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For multiplying (xvii.) by 1, X2,... xq respectively and adding, we find

1 + Qu2=0;

or if R be a maximum or minimum value of u, Q= −1/R2. Now compare these equations with those we obtain for the directions and magnitudes of axes of the ellipsoid of residuals:

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Now eliminate the 's from (xviii.) and the a's from (xx.)

and we have precisely the same determinant to find

AR and e/R'. Hence for the semi-axes or max.-min. values :

AR2=e1/R/2

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Equations (xviii.) and (xx.) will now give the same values for the ratios of the x's and of the x's, or for any axis:

x1/x1'=x2/x2'=x3/xz'= ... =xq/xq' . (xxii.)

But (xxii.) combined with (xxi.) gives us :

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X2=

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VAR (xxiii.) In other words, if we define points given by (xxiii.) to be corresponding points,―i. e., if corresponding points lie on the same line at distances inversely as each other from the origin,then the ends of the principal axes of the two ellipsoids are corresponding points. Thus the principal axes of the correlation ellipsoid coincide with those of the ellipsoid of residuals in direction, and a minimum axis of the one is a maximum axis of the other and vice versa. We therefore conclude:

(i.) That the best fitting plane to a system of points is perpendicular to the least axis of the correlation ellipsoid, and that if 2 Rmin. be the length of this axis the mean square residual = WAXR where A is the well-known determinant of the correlation coefficients.

min.

(ii.) The best-fitting straight line to a system of points coincides in direction with the maximum axis of the correlation ellipsoid, and the mean square residual

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= √o3x1+o2x2+σ2±2+ +0x-A. R2max., (xxiv.) where 2 R max. is the length of the maximum axis. -We have thus the properties of the best-fitting plane and line in terms of the correlation ellipsoid, which is the one generally adopted for variation problems. At the same time our investigation shows us that the q directions of independent variation and the standard-deviations of the independent variables may be found from the ellipsoid of residuals, which will usually be a process involving much simpler arithmetic. (7) Numerical Illustrations.

Case (i.). Find the best fitting straight line to the following system of points supposed of equal weight:

x= 0 y=5.9

x=44 y=3.7

x=5.2

y=2.8

x=6.1 y=2.8

x= -9 y=5.4

x=1.8

3=44

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x=6.5

| x=3·3

y=2.4

y=3.5

x=7·4 y=1.5

We have at once:

x=3.82
c=23748

y=3.70

av=131225

:

Tzy=-9765

tan 20=2rxyoxoy/(σx2—σy2) = −1·5535.

Hence tan 0=-54556, or the best-fitting line passes through the point 3.82, 3.70 at a slope of 546. This line is shown in the accompanying diagram by AB. The mean

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square residual is 2484. Had we made S(y-y')2 a minimum, the slope of the "best-fitting" line would have been 5396 and the mean square residual 2828; had we made S(x-x')2 a minimum, the slope of the "best-fitting" line would have been 5659, and the mean square residual 5118. These lines are of course the regression lines of slopes rayoy/σr and (rayor) to the horizontal and mean square vertical and horizontal residuals of oy I-r and or 1-72 respectively. Illustration (2).-The following system gives four values

of a certain function z:

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Let us find the best-fitting plane, treating these as four points in three-dimensioned space. We have at once

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y= 5,

Z=209.5

a = 500275

rzyσyσz=182.5, Txz ̄xσz=−30.5, rxyoxσy=().

Thus the ellipsoid of residuals is :

x2+25y2+2502·75z2+365yz-61æz=e1.

The equations to find the direction-cosines are:

(2+212) 4+0.4-61.4=0,

0.4+ (50+22)42+3654=0

−611, +3654,+ (5005·5+2 2)4= =0.

Whence writing 2 =x (n=number of points=4) the cubic

for X

Χ

is:

n

C=x2+5057·5x2 + 123,440x + 48,050=0.

We want the least root:

x=0, C=+; x=-5, C=; x=-100, C=+; x=−∞, C. Thus the required root lies between 0 and -5. It is easily found to be

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or:

38 02187 (x-3)-7-35823 (y-21)+z-209.5=0,

z+38·02187x--7-35823y-169-03778=0.. (xxv.)

If we find the values for z for given x and y, say those of the four points, which are

2117, 135-7, 283-3, 207-3,

we should not be impressed by the goodness of the fit. But

the small value of the mean square residual shows how close to each of the points the plane really goes when we measure its distance from a point not by the vertical intercept, but by the perpendicular from the point on the plane. Thus the vertical distance from =4, y=16, z=127, to the plane is 8.7, but the perpendicular distance is only 1988.

If x and x3 be the other two roots of the cubic C we easily X3 find:

X2X3=121,442 65, X2+x3=-5057-10434.

Thus we have the quadratic to find X and X3

or:

x2+5057-10434x+121,442.65=0,

X2=-24-12895, X3=-5032-97445.

X3 gives the least axis of the ellipsoid of residuals; hence the direction-cosines of this axis are given by

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We have accordingly for the equation of the best-fitting straight line to the four points:

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The mean square residual for this line ' is given by (xiv.)

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This again is remarkably small, considering how far our four points are from being co-linear.

The reader will easily prove directly that the best line (xxvi.) really lies in the best plane (xxv.).

These two illustrations may suffice to show that the methods of this paper can be easily applied to numerical problems; the labour is not largely increased if we have a considerable number of points. It becomes more cumbersome if we have four, five, or more variables or characters which involve the determination of the least (or greatest) root (as the case may be) or an equation of the fourth, fifth, or higher order. Still, the coefficients being numerical and all the roots real and negative, it is not very difficult to localize them.

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