Page:EB1911 - Volume 22.djvu/402

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if the circle were fixed, and the band thrown on it at random. Now let A (fig. 2) be a position of the random point; the favourable cases are when HK, the bisector of the band, meets a circle, centre A, radius ½c; and the whole number are when HK meets a circle, centre O, radius r + ½c; hence the probability is

p = 2π ⋅ ½c/2π(r + ½c) = c/r + ½c.

EB1911 Probability - band and circle.jpg
Fig. 2.

This is constant for all positions of A; hence, equating these two values of p, the mean value required is M(S) = c(2r + c)−1πr 2.

The mean value of the portion of the circumference which falls on the band is the same fraction c/(2r + c) of the whole circumference.

If any convex area whose surface is Ω and circumference L be thrown on the band, instead of a circle, the mean area covered is

M(S) = πc(L + πc)−1Ω.

For as before, fixing the random point at A, the chance of a random point in Ω falling on the band is p = 2π ⋅ ½c/L′, where L′ is the perimeter of a parallel curve to L, at a normal distance ½c from it. Now

L′ = L + 2π ⋅ ½c.

M(S)/Ω = πc/L + πc.

EB1911 Probability - circle with line.jpg
Fig. 3.

84. Buffon's problem may be easily deduced in a similar manner. Thus, if 2r = length of line, a = distance between the parallels, and we conceive a circle (fig. 3) of diameter a with its centre at the middle O of the line,[1] rigidly attached to the latter, and thrown with it on the parallels, this circle must meet one of the parallels; if it be thrown an infinite number of times we shall thus have an infinite number of chords crossing it at random. Their number is measured by 2π ⋅ ½a, and the number which meet 2r is measured by 4r. Hence the chance that the line 2r meets one of the parallels is p = 4r/πa.

EB1911 Probability - area and line.jpg
Fig. 4.

85. To investigate the probability that the inclination of the line joining any two points in a given convex area Ω shall lie within given limits. We give here a method of reducing this question to calculation, for the sake of an integral to which it leads, and which is not easy to deduce otherwise.

First let one of the points A (fig. 4) be fixed; draw through it a chord PQ = C, at an inclination θ to some fixed line; put AP = r, AQ = r′; then the number of cases for which the direction of the line joining A and B lies between θ and θ + dθ is measured by ½(r2 + r′2)dθ.

Now let A range over the space between PQ and a parallel chord distant dp from it, the number of cases for which A lies in this space and the direction of AB from θ to θ + dθ is (first considering A to lie in the element drdp)

Let p be the perpendicular on C from a given origin O, and let ω be the inclination of p (we may put dω for dθ), C will be a given function of p, ω; and, integrating first for ω constant, the whole number of cases for which ω falls between given limits ω′, ω″ is


the integral ∫C3dp being taken for all positions of C between two tangents to the boundary parallel to PQ. The question is thus reduced to the evaluation of this double integral, which, of course, is generally difficult enough; we may, however, deduce from it a remarkable result; for, if the integral ⅓∬C3dpdω be extended to all possible positions of C, it gives the whole number of pairs of positions of the points A, B which lie inside the area; but this number is Ω2; hence

∬C3dpdω = 3Ω2;

the integration extending to all possible positions of the chord C,—its length being a given function of its co-ordinates p, ω.[2]

Cor. Hence if L, Ω be the perimeter and area of any closed convex contour, the mean value of the cube of a chord drawn across it at random is 3Ω2/L.

EB1911 Probability - two convex boundaries.jpg
Fig. 5.

86. Let there be any two convex boundaries (fig. 5) so related that a tangent at any point V to the inner cuts off a constant segment S from the outer (e.g. two concentric similar ellipses); let the annular area between them be called A; from a point X taken at random on this annulus draw tangents XA, XB to the inner. The mean value of the arc AB, M(AB) = LS/A, L being the whole length of the inner curve ABV.

The following lemma will first be proved:—

EB1911 Probability - convex arc.jpg
Fig. 6.

If there be any convex arc AB (fig. 6), and if N1 be (the measure of) the number of random lines which meet it once, N2 the number which meet it twice,

2 arc AB = N1 + 2N2.

For draw the chord AB; the number of lines meeting the convex-figure so formed is N1 + N2 = arc + chord (the perimeter); but N1 = number of lines meeting the chord = 2 chord;

∴ 2 arc + N1 = 2N2 + 2N2, ∴ 2 arc = N1 + 2N2.

Now fix the point X, in fig. 5, and draw XA, XB. If a random line cross the boundary L, and p1 be the probability that it meets the arc AB once, p2 that it does so twice,

2AB/L = p1 + 2p2;

and if the point X range all over the annulus, and p1, p2 are the same probabilities for all positions of X,

2M(AB)/L = p1 + 2p2.

EB1911 Probability - two convex boundaries with random line.jpg
Fig. 7.

Let now IK (fig. 7) be any position of the random line; drawing tangents at I, K, it is easy to see that it will cut the arc AB twice when X is in the space marked α, and once when X is in either space marked β; hence, for this position of the line, p1 + 2p2 = 2(α + β)/A = 2S/A, which is constant; hence M(AB)/L = S/A.

Hence the mean value of the arc is the same fraction of the perimeter that the constant area S is of the annulus.

If L be not related as above to the outer boundary, M(AB)/L = M(S)/A, M(S) being the mean area of the segment cut off by a tangent at a random point on the perimeter L.

The above result may be expressed as an integral. If s be the arc AB included by tangents from any point (x, y) on the annulus,

sdxdy = LS.

It has been shown (Phil. Trans., 1868, p. 191) that, if θ be the angle between the tangents XA, XB,

θdxdy = π(A − 2S).

The mean value of the tangent XA or XB may be shown to be M(XA) = SP/2A, where P = perimeter of locus of centre of gravity of the segment S.

87. When we go on to species of three dimensions further speculative difficulties occur. How is a random line through a given point to be defined? Since it is usual to define a vector by two angles (viz. φ the angle made with the axis X by a vector r in the plane XY, and θ (or ½πθ) the angle made by the vector ρ with r in the plane containing both ρ and r and the axis Z) it seems natural to treat the angles φ and θ as the equiprobable variables. In other words, if we take at random any meridian on the celestial globe and combine it with any right ascension the vector joining the centre to the point thus assigned is a random line.[3] It is possible that for some purposes this conception may be appropriate. For many purposes surely it is proper to assume a more symmetrical distribution of the terminal points on the surface of a sphere, a distribution such that each element of the surface shall contain an approximately equal number of points. Such an assumption is usually made in the kinetic theory of molecules with respect to the direction of the line joining the centres of two colliding spheres in a “molecular chaos.”[4] It is safe to say with Czuber, “No discussion can remove indeterminateness.” Let us hope with him that “though this branch of probability can for the present claim only a theoretic interest, in the future it will perhaps also lead to practical results.”[5]

88. Illustrations of probability and expectation.—The close relation between probability and expectation is well 'illustrated by geometrical examples. As above stated, when a given space S is included within a given space A, if p is the probability that a point

  1. The line might be anywhere within the circle without altering the question.
  2. This integral was given by Morgan Crofton in the Comptes rendus (1869), p. 1469. An analytical proof was given by Serret, Annales scient. de l'école normale (1869), p. 177.
  3. Cf. Bertrand, op. cit. § 135.
  4. See e.g. Watson, Kinetic Theory of Gases, p. 2; Tait, Trans. Roy. Soc., Edin. (1888), xxxiii. 68.
  5. Wahrscheinlichkeitstheorie, p. 64.