where bA, b\b\are the regressions in the two cases. To which distribution are we, in such a case, to attribute the greater correlation ? Bravais’ coefficient solves the difficulty, we may say, in one way, by taking the geometrical mean of the" two regressions as the measure of correlation. It will still remain valid for non-normal correlation. But there are other and less arbitrary interpretations even in the general case.
Suppose that instead of measuring and in arbitrary units we measure each in terms of its own standard deviation. Then let us write
and solve for p by the method of least squares. We have omitted a constant on the right-hand side, since it would vanish as before. We have, at once, P = S (xy) <r2 S(y2) *1 r (6),
That is to say, if we measure x and y each in terms of its own standard deviation, r becomes at once the regression of x on y, and the regression of y on x. The regressions being, in fact, the fundamental physical quantities, r is a coefficient of correlation because it is a coefficient of regression.*
Again, let us form the sums of the squares of residuals in equations (1) and (5). Inserting the values of 61, &2, and p, we have— S (x—biy)z = N V ( l- r 2) S(y — b2x)2 = N<t22(i —r2) H ( l - r 2) j ( 7).
Any one of these quantities, being the sum of a series of squares, must be positive. Hence r cannot be greater than unity. If r be equal to unity, or if the correlation be perfect, all the above three sums become zero. But can only vanish if s ( - ± - T x y —+ — = 0 G1 g2 in every case, or if the relation hold good,
- That the regression becomes the coefficient of correlation when each deviation
is measured in terms of its standard-deviation in the case of normal correlation has been pointed out by Mr. Francis Galton. Vide Pearson * Phil. Trans./ A, vol. 187, p. 307, note.
