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SS2 =nX2¢"-'S¢ =n ' a"-'Sa+(n— I) an-'X See also:aba, :.na2=n'+(n—1)A''; but VI = (n— 0-2; ..nr2=nl+(n In the same way we can show generally that and thus the required mean value of the rth See also:part is a,¢=¢n-'{n' F(n—i)-' ~(n—2)'' (n—r I-i)''). he places among the " misapplications of the calculus which have made it the real opprobrium of See also:mathematics " (See also:Logic, See also:Book III, ch. xviii. § 3). Cf. See also:Bertrand, Calcul See also:des probabilites; See also:Venn, Logic of See also:Chance, ch. xvi. § 5-7; v. Kries, Princspien der Wahrscheinlichkeitsrechnung, ch. ix., See also:preface, § v., and ch. ,iii. §§ 12, 13; See also:Laplace's See also:general reflections on this See also:matter seem more valuable than his calculations: " Tant de passions at d'interets particuliers y melent si souvent leur See also:influence qu'il est impossible de soumettre au calcul See also:cette probabilit6," op. cit. Introduction (Des Choix et decisions des assemblees). " As to the possibility of See also:mistake in this respect, see See also:Proctor, How to See also:play See also:Whist, p. 121. Bertrand, loc. cit. - 8 Loc. cit. i 43. 384 Thus each See also:judge implicitly assigns the probabilities n2n\n+n I I,n(n+n I I+n I 2/ to the causes as they stand on his See also:list, beginning from the lowest. The values assigned for the See also:probability of each alternative cause may be treated as so many equally authoritative observations representing a quantity which it is required to determine. According to a general See also:rule given below the observations are to be added and divided by their number; but here if we are concerned only with the relative magnitudes of the probabilities in favour of each alternative it suffices to compare the sums of the observations. We thus arrive at Laplace's rule. Add the See also:numbers found on the different lists for the cause A, for the cause B, and so on; that cause which has the greatest sum is the most probable. 53. Probability of Future Effects deduced from Causes.—Another class of problems which it is usual to See also:place in a See also:separate See also:category are those which require that, having ascended from an observed event to probable causes, we should descend to the probability of See also:collateral effects. But no new principle is involved in such problems. The See also:reason may be illustrated by the following modification of the problem about digits which was above set 2 to illustrate the method of deducing the probability of alternative causes. What is the probability that if to the second See also:digit which contributed to the effect there described there is added a third digit taken at See also:random, the sum of the second and third will be greater than to (or any other assigned figure)? The probabilities—the a posteriori probabilities derived from the observed event (that the sum of the first and second digit exceeds 9)—each multiplied by 45, of the alternatives constituted by the different values o, i, 2, . . . 8, 9 of the second figure are written in the first of the subjoined rows.
o I 2 3 4 5 6 7 8 9
o o I 2 3 4 5 6 7 8
o o 2 6 12 20 30 42 56 72
Below each of these probabilities is written the probability, X io that if the corresponding cause existed the effect under See also:consideration would result. The product of the two probabilities pertaining to each alternative way of producing the event gives the probability of the event occurring in that way. The sum of these products which are written in the third See also:row divided by 45 X10, viz. It$ =?6, is the required probability. It may be expected that actual trial would verify this result.
54• Rule of See also:Succession."—One See also:case of inferred future effects, sometimes called the " rule of succession," claims See also:special See also:notice as having been thought to furnish a test for the cogency of See also:induction. A See also: The probability that this cause, if it exists, will produce the effect in question, the extraction of a white ball at the (n+I)th trial, is p. The probability of the event, obtained by summing the probabilities of all the different ways in which it may occur, is accordingly mpn+i/z p", where p both in the numerator and the denominator is to receive all possible values (between o p and I. In the limit we have f IPn+Idp/J 'PndP = (n +I)/ (n+2). In particular if n = i, the probability that an event which has been observed once will recur on a second trial is 1. These results are perhaps not so absurd as they have seemed to some critics, when the principle of " See also:cross-series ' * is taken into See also:account. Among authorities who seem to attach importance to the rule of succession, in addition to the classical writers on Probabilities, may be mentioned See also:Lotze 6 and Karl See also:Pearson.' See also:Section III.—Calculation of Expectation. 55. Analogues of Preceding Problems.—T his section presents problems analogous. to the preceding. If n balls are extracted 2 Below, pars. 135, 136. A difficulty raised by See also:Cournot with respect to the determination of several quantities which are connected by an See also:equation does not here arise. The See also:system of values determined for the several causes fulfils by construction the See also:condition that the sum of the values should be equal to unity. 2 Above, See also:par. 44. 2 It comes to the same to suppose the See also:total number of balls in the mixture to be N; and to assume that the number of white balls is a priori equally likely to have any one of the values 1, 2, . . . N-1, N. 6 Above, par. 5. 6 Logic, bk. ii. ch. ix. § 5. 6 See also:Grammar of See also:Science, ch. iv. § 16. Cf. the See also:article in Mind above referred to, ix. 234. [METHODS .OF CALCULATION from an See also:urn containing black and white balls mixed up in the proportions p: (1-p), each ball being replaced after extraction, the expected number of white balls in the set of n is by See also:definition np 7 It may be instructive to verify the consistency of first principles by demonstrating this axiomatic proposition .9 Consider the respective probabilities that in the series of n trials there will occur no white balls, exactly one white ball, exactly two white balls, and so on, as shown in the following See also:scheme: No. of white balls o, I, 2, C probabi ity. (I-p)"> n n(I-P)n-I", (n 2)!2!(I_P)n-2Y ..r To calculate the expectation of white balls it is proper to multiply i by the probability that exactly one white ball will occur, 2 by the probability of two white balls, and so on. We have thus for the required expectation (n" !i)!(I —p)n-Ip+(n–2)1(1 —p)n—2p2+. . =np[(I—p)+p1' +P1'=np. The expectation in the case where the balls are not replaced—not similarly axiomatic—may be found by approximative formulae 56. See also:Games of Chance.—With reference to the topic which occurred next under the See also:head of probabilities, a distinction must be drawn between the number of trials which make it an even chance that all the faces of a See also:die will not have turned up at least once, and the number of trials which are made on an See also:average before that event occurs. We may pass from the probability to expectation in such cases by means of the following theorem. If s is the number of trials in which on an average success (such as turning up every See also:face of a die at least once) is obtained, then s= i +fi+See also:f2+... ; where fr denotes the probability of failing in the first r trials. For the required expectation is equal to I Xprobability of succeeding at the first trial + 2 X probability of succeeding at the second trial+&c. Now the probability of succeeding at the first trial is i -1'i; the probability of succeeding at the second trial. (after failing at the first) is fi(I—f2); the probability of succeeding at the third trial is similarly f2(' —is); and so on. Substituting these values for the expression for the expectation, we have the proposition which was to /be\ proved. In the proposed problem fn= 6 c16\1"—15161"X2o(6)"—15(6)+6(6") Assigning to n in each of /these terms, every value from i to oo we have 6•§/(i -1), =30, for the sum of the first set, with corresponding expressions for the sets formed from the following terms. Whence s = I +30 - 30 +20 -126 + = 141. By parity of reasoning it is proved that on an average 7tk§ cards10 must be dealt before at least one card of every suit has turned up.0 57. See also:Dominoes are taken at random (with replacement after each extraction) from the set of the See also:kind described in a preceding See also:paragraph.12 What is the difference (irrespective of sign) to be expected between the two numbers on each domino? The digit 9, according as it is combined with itself, or any smaller digit, gives the sum of See also:differences 0 +I+2-1-...+9. The digit 8 combined with itself or any smaller digit gives the sum of differences o + 1 + 2 + ... + 8 and so on. The sum of the differences is E z r. r+r, where r has every integer value from i to 9 inclusive, = - 9(9+I)-(9+2) 2,3 =16$. And the number of, the differences is Io + 9 + 8 + . . . + 2 + I = 55. There-fore the required expectation is 165/55 =3. 58. Digits taken at Random.—The last question is to be distinguished from the following. What is the difference (irrespective of sign) between two digits, taken at random from mathematical tables, or the expansion of an endless See also:constant like 7r? The combinations of different digits will now occur twice as often as the repetitions of the same digit. The sum of the differences may, now be obtained from the consideration that the sum of the See also:positive differences must be equal to sum of the negative differences when the null differences are distributed equally between the positive and the negative set. The sum of the positive set is, as before, 7 See the See also:introductory remarks headed " Description and See also:Division of the Subject." s Cf. above, par. 25. 9 See Pearson, Phil. Trans. (1895), A. I° See also:Whitworth, Exercises, No. 502. u Ibid. No. 504, cf. above, par. 29. 12 Ibid. par. 36. /. +(nh—r) !(r-I)1(1-p)n-rpr+... +np5 =np [(I -p)"—1+(n—I) (I —p)"—2C+.. See also:fit +(n —(nr) (Ir)- 1)1(1 —p)" rp I+ 1 ... -i-P 1 165. But the denominator of this numerator is not the same as before, but less by See also:half the number of null differences, that is 5. We thus obtain for the required expectation 165/5o=3 3.
59. A See also:simple verification of this prediction may thus be obtained. In a table of logarithms See also:note any two digits so situated as to afford no presumption of See also:close correlation; for instance, in the last place of the See also:logarithm of 10009 the digit 7 and in the last place of the logarithm of 10019 the digit 4, and take the difference between these two, viz. 3, irrespective of sign. Proceed similarly with the similarly situated pair which See also:form the last places of the logarithms of 10029 and 10039; for which the difference is 1, and so on. The mean of the differences thus found ought to be approximately 3.3. Experimenting thus on the last digits of logarithms, in See also:Hutton's tables extending to seven places, from the logarithm of 10009 to the logarithm of 10909, the writer has found for the mean of 250 differences, 3.2.
6o. Points taken at Random.—By parity of reasoning it may be shown that if two different milestones are taken at random on a road n See also:miles See also:long (there being a See also: If instead of finite differences as in the last two problems the intervals between the numbers or degrees which may be selected are indefinitely small, we have the theorem that the mean distance between two points taken at random on a finite straight See also:line is a third of the length of that strat line. 62. The fortuitous division of a straight line is happily employed by See also:Professor See also:Morgan Crofton to exhibit Laplace's method of deter- See also:mining for mining the See also:worth of several candidates by combining Rules For the votes of See also:electors. There is a close relation between t:tecttana, this method and the method above given for deter- mining the probabilities of several alternatives by combining the judgments of different See also:judges.' But there is this difference--that the several estimates of worth, unlike those of probability, are not subject to the condition that their sum should be equal to a constant quantity (unity). The quaesila are now expectations, not probabilities. Professor Morgan Crof ton's version 2 of the See also:argument is as follows. Suppose there are n candidates for an See also:office ; each elector is to arrange them in what he believes to be the See also:order of merit ; and we have first to find the numerical value of thq merit he thus implicitly attributes to each See also:candidate. Fixing on some limit a as the maximum of merit, n arbitrary values less than a are taken and then arranged in order of magnitude—least, second, third, . . . greatest ; to find the mean value of each. A X Y B Take a line AB=a, and set off n arbitrary lengths AX, AY, AZ . . beginning at A; that is, n points are taken at random in AB, Now the mean values of AZ, XY, YZ, . are all equal; for if a new point P be taken at random, it is equally likely to be 1st, 2nd, 3rd, &c., in order beginning from A, because out of n+r points the chance of an assigned one being 1st is (n-1-W1; of its being 2nd (n+i)-'; and so on. But the chance of P being 1st is equal to the mean value of AX divided by AB; of its being 2nd M(XY)=AB; and so on. Hence the mean value of AX is AB (n+i)-'; that of AY is 2AB (n+i)-l; and so on. Thus the mean merit assigned to the several candidates is a(n+i)-1, 2a(n+0-', 3a(n+1)-1...na(n+1)-1. Thus the relative merits may be estimated by See also:writing under the names of the candidates the numbers 1, 2, 3, . n. The same being done by each elector, the probability will be in favour of the candidate who has the greatest sum. Practically it is to be feared that this See also:plan would not succeed, because, as Laplace observes, not only are electors swayed by many considerations See also:independent of the merit of the candidates, but they would often place See also:low down in their list any candidate whom they judged a formidable competitor to the one they preferred, thus giving an unfair See also:advantage to candidates of mediocre merit. 63. This objection is less appropriate to competitive See also:examinations, to which the method may seem applicable. But there is a more fundamental objection in this case, if not indeed in every case, to the reasoning on which the method rests: viz. that there is sup-posed an a priori See also:distribution of values which is in general not supposable; viz. that the several estimates of worth, the marks given to different candidates by the same examiner, are likely to See also:cover evenly the whole of the See also:tract between the minimum and maximum, e.g. between o and See also:loo. Experience, fortified by theory, shows that very generally such estimates are not thus indifferently disposed, but rather in an order which will presently be described as the normal See also:law of See also:error.' The theorem governing the case would therefore seem to be not that which is applied by Laplace and Morgan Crofton, but that which has been investigated by Karl Pearson,' a theorem which does not lend itself so readily to the purpose in See also:hand.' Above, par. 52. 2 Loc. cit. § 45. 3 See See also:Edgeworth, " Elements of Chance in Examinations," Journ. Stat. See also:Soc. (1890). Cf. below, par. 124. ' Biometrika, i. 390. s See also:Moore, of See also:Columbia University, New See also:York, has attempted to 64. Expectation of Advantage.—The general examples of expectation which have been given may be supplemented by some appropriate to that special use of the See also:term which Laplace has sanctioned when he considers the subject of expectation as a " See also:good "; in particular See also:money, or that for the See also:sake of which money is desired, "moral " advantage, in more See also:modern phrase utility or See also:satisfaction. 65. Pecuniary Advantage.—The most important calculations of pecuniary expectation relate to annuities and See also:insurance; based largely on See also:life tables from which the expectation of life itself, as well as of money value at the end, or at any See also:period, of life is predicted. The reader is referred to these heads for See also:practical exemplifications of the calculus. It must suffice here to point out how the calculations are facilitated by the See also:adoption of a law of frequency, the Gompertz or the Gompertz-Makeham law, which on the one hand can hardly be ranked with hypotheses resting on a See also:vera causa, yet on the other hand is not purely empirical, but is recommended, as germane to the subject-matter, by colourable suppositions' 66. There is space here only for one or two simple examples of money as the subject of expectation. Two persons A and B throw a die alternately, A beginning, with the understanding that the one who first throws an See also:ace is to receive a See also:prize of £I. What are their respective expectations?7 The chance that the prize should be won at the first throw is i, the chance that it should be won at the second throw is g T at the third throw (g )2A, at the See also:fourth throw (g)3 }, and so on. Accordingly the expectation of A =£I X { 1-E(U2+(8)4+ . . . 1 ; expectation of B =£I X}• 11-F(l)2+(1)'+ . . .1. Thus A's expectation is to B's as I : S. 'But their expectations must together amount to £1. Therefore A's expectation is See also:Tar of a See also:pound, B's 67. There are n tickets in a bag, numbered 1, 2, 3, . . n. A See also:man draws two tickets at once, and is to receive a number of sovereigns equal to the product of the numbers drawn. What is his expectation?' It is the number of pounds divided by an improper fraction of which the denominator is the number of possible products, §n(n—1), and the numerator is the sum of all possible products =1{(I+2+3. . . +n)2—('2+22+. . +n}. Whence the required number (of pounds) is found to be 3(n+1) (3n+2). The result may be contrasted with what it would be if the two tickets were not to be drawn at once, but the second after replacement of the first. On this supposition the expectation in respect of one of the tickets separately is j (n-(-1). Therefore, as the two events are now independent, the expectation of the product,9 being the product of the expectations, is { z (n+i) }2. 68. See also:Peter throws three coins, See also:Paul two. The one who obtains the greater number of heads wins £I. If the number of heads are equal, they play again, and so on, until one or other obtains a greater number of heads. What are their respective expectations?" At the first trial there are three alternatives: (a) Peter obtains more heads than Paul, (3) an equal number, (y) fewer. The cases in favour of a are (I) Peter obtains three heads, (2) Peter, two heads, while Paul one or none, (3) Peter one head, Paul none. The cases in favour of /3 are (I) two heads for both, or (2) one head, or (3) none, for both. The remaining case favours y: The probability of a is 8-F14+14=1. The probability of /3 is e,++=. The probability of y is i —fie = 3. Alternative / is to be split up into three a', /3', y', of which the probabilities (when / has occurred') are as before, ;g, §g, {s. 13' is similarly split up, and so on. Thus Peter's expectation is 1B3{i+,+()+ . . . ]i =11£I. Paul's expectation is £1. An urn contains m balls marked 1, 2, 3, . . . m. Paul extracts successively the m balls, under an agreement to give Peter a See also:shilling every See also:time that a ball comes out in its proper order. What is Peter's expectation? The expectation with respect to any one ball ism , and therefore the expectation with respect to all is I (shilling)." 69. Advantage subjectively estimated.—Elaborate calculations are paradoxically employed by Laplace and other mathematicians to determine the expectation of subjective advantage in various cases of See also:risk. The calculation is based on See also:Daniel See also:Bernoulli's See also:formula which may be written thus: If x denote a man's See also:physical See also:fortune, and y the corresponding moral fortune y =k See also:log (x/h), k, h being constants. x and y are alwayspositive,andx>h; forevery trace Karl Pearson's theory in the See also:statistics See also:relating to the efficiency of See also:wages (Economic See also:Journal, Dec. 1907; and Journ. Stat. Soc., Dec. 1907). 6 Cf. below, par. 169. 7 Whitworth, Choice and Chance, question 126. 3 Whitworth, Exercises, No. 567. s According to the principle above enounced, par. 15. Io Bertrand, id. § 44, prob. xlvii. "Bertrand, id. § 39, prob. xliii. It is not to be objected that the probabilities on which the several expectations are calculated are not independent (above, par. 16). II man must possess some fortune, or its See also:equivalent, in order to live. To estimate now the value of a moral expectation. Suppose a See also:person whose fortune is a to have the chance p of obtaining a sum a, q of obtaining p, r of obtaining -y, &c., and let p+q+r+ . . . =I, only one of the events being possible. Now his moral expectation from the first chance—that is, the increment of his moral fortune multiplied by the chance—is pk I log a h as 1—7—log h = pk log (a +a) — pk log a. Hence his whole moral expectation is E=kp log (a+a)+kq log(a+#)+krlog(a+y)+. . . —k log a; and, if Y stands for his moral fortune including this expectation, that is, k log (a/h) + E, we have Y=kp log(a+a)+kq log(a+p)+. . . —k log h. To find X, the physical fortune corresponding to this moral one, we have Y =k log X —k log h. Hence X = (a+a)P(a+13)Q(a+7)r, and X —a will be the actual or physical increase of fortune which is of the same value to him as his expectation, and which he may reasonably accept in lieu of it. The mathematical value of the same expectation is2 pa+qR+ry+. . 70. Gambling and Insurance.—These formulae are employed, often with the aid of refined mathematical theorems, to demonstrate received propositions of See also:great practical importance: that in general gambling is disadvantageous, insurance beneficial, and that in speculative operations it is better to subdivide risks—not to " have all your eggs in one See also:basket." 71. These propositions may be deduced by the use of a formula which perhaps keeps closer to the facts: viz. that utility or satisfaction is a See also:function of material goods not definitely ascertainable, defined only by the conditions that the function continually in-creases with the increase of the variable, but at a continually decreasing See also:rate (and some additional postulate as to the See also:lower limit of the variable), say y=/, (x) (if x as before denotes physical fortune, and y the corresponding utility or satisfaction) ; where all that is known in general of-~ is that ,/'(x) is positive, >G"(x) is negative; and 1'(x) is never less, x is always greater than zero. Suppose a gambler whose (physical) fortune is a, to have the chance p obtaining a sum a and the chance q(=1— ) of losing the sum /3. If the See also:game is See also:fair in the usual sense ofp the term pa=q$. Accordingly the prospective psychical advantage of the party is pl,l,(a+a)+qga—$)=p'G(a+a) +q+G{a — (p/q)a), say ya. When a is zero the expression reduces to the first See also:state of the man, ¢(a), say yo. To compare this state with what it becomes by the gambling transaction, let a receive continually small increments of Aa. When a is zero the first See also:differential coefficient of (ya—yo), viz. pIP!'(a)—p','(a),=o. Also the second differential coefficient, viz. pIP"(a) +Q (a) , is negative, since by See also:hypothesis ¢" is continually negative. And as a continues to increase from zero,the second differential coefficient of (ya—yo), viz. (a+a)+4' (a+qa), continues to be negative. Therefore the increments received by the first differential coefficient of (ya—y0) are continually negative; and therefore (ya—yo) is continually negative; ya<yo for finite values of a (not exceeding qa/p).' 72. To show the advantage of insurance, let us suppose with Morgan Crofton 6 that a See also:merchant, whose fortune is represented by 1, will realize a sum a if a certain See also:vessel arrives safely. Let the probability of this be p. To make up exactly for the risk run by the insurance See also:company, he should pay them a sum (I —p)e. If he does, his moral fortune becomes, according to the formula now proposed (1 +pe), since his physical fortune is increased by the secured sum e, minus the See also:payment (1—p)e; while if he does not insure it will be p¢(1+e)+(I —p)/' (I). We have then to compare '(I+pe), say Yi, with ptk(I+s)+(I—p),p(I), say y2. BY reasoning analogous to that of the preceding paragraph it appears that (y2—yi) is zero when a=o and continually diminishes as e increases up to any assigned finite (admissible) value. Similarly it may be shown that it is better to expose one's fortune in a number of separate sums to risks independent of each other than 1 It is important to remark that we should be wrong in thus adding the expectations if the events were not mutually exclusive. For the mathematical expectations it is not so. 2 This paragraph is taken from Morgan Crofton's article on " Probability, in the 9th edition of the Ency. Brit. 2 Cf. See also:Marshall, Principles of See also:Economics, Mathematical Appendix, note ix. ' Or should we rather say, not exceeding the limit at which '(a—pa/q) becomes o ? (The value of I(o) may be regarded as —oo.) Neither of the proposed limitations materially affects the validity of the theorem. 6 Loc. cit. par. 25.to expose the whole to the same danger. Suppose a merchant, having a fortune, has besides a sum a which he must receive if a See also:ship arrives in safety. Then, if the chance of the ship arriving = p, and q = I— p, his prospective advantage is p¢ (1+e)+0 (I). Now instead of exposing the lump sum a to a single risk, let him subdivide a into n equal parts, each exposed to an independent equal risk (q) of being lost. As n is made larger 6 it be-comes more and more nearly a certainty that he will realize Pe out of the total e exposed to risk. Therefore his condition (in respect of the sort of advantage which is under consideration) will be approximately ,p(1+pe). Then we have to compare ¢ (1+pe), say yI, with p~(I+e)-{-q¢ (I), say Y2. By reasoning analogous to that which has been above employed—observing that (p—p2) 'G"(I) is negative for all possible values of p—we conclude that yz <yi. 73. The See also:Petersburg Problem.—The See also:doctrine of " moral fortune " was first formulated by Daniel Bernoulli 7 with reference to their celebrated " Petersburg Problem," which is thus stated by Todhunter3: " A throws a See also:coin in the See also:air: if head appears at the first throw he is to receive a shilling from B, if head does not appear until the second throw he is to receive 2s., if head does not appear until the third throw he is to receive 4s., and so on, required the expectation of A." So many lessons are presented by this problem that there has been See also:room for disputing what is the See also:lesson. Laplace and other high authorities follow Daniel Bernoulli. See also:Poisson finds the explanation in the fact that B could not be expected to pay up so large a sum. Whitworth, who regards the disadvantage of gambling as consisting mainly in the danger of becoming " cleaned out, '9 finds this moral in the Petersburg problem. All have not noticed what some regard as the See also:principal lesson to be obtained from the See also:paradox: viz. that a transaction which cannot be regarded as one of a series—at least a " cross-series " 10—is not subject to the general rule for expectations of advantage whether material or moral." Section IV.—Geometrical Applications. 74. Under this head occur some interesting illustrations of principles employed in the preceding sections; in particular of a priori probabilities and of the relation between probability and expectation. 75. Illustrations of a priori Probabilities.—The See also:assumption.' which has been made under preceding heads that the probability of certain alternatives is approximately equal appears to See also:rest on See also:evidence of much the same See also:character as the assumption which is made under this head that one point in a line, See also:plane or See also:volume is as likely to occur as another, under certain circumstances. Thus consider the proposition: if a given See also:area S is included within a given area A, the chance of a point P, taken at random on A, falling on S is S/A. In a great variety of circumstances such a See also:size can be assigned to the spaces, and ' taking at random " can be so defined that the proposition is more or less directly based on experience. The fact that the points of incidence are equally distributed in space is observed, or connected by inference with observation, in many cases, e.g. raindrops and molecules. There is a solid substratum of evidence for the premiss employed in the See also:solution of problems like the following: On a See also:chess-See also:board, on which the See also:side of every square is a, there is thrown a coin of See also:diameter b(b <a) so as to be entirely on the board, which may be supposed to have no border. What is the probability that the coin is entirely on one square?i2 The area on which the coin can fall is (8a—b)2. The portion of the area which is favourable to the event is 64 (a—b)2. There-fore the required probability is (a—b)2/(a—eb)2. 76. Random Lines.—Speculative difficulties recur when we have to define a straight line taken at random in a plane; for instance, in the following problem proposed by See also:Buffon.13 A See also:floor is ruled with equidistant parallel lines; a See also:rod, shorter than the distance between each pair, being thrown at random on the floor, to find the chance of its falling on one of the lines. The problem is usually solved as follows:
Let x be the distance of the centre of the rod from the nearest line, 0 the inclination of the rod to a perpendicular to the See also:parallels, 2a the See also:common distance of the parallels, 2C the length of rod; then, as all values of x and B between their extreme limits are equally probable, the whole number of cases will be represented by
f i,2dxd° =Ira. O. 1r/2
e See above, par. 25 (See also: 20. I2 Whitworth, xercises, No. 500. is Cf. Morgan Crofton, loc. cit. Now if the rod crosses one of the lines we must have c>x/See also:cos 0; so that the favourable cases will be measured by a/ 2 c cos e f~/2de u dx=2c. Thus the probability required is p=2c/7ra. It may be asked—why should we take the centre of the rod as the point where distance from the nearest line has all its values equally probable? Why not one extremity of the line, or some other point suited to the circumstances of See also:projection? Fortunately it makes no difference in the result to what point in the rod we assign this pre-See also:eminence. 77. The See also:legitimacy of the assumption obtains some verification from the success of a test suggested by Laplace. If a rod is actually thrown, as supposed in the problem, a great number of times, and the frequency with which it falls on one of the parallels is observed, that proportionate number thus found, say p, furnishes a value for the constant a. For ir ought to equal 2c/pa. The experiment has been made by Professor See also:Wolf of See also:Frankfort. Having thrown a See also:needle of length 36 mm. on a plane ruled with parallel lines at a distance from each other of 45 mm. 5000 times, he observed that the needle crossed a parallel 2532 times. Whence the value of v is deduced 3.1596, with a probable error' t •05. 78. More hesitation may be See also:felt when we have to define a random chord of a circle,' for instance, with reference to the question, what is the probability that a chord taken at random will be greater than the side of an equilateral triangle? For some purposes it would no doubt be proper to assume that the chord is constructed by taking any point on the circumference and joining it to another point on the circumference, the points from which one is taken at random being distributed at equal intervals around the circumference. On this understanding the probability in question would be i. But in other connexions, for instance, if the chord is obtained by the intersection with the circle of a rod thrown in random See also:fashion, it seems preferable to consider the chord as a case of a straight line falling at random on a plane. Morgan Crofton' himself gives the following definition of such a line: If an See also:infinite number of straight lines be drawn at random in a plane, there will be as many parallel to any given direction as to any other, all directions being equally probable; also those having any given direction will be disposed with equal frequency all over the plane. Hence, if a line be determined by the co-ordinates p, w, the perpendicular on it from a fixed origin 0, and the inclination of that perpendicular to a fixed See also:axis, then, if p, w be made to vary by equal infinitesimal increments, the series of lines so given will represent the entire series of random straight lines. Thus the number of lines for which p falls between p and p+dp, and w between w and w+See also:dw, will be measured by dpdw, and the integral rfdpdw, between any limits, See also:measures the number of lines within those limits. 79. Authoritative and useful as this definition is, it is not entirely See also:free from difficulty. It amounts to this, that if we write the equation of the random line x cos a+y See also:sin a—p=o, we ought to take a and p as those variables, of which, the equicrescent values are equally probable—the equiprobable variables, as we may say. But might we not also write the equation in either of the following forms (I) x/a+y/b—1=o, (2) ax-i-by — t = o, and take a and b in either system as the equiprobable variables? To be sure, if the equal distribution of probabilities is extended to infinity we shall be landed in the absurdity that of the random lines passing through any point on the axis of y a proportion differing infinitesimally from unity—l00%—are either (1) parallel or (2) perpendicular to the axis of x. But the See also:admission of infinite values will render any scheme for the equal distribution of probabilities absurd. If Professor Crofton's constant for example, becomes infinite, the origin being thus placed at an infinite distance, all the random chords intersecting a finite circle would be parallel! 80. However this may be, Professor Crofton's conception has the distinction of leading to a series of interesting propositions, of which specimens are here subjoined.' The number of random lines which meet any closed See also:convex See also:contour of length L is measured by L. For, taking 0 inside the contour, and integrating first for p, from o to p, the perpendicular on the tangent to the contour, we have fpdw; taking this through four right angles for w, we have ' As recorded by Czuber, Geometrische Wahrscheinlichkeiten, p. 90. 2 Cf. Bertrand, Calcul des probabilites, pp. 4 seq. The matter has been much discussed in the Educational Times. See Mathematical Questions . . . from the Educational Times [a reprint], See also:xxix. 17-20, containing references to earlier discussions, e.g. x. 33 (by Woolhouse). ' Loc. cit. § 75. 4 The whole of p. 787 of Morgan Crofton's article is often referred to, and parts of pp. 786, 788 are transferred here.by See also:Legendre's theorem on rectification, N being the measure of the number of lines, N =fo'rpd.=L.6 Thus, if a random line meet a given contour, of length L, the chance of its See also:meeting another convex contour, of length 1, See also:internal to the former is p=l/L. If the given contour be not convex, or not closed, N will evidently be the length of an endless See also:string, drawn tight around the contour. 81. If a random line meet a closed convex contour of length L, the chance of it meeting another such contour, See also:external to the former, is p= (X—Y)/L, where X is the length of an endless See also:band enveloping both contours, and See also:crossing between them, and Y that of a band also enveloping both, but not crossing. This may be shown by means of Legendre's integral above; or as FIG. I. follows: See also:Call, for shortness, N(A) the number of lines meeting an area A; N(A, A') the number which meet both A and A'; then (fig. 1) N (SROQPH) +N(S'Q'OR'P'H') = N (SROQPH +S'Q'OR'P'H') +N(SROQPH, S'Q'OR'P'H'), since in the first member each line meeting both areas is counted twice. But the number of lines meeting the non-convex figure consisting of OQPHSR and OQ'S'H'P'R' is equal to the band Y, and the number meeting both these areas is identical with that of those meeting the given areas t2, t2'; hence; X=Y+N(tl, 12'). Thus the number meeting both the given areas is measured by X—Y. Hence the theorem follows. } 82. Two random chords cross a given convex boundary, of length L, and area 12; to find the chance that their intersection falls inside the boundary. Consider the first chord in any position; let C be its length; considering it as a closed area, the chance of the second chord meeting it is 2C/L; and the whole chance of its coordinates falling in dp, du) and of the second chord meeting it in that position is 2C dpdu, = 22Cdpdw. L fJdpdw L But the whole chance is the sum of these chances for all its positions; . prob. =2L- fJCdpdw. Now, for a given value of w, the value of f Cdp is evidently the area 12; then, taking w from it to o, we have required probability =2a12L-2. The mean value of a chord drawn at random across the boundary is M ffCdpdw—~B ffdpdw L 83. A straight band of breadth c being traced on a floor, and a circle of See also:radius r thrown on it at random; to find the mean area of the band which is covered by the circle. (The cases are omitted where the circle falls outside the band.)" If S be the space covered, the chance of a random point on the circle falling on the band is p = M (S)/ve', this is the same as 6 This result also follows by considering that, if an infinite plane be covered by an infinity of lines drawn at random, it is evident that the number of these which meet a given finite straight line is proportional to its length, and is the same whatever be its position. Hence, if we take 1 the length of the line as the measure of this number, the number of random lines which cut any See also:element ds of the contour is measured by ds, and the number which meet the See also:con-tour is therefore measured by 4L, half the length of the boundary. If we take 21 as the measure for the line, the measure for the contour will be L, as above. Of course we have to remember that each line must meet the contour twice. It would be possible to rectify any closed See also:curve by means of this principle. Suppose it traced on the See also:surface of a circular disk, of circumference L, and the disk thrown a great number of times on a system of parallel lines, whose distance asunder equals the diameter, if we See also:count the number of cases in which the closed curve meets one of the parallels, the ratio of this number to the whole number of trials will be ultimately the ratio of the circumference of the curve to that of the circle. [Morgan Crofton's note.] 4 Or the floor may be supposed painted with parallel bands, at a distance asunder equal to the diameter; so that the circle must fall on one. 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)-'rr'. 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 St and circumference L be thrown on the band, instead of a circle, the mean area covered is M (S) = re (L +See also:rc)-'S2. For as before, fixing the random point at A, the chance of a random point in S2 falling on the band is p=2r. =c/L', where L' is the perimeter of a parallel curve to L, at a normal distance 2c from it. Now L'=L+2r. lc. M(S) — rc — L+re. 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 See also:middle 0 of the line," 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 3' measured by 2r. la, 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 ra. 85. To investigate the probability that the inclination of the line joining any two points in a given convex area S2 shall See also: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 B 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 0 and B+dB is measured by l(See also:r2+r'2)d0. 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 0 to o+dB is (first considering A to lie in the element drdp) /' ,dpd& f0C(r2+r'2)dr = IC'dpdo. Let p be the perpendicular on C from a given origin 0, and let w be the inclination of p (we may put de) for do), C will be a given function of p, w; and, integrating first for w constant, the whole number of cases for which w falls between given limits w', w" is if: df C3dp; the integral f 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 See also:double integral, which, of course, is generally difficult enough; we may, however, deduce from it a remarkable result; for, if the integral iffC3dpdw 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 522; hence ffC3dpdw=3122, the integration extending to all possible positions of the chord C,—its length being a given function of its co-ordinates p. w.2 The line might be anywhere within the circle without altering the question. This integral was given by Morgan Crofton in the Comptes rendus (1869), p. 1469. An See also:analytical See also:proof was given by Serret, Annaks stied. de 1'ecole normale (1869), p. 177. See also:Cox. Hence if L, S2 be the perimeter and area of any closed convex contour, the mean value of the See also:cube of a chord drawn across it at random is 3S2'/L. 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 See also:outer (e.g. two concentric similar ellipses) ; let the See also: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:-- If there be any convex arc AB (fig. 6), and if NI be (the measurr of) the number of random lines which meet it once, N2 the number which meet it twice, A ^ g 2 arc AB = N,+2N2• FIG. 6. For draw the chord AB; the number of lines meeting the convex figure so formed is NI+N2=arc+chord (the perimeter) ; but Ni = number of lines meeting the chord= 2 chord; :. 2 arc + Ni = 2N2 + 2N2, 2 arc = N,+2N2. Now See also:fix the point X, in fig. 5, and draw XA, XB. If a random line cross the boundary L, and pi be the probability that it meets the arc AB once, p2 that it does so twice, 2AB/L = pi +2P2 ; and if the point X range all over the annulus, and p2 are the same probabilities for all positions of X, 2M (AB) /L = +2f)2. Let now IK (fig. 7) be any position of the random line; drawing tangents at I, K, FIG. 7. it is easy to see that it will cut the arc AB twice when X is in the space marked a, and once when X is in either space marked 0; hence, for this position of the line, pi+2P2=2(a+19)/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, ffsdxdy=LS. It has been shown (Phil. Trans., 1868, p. 191) that, if 0 be the See also:angle between the tangents XA, XB, f fodxdy = r (A—2S). The mean value of the tangent XA or XB may be shown to be M(XA)=SP/2A, where P=perimeter of See also:locus of centre of gravity of the segment S. 87. When we go on to See also: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. 4 the angle made with the axis X by a vector r in the plane XY, and 0 (or 2r—o) the angle made by the vector p with r in the plane containing both p and r and the axis Z) it seems natural to treat the angles ' and o as the equiprobable variables. In other words, if we take at random any See also:meridian on the See also:celestial globe and combine it with any right See also:ascension the vector joining the centre to the point thus assigned is a random line.' It is possible that for some purposes this conception may be appropriate. For many purposes surely it is proper to assume a more symmetrical distribu-. tion of the terminal points on the surface of a See also: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 See also:spheres in a " molecular See also:chaos."4 It is safe to say with Czuber, " No discussion can remove indeterminateness." Let us See also:hope with him that "though this See also:branch of probability can for the See also:present claim only a theoretic See also:interest, in the future it will perhaps also See also:lead to practical results."' 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 'Cf. Bertrand, op. cit. § 135. See e.g. See also:Watson, Kinetic Theory of Gases, p. 2; See also:Tait, Trans. See also:Roy. Soc., Edin. (1888), xxxiii. 68. 5 Wahrscheinlichkeitstheorie, p. 64. 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 lc; and the whole number are when HK meets a circle, centre 0, radius r+lc; hence the probability is 2r • IC c p—2r(r + Zc) -2r +G P, taken at random on A, falling on S, p=S/A. If now the as M' nearly =M. For example, suppose two points X, Y are 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 SI, S2, S3 ..., the chance of any one Si being taken, and of P falling on is pi =n-'Si/A; or now the whole probability p=P1+P2+P3+ . - at once to the above expression. The chance of two points falling on S is, in the same way, P= M (S2) /A?, and so on. In such a case, if the probability be known, the mean value follows, and See also:vice versa. Thus, we might find the mean value of the nth See also:power of the distance XY between two points taken at random in a line of length 1, by considering the chance that, if n more points are so taken, they shall all fall between X and Y. This chance is lt'I (XY)"/l" = 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) " = 2l" (n + I)-1(n +2)-1. A line 1 is divided into n segments by n-I 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"; hence the chance that this cccurs, however the line is divided, is n•l-"M (See also:abc ...). Now the whole number of different orders in which the whole 2n-I points may occur is (2n--I)!; 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-l)!. Hence the required mean value of the product is M(abc....)_ (n-I)! ln. (2n-1)! 89. Additional examples of the relation. between probability and expectation appear in the following series of propositions: (I) 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; Ma, M6 the same mean when both points are taken in A and both in B respectively; then (A+B)2M' =2ABM +A2M,+B2Mb. if the space A=B, 4M'=2M+M,+Mo; if, also, Ma=Mb, then 2M'= (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 M1 be the mean value when both points fall in the triangle ACD, and M2 the value when one falls in ACD and the other in See also:BCD; then 2M=M1+M2. But Mi=4,M; and M2=GG'C, where G, G' are the centres of gravity of ACD, BCD; hence M2=9ABC, and M=11ABC. (4) From this mean value we pass to probabilities. The chance that a new point Z falls on the triangle XYC is 2 ; and the chance that three points X, Y, Z taken at random form, with a vertex C, a re-entrant See also:quadrilateral, is 1-. 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 a, and M' be the same mean when the n points are taken on A+a, and M the same mean when one point falls on a and•the remaining n-I on A; then, the sum of all the cases being M'(A+a)", and this sum consisting of the cases (I) when all the points are on A, (2) when one is on a the others on A (as we may neglect all where two or more fall on a), we have M'(A+a)" = MA"-I-nMia"-1; :. (M'—M)A=nA(Mi—M), taken in a line of length 1, to find the mean value M of (XY)". If 1 receives an increment dl, ldM=2dl(Mi-M). Now Mi here= the mean nth power of the distance of a single point taken at random in l from one extremity of 1; and this is l"(n+I) (as is shown by finding the chance of n other points falling on that distance) ; hence ldM =2d1{l"(n+1)-1—M}; . ldM +2 Mdl =2(n+ 1)-i1ndl, l-1.d. M12=2(n+I) 1l"dl: M12=2(n+I)-1fl"+ldl2ln+2/(n+I)(n+2) +C ; :. N =21"/(n+1) (n+2), C being evidently o. 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+a, and p' the same when one point falls on a and the rest on A, then, since the numbers of favourable cases are respectively p'(A+a)", pA", npiaA"-1, we find (p'—p)A=na(pl—p). Hence if p' = p then Pi = 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 See also:ring to the circle, we have evidently 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 quadri- g lateral, if we add to the triangle a small band between the See also: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 1 4 ST2 1 -_ 1a z .5• That of its being convex is 3. 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 I; and accordingly M = -hA. 93. The preceding problem is a particular case of a more general problem investigated by See also: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 FIG. mean value of the triangle thus formed. 9. Adding a concentric ring a, we have since M': M as the areas of the circles, M' =M(A+a)/A. AMa/A=3a(M1—M); :.M=IM1, where MI is the value of M when one of the points is on the circumference. Take 0 fixed; we have to find the mean value of OXY (fig. 9). Taking (p, 0) (p', 8') as co,ordinates of X, Y, M1= (aa2)-2ffpdpdoffp'dp'do. (OXY) M1=(7r4a2) ffffapP sin (o-O')ppdpdp'dbdo' (20a4)-1. ff1r3r'3. sin (o-O')dedo', 01+02, where pi=prob. (WXYZ re-entrant), X, Y, Z in one triangle; P2 = do., X in one triangle, Y in the other, Z in either. But pi=t. Now to find P2; 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 easyy to see that WGG'=}ABC=];, putting ABC=1. Now if 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 p2=2M(IYW+IXW)=g. Hence the probability of a re-entrant quadrilateral is which leads 390 putting r = OH, r' = OK; as r = 2a sin 8, r' = 2a sin 0', * ,r B M,=-sa -- f 0 sin'B sin'8'sin (0-0')dOd9'. Professor Sylvester has remarked that this double integral, by means of the theorem i:1 f(x, y)dxdy=Pp f(a-y, a-x)dxdy, is easily shown to be identical with 2 f o f: sine sin'O' cos See also:Wade' = if: sin 'See also:ode = M, 36r' M =4girrz See also:rat. From this mean value we pass to the probability that four points within a circle shall form a re-entrant figure, viz. Additional information and CommentsThere are no comments yet for this article.
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