cases are 
h = (i^y [^n q.o + dJ'q.. + 0,, q,, + d^q^ 
- 6^i,r {du'q,, + d-Jq,, + 20n0,,q,,) 
+ I20jr-"- {0,,q,, + 0^q,:) - 80,,Vq,,}, 
- H0,,r {0n'qn + ^^,,0^. (^n?., + 0,2q,^ + 0-^qu) 
+ 24.6 Jr' {0n%;, + 0,.?q,, + 20,, 0,,q,,) 
- S20,i>- (0,,q,, + 0.,,q,:,) + lQi^0,,'qu] (vi). 
These expressions simplify for various cases, but it is clear that for the general 
case of unknown type of distribution we shall have to find very high product moments 
from the observations in order to use our generalised Tchebychefif's Theorem. 
Otherwise we shall have to make assumptions as to the relations between high order 
and low order q's. 
Since generally q.,,^ = ^o, = 1 and q^ = r, we have 
^^ = 1 .l,..^^>• + ^---^'•^'■')• 
This suggests that for all cases we are likely to get simplified results, if we take 
0,1 — 0,20. = 012 = 1 when we find I, = 2. In other words, simplification arises if we make 
