-PROF. KARL PEARSON ON SKEW VARIATION. 
447 
condition 5/3 2 — 6 /9 : — 9 = 0 is satisfied within the limits of random sampling. Tts 
possibilities extend from & = 0 to & = oo. When f3 l = 0 , 
V = 
, the rectangle. 
Now consider what happens for any frequency curve of the limiting character 
when both /3 : and /3 2 become infinite, say, in the ratio fi 2 — pfi v Then 
6(p-l) 
3 — 2 p ’ 
and accordingly r will be finite if p is finite, except along the Type III. line. 
Accordingly for /3 1 = co 5 e will be zero. Thus the ratio of /3 2 to (3 1 is from their 
values, 
r ±2 
v + 3 
which agrees with the above result for r. 
For the special case when r — 2 , we have p = |, which agrees with the limiting 
ratio of (3 2 //3 1 along the R-line. 
Now when e = 0 we have from 
m 2 — (r—2) m + e— ?■+1 = 0, 
m = \ (r— 2 + \/(r—2) 2 + 4?‘—4), 
= T(r— 2 + r) = r— 1 or — 1 . 
Thus from the equations on page 445, 
f _ N F (m ] + ?n 2 + 2) _/ + 1 x y ?t| / m 2 + 1 ) _ x \ 
5 F (wij + 1) r ( m 2 + 1) \wq + m 2 +2 6 / \mj + m 2 + 2 b / 
_N(m 2 +l) 1 /r a;Y _1 / .r\ _1 
b T(m 2 + 2)\r b) \ h) ’ 
= N (w'ig+l) ^ _ a\ r_1 
a. \ ^b) 
if we chancre the sense of the axis of x and take x from 0 to +b. 
Now in order that <7 should he finite it is needful that b should be infinite when 
m 2 = — 1 , for 
o - 2 = b 2 (m 2 +l)/{r (r+l)}. 
But if b be infinite, y — 0 owing to the factor m 2 + f, for every value of x, except x = 0 . 
Hence the frequency is a concentrated lump at x = 0 , and this involves of itself cr = 0 . 
3 P 2 
