434 
PROF. KARL PEARSON ON SKEW VARIATION. 
when both \\ 2 are to be made vanishingly small, being m 1 + 1 and m 2 +1 respectively. 
Thus the limit 
= i_= -L = o. 
1 / x 1 + 1 /\ 2 03 
Hence y 0 vanishes or y is zero at all points, but x = — oq and x — a 2 where it is 
undetermined. 
Since ftq/cq = m 2 /a 2> we have aq = a 2 , and the frequency really consists of two 
concentrated groups at — cq and a 2 , or at ±^b. 
If fx i and iu'\ he the distances of the centroid from the two ends of the range, 
/ 
M l 
// 
M l 
n 
n' 
where ft' and n" are the frequencies^ concentrated at the range terminals. But 
/j.\ = b (mj+ l)/(m 1 +m 2 + 2), or we have = (ftq + l)/(m 2 +l) = Xj/X 2 , or is the 
finite quantity which marks the ratio of the vanishing of ftq+1 and m 2 +l ; this, 
therefore, is equal to n"ln'. 
Clearly 
n'i = (■ n"-n')/(n"+n ') 
/4 = (n"+n')/(n" + n') ^ = W, 
tx 3 = (n"—n')/(n"+n')^~ > 
and 
Thus 
m' 4 = (ft"+ *')/(*"+ *0 T7T = tV& 4 , 
Io 
/x 2 - b 2 {n'n")/{n' + n"Y, 
M3 = b 3 n'n"{n"-n')/{n'+n"Y, 
^ = Vn'n" {n ,2 +n m -nW)l{n'+n")\ 
o _ n" ,ri o „ _ ft" ri , 
Pi — — H—— A P 2 — 4—7> “ 1 5 
ft ft ft ft 
giving as verification /3 2 —/3 X — 1 = 0. 
Thus the whole problem is solved if we know the magnitude of the two frequencies 
ft' and ft" concentrated at b and + 
As special cases the point on the da - axis gives /3 1 = 0, /3 2 = 1, and represents two 
equal concentrated frequency lumps ft' = ft" = g-N. The point at co on the line 
/3 2 —/?i — 1 = 0, or /3 l — /3 2 = co represents a single frequency lump, for which ft' = 0, 
ft" = N. I speak of these concentrated frequency lumps lying on the line /3 2 — 1 = 0, 
