Page:Logic of Chance (1888).djvu/67

This page has been proofread, but needs to be validated.

sect. 10.

Arrangement and Formation of the Series.

37

Now if all deflections from a mean were brought about in the way just indicated (an indication which must suffice for the present) we should always have one and the same law of arrangement of frequency for these deflections or errors, viz. the exponential^[1] law mentioned in §5.

§10. It may be readily admitted from what we know about the production of events that something resembling

↑ A definite numerical example of this kind of concentration of frequency about the mean was given in the note to §4. It was of a binomial form, consisting of the successive terms of the expansion of $(1+1)^{m}$ . Now it may be shown (Quetelet, Letters, p. 263; Liagre, Calcul des probabilités, §34) that the expansion of such a binomial, as $m$ becomes indefinitely great, approaches as its limit the exponential form; that is, if we take a number of equidistant ordinates proportional respectively to 1, $m$ , ${\tfrac {m(m-1)}{1\cdot 2}}$ &c., and connect their vertices, the figure we obtain approximately represents some form of the curve $y=Ae^{-hx^{2}}$ , and tends to become identical with it, as $m$ is increased without limit. In other words, if we suppose the errors to be produced by a limited number of finite, equal and independent causes, we have an approximation to the exponential Law of Error, which merges into identity as the causes are increased in number and diminished in magnitude without limit. Jevons has given (Principles of Science, p. 381) a diagram drawn to scale, to show how rapid this approximation is. One point must be carefully remembered here, as it is frequently overlooked (by Quetelet, for instance). The coefficients of a binomial of two equal terms—as $(1+1)^{m}$ , in the preceding paragraph—are symmetrical in their arrangement from the first, and very speedily become indistinguishable in (graphical) outline from the final exponential form. But if, on the other hand, we were to consider the successive terms of such a binomial as $(1+4)^{m}$ (which are proportional to the relative chances of 0, 1, 2, 3, . . . failures in $m$ ventures, of an event which has one chance in its favour to four against it) we should have an unsymmetrical succession. If however we suppose $m$ to increase without limit, as in the former supposition, the unsymmetry gradually disappears and we tend towards precisely the same exponential form as if we had begun with two equal terms. The only difference is that the position of the vertex of the curve is no longer in the centre: in other words, the likeliest term or event is not an equal number of successes and failures but successes and failures in the ratio of 1 to 4.

[1] A definite numerical example of this kind of concentration of frequency about the mean was given in the note to §4. It was of a binomial form, consisting of the successive terms of the expansion of $(1+1)^{m}$ . Now it may be shown (Quetelet, Letters, p. 263; Liagre, Calcul des probabilités, §34) that the expansion of such a binomial, as $m$ becomes indefinitely great, approaches as its limit the exponential form; that is, if we take a number of equidistant ordinates proportional respectively to 1, $m$ , ${\tfrac {m(m-1)}{1\cdot 2}}$ &c., and connect their vertices, the figure we obtain approximately represents some form of the curve $y=Ae^{-hx^{2}}$ , and tends to become identical with it, as $m$ is increased without limit. In other words, if we suppose the errors to be produced by a limited number of finite, equal and independent causes, we have an approximation to the exponential Law of Error, which merges into identity as the causes are increased in number and diminished in magnitude without limit. Jevons has given (Principles of Science, p. 381) a diagram drawn to scale, to show how rapid this approximation is. One point must be carefully remembered here, as it is frequently overlooked (by Quetelet, for instance). The coefficients of a binomial of two equal terms—as $(1+1)^{m}$ , in the preceding paragraph—are symmetrical in their arrangement from the first, and very speedily become indistinguishable in (graphical) outline from the final exponential form. But if, on the other hand, we were to consider the successive terms of such a binomial as $(1+4)^{m}$ (which are proportional to the relative chances of 0, 1, 2, 3, . . . failures in $m$ ventures, of an event which has one chance in its favour to four against it) we should have an unsymmetrical succession. If however we suppose $m$ to increase without limit, as in the former supposition, the unsymmetry gradually disappears and we tend towards precisely the same exponential form as if we had begun with two equal terms. The only difference is that the position of the vertex of the curve is no longer in the centre: in other words, the likeliest term or event is not an equal number of successes and failures but successes and failures in the ratio of 1 to 4.

[1]