P, taken at random on A, falling on S, p = S/A. If now the
space S be variable, and M(S) be its mean value
p = M(S)/A.
For, if we suppose S to have n equally probable values S_{1}, S_{2}, S_{3} . . ., the chance of any one S_{1} being taken, and of P falling on S_{1}, is
p_{1} = n^{−1}S_{1}/A;
now the whole probability p = p_{1} + p_{2} + p_{3} + . . . , which leads at once to the above expression. The chance of two points falling on S is, in the same way,
p = M(S^{2})/A^{2},
and so on.
In such a case, if the probability be known, the mean value follows, and vice versa. Thus, we might find the mean value of the nth power of the distance XY between two points taken at random in a line of length l, by considering the chance that, if n more points are so taken, they shall all fall between X and Y. This chance is
M(XY)^{n}/l^{n} = 2(n + 1)^{−1}(n + 2)^{−1};
for the chance that X shall be one of the extreme points, out of the whole (n + 2), is 2(n + 2)^{−1}; and, if it is, the chance that the other extreme point is Y is (n + 1)^{−1}. Therefore
M(XY)^{n} = 2l^{n}(n + 1)^{−1}(n + 2)^{−1}.
A line l is divided into n segments by n − 1 points taken at random; to find the mean value of the product of the n segments. Let a, b, c, . . . be the segments in one particular case. If n new points are taken at random in the line, the chance that one falls on each segment is
1.2.3 . . . nabc . . . /l^{n};
hence the chance that this occurs, however the line is divided, is
n!l^{−n}M(abc . . .).
Now the whole number of different orders in which the whole 2n − 1 points may occur is (2n − 1)!; out of these the number in which one of the first series falls between every two of the second is easily found by the theory of permutations to be n!(n − 1)!. Hence the required mean value of the product is
M(abc. . .) = (n − 1)!(2n − 1)!l^{n}.
89. Additional examples of the relation between probability and expectation appear in the following series of propositions: (1) If M be the mean value of any quantity depending on the positions of two points (e.g. their distance) which are taken, one in a space A, the other in a space B (external to A); and if M′ be the same mean when both points are taken indiscriminately in the whole space A + B; M_{a}, M_{b}, the same mean when both points are taken in A and both in B respectively; then
(A + B)^{2}M′ = 2ABM + A^{2}M_{a} + B^{2}M_{b}.
If the space A = B, 4M′ = 2M + M_{a} + M_{b}; if, also, M_{a} = M_{b}, then 2M′ = M + M_{a}.
(2) The mean distance of a point P within a given area from a fixed straight line (which does not meet the area) is evidently the distance of the centre of gravity G of the area from the line. Thus, if A, B are two fixed points on a line outside the area, the mean value of the area of the triangle APB = the triangle AGB. From this it will follow that, if X, Y, Z are three points taken at random in three given spaces on a plane (such that they cannot all be cut by any straight line), the mean value of the area of the triangle XYZ is the triangle GG′G″, determined by the three centres of gravity of the spaces.
(3) This proposition is of use in the solution of the following problem:—
Two points X, Y are taken at random within a triangle. What is the mean area M of the triangle XYC, formed by joining them with one of the angles of the triangle?
Bisect the triangle by the line CD; let M_{1} be the mean value when both points all in the triangle ACD, and M_{2} the value when one falls in ACD and the other in BCD; then 2M = M_{1} + M_{2}. But M_{1} = ½M; and M_{2} = GG′C, where G, G′ are the centres of gravity of ACD, BCD; hence M_{2} = 29ABC, and M = 427ABC.
(4) From this mean value we pass to probabilities. The chance that a new point Z falls on the triangle XYC is 427; and the chance that three points X, Y, Z taken at random form, with a vertex C, a re-entrant quadrilateral, is 49.
90. The calculation of geometrical probability and expectation is much facilitated by the following general principle: If M be a mean value depending on the positions of n points falling on a space A; and if this space receive a small increment α, and M′ be the same mean when the n points are taken on A + α, and M the same mean when one point falls on α and the remaining n − 1 on A; then, the sum of all the cases being M′(A + α)^{n}, and this sum consisting of the cases (1) when all the points are on A, (2) when one is on α the others on A (as we may neglect all where two or more fall on α), we have
M′(A + α)^{n} = MA^{n} + nM_{1}α^{n−1};
∴ (M′ − M)A = nA(M_{1} − M),
as M′ nearly = M. For example, suppose two points X, Y are taken in a line of length l, to find the mean value M of (XY)^{n}. If l receives an increment dl, ldM = 2dl(M_{1} − M). Now M_{1} here = the mean nth power of the distance of a single point taken at random in l from one extremity of l; and this is l^{n}(n + 1)^{−1} (as is shown by finding the chance of n other points falling on that distance); hence
ldM = 2dl{l^{n}(n + 1)^{−1} − M};
∴ ldM + 2Mdl = 2(n + 1)^{−1}l^{n}dl,
or
l^{−1}.d.Ml^{2} = 2(n + 1)^{−1}l^{n}dl:
∴ Ml^{2} = 2(n + 1)^{−1}∫l^{n+1}dl = 2l^{n+2}/(n + 1)(n + 2) + C;
∴ N = 2l^{n}/(n + 1)(n + 2),
C being evidently 0.
91. The corresponding principle for probabilities may thus be stated: If p is the probability of a certain condition being satisfied by the n points within A in art. 90, p′ the same probability when they fall on the space A + α, and p′ the same when one point falls on α and the rest on A, then, since the numbers of favourable cases are respectively p′(A + α)^{n}, pA^{n}, np_{1}α^{n−1}, we find,
(p′ − p)A = nα(p_{1} − p).
Fig. 8. |
Hence if p′ = p then p_{1} = p. For example, if we have to find the chance of three points within a circle forming an acute-angled triangle, by adding an infinitesimal concentric ring to the circle, we have evidently p′ = p; hence the required chance is unaltered by assuming one of the three points taken on the circumference. Again, in finding the chance that four points within a triangle shall form a convex quadrilateral, if we add to the triangle a small band between the base and a line parallel to it, the chance is clearly unaltered. Therefore we may take one of the points at random on the base (fig. 8), the others X, Y, Z within the triangle. Now the four lines from the vertex B to the four points are as likely to occur in any specified order as any other. Hence it is an even chance that X, Y, Z fall on one of the triangles ABW, CBW, or that two fall on one of these triangles and the remaining one on the other. Hence the probability of a re-entrant quadrilateral is
½p_{1} + ½p_{2},
where
p_{1} = | prob. (WXYZ re-entrant), | X, Y, Z in one triangle; |
p_{2} = | do., | X in one triangle, Y in the other, Z in either. |
But p_{1} = 49. Now to find p_{2}; the chance of Z falling within the triangle WXY is the mean area of WXY divided by ABC. Now by par. 89, for any particular position of W, M(WXY) = WGG′, where G, G′ are the centres of gravity of ABW, CBW. It is easy to see that WGG′ = 19ABC = 19, putting ABC = 1. Now if Z falls in CBW, the chance of WXYZ re-entrant is 2M(IYW), for Y is as likely to fall in WXZ as Z to fall in WXY; also if Z falls in ABW the chance of WXYZ re-entrant is 2M(IXW). Thus the whole chance is p_{2} = 2M(IYW + IXW) = 29. Hence the probability of a re-entrant quadrilateral is
12⋅49 + 12⋅29 = 13.
That of its being convex is ⅔.
Fig. 9. |
92. From this probability we may pass to the mean value of the area XYZ, if M be this mean, and A the given area, the chance of a fourth point falling on the triangle is M/A; and the chance of a re-entrant quadrilateral is four times this, or 4M/A. This chance has just been shown to be 13; and accordingly M = 112A.
93. The preceding problem is a particular case of a more general problem investigated by Sylvester. For another instance, let the given area A be a circle; within such three points are taken at random; and let M be the mean value of the triangle thus formed. Adding a concentric ring a, we have since M': M as the areas of the circles, M′ = M(A + α)/A.
AMα/A = 3α(M_{1} − M); ∴ M = ¾M_{1},
where M_{1} is the value of M when one of the points is on the circumference. Take O fixed; we have to find the mean value of OXY (fig. 9). Taking (ρ, θ)(ρ′, θ′) as co-ordinates of X, Y,
M_{1} = (πα^{2})^{−2}∬ρdρdθ∬ρ′dρ′dθ. (OXY).
∵ M_{1} = (π^{4}α^{2})^{−1}∬∬½ρρ′ sin (θ − θ′)ρρ′dρdρ′dθdθ′
= (π^{2}α^{4})^{−1}⋅½∬19r^{2}r′^{2}⋅sin (θ − θ′)dθdθ′,