Miscellanea 
141 
Hence dividing each equation by the preceding one : 
«2 _ Ml 
n (n - \) n 
1i 
n {n - \) (n - 2) n(n ■ 
n (n - 1 ) n 
1) _ w(w - - 2)(w - 3) w(w-l)(w-2) 
«3 "j?! 
?2- 
n (n - 1) (n - 2) w (»? - 1) 
Writing q^q^ = and qi + q^ = Pi, we obtain 
ftg - («- - 1) fliPj + n {n - I) P2 = 0, 
a^- (n ~ 2) a^Py + (n - 1) {n - 2) a^P^ 
- (w - 3) «3Pi + (w - 2) (K - 3) a^Pj 
three equations to determine ?(, P, and Pj. 1 
Eliminating - Pj and P2 we find : 
(n - 1) fli n{n - 1) 
(« - 2) fflj - 1) (m - 2) «! = 0, 
(71 - 3) ag (w - 2) (w - 3) fflj 
which expanded gives us the cubic for n : 
.(xii) 
+ 11^ (- 12aj02ff3 + la^ + ittj^a^ - Sa^a^ + 4(7,^) 
+ ?i(22aia2a3 - 16«2' - 5aj^a^ + 2a2a4 - ^a-/) + (- ISajaaag + 12a2^ + 2ay^a^) = 0. ...(xii)'"^ 
A root of this cubic substituted in the first two equations of (xi) will give Pj and Pj and then 
the quadratic 
- PiJ) + P2 = 0 (xiii) 
will determine the two values q^ and q^ corresponding to the value of The first two equations 
of (viii) then complete the solution by providing Xj and Xj. 
Until the roots of the cubic (xii)'''^ have been discussed we can only assume that three 
solutions are possible. As a matter of fact in the examples so far dealt with some of these solutions 
have usually to be discarded. 
For the special case of Poisson's limit to the binomial, we make n indefinitely large, q inde- 
finitely small, and nq = m finite. Hence equations (viii) become 
l = Xi + X2, 
«j = Xj + Xj 1/1.2 , 
= XiMj^ + XoWj^, 
= XiWij* + XoWio*', 
leading to 
Thus we find : 
subject to the condition* 
ff.4 (flj - 
Hence »ij and are roots of 
m? (a2 - o-i 
Qi = («3 - aia2)/("2 - 
Q2 = («3»i - f'2-)/(ff2 - "l'^). 
•(XV) 
a, a,) + a,a, - = 0. 
.(xvi) 
* Of course equations of condition hold for the 5th and higher moments in the case of the two 
binomial components. But they are of small service as the probable errors of these high moments are 
usually very considerable. 
