the n-th places, whose total length is equal to 1/2 n + 1 , <strong>and</strong> thus tends to 0 together with n,would have differed from zero <strong>and</strong> the function F(z), contrary to the dem<strong>and</strong> stated, wouldnot have been continuous in the narrow sense.Without dwelling in more detail on the investigation of the connection between theproperties of continuous functions F(x) with the probabilities of the various binary (ordecimal) digits of the number x, I pass on to considering discontinuous functions.3.2.6. Arithmetization of Totalities of the Fourth Type by Discontinuous FunctionsWe saw that F(z) is an arbitrary monotone function obeying the condition F(1) – F(0) = 1.It is known from the theory of functions that, if F has points of discontinuity, i.e., such pointsa whereF(a + 0) – F(a – 0) > 0,their totality is countable.I showed that the left side of this inequality is the limit of the probability of the inequalities(26). Since the proposition x = a is the combination of all the propositions (26), it will, on thestrength of the generalized addition (multiplication) theorem, have probabilityh o = F(a + 0) – F(a – 0).Let us isolate all the countable totality of the points of discontinuity, a 1 , a 2 , …, a n , …Denoteh n = F(a n + 0) – F(a n – 0),<strong>and</strong> compile such a function F 1 (z), for which the same equality will hold <strong>and</strong> the sum of itsvariations at all the other points will be zero. Then the functionF 2 (x) = F(x) – F 1 (x)will be continuous 31 .First suppose thatF 2 (x) = 0, i.e., that h n = 1, 1 n < .Here, we have only a finite or a countable totality of elementary propositions x = a 1 , x = a 2 ,… <strong>and</strong> of all of their possible joins, – that is, a totality of the fourth type. Any propositiona < x < b (33)has probability equal to the sum of the probabilities of all the elementary propositions x = a nsatisfying the inequalities (33). It is self-evident that the symbols < <strong>and</strong> are now equivalentonly if they are applied to values differing from {those at} the points of discontinuity.This case most essentially differs from the instance of a continuous F(x). Here, generally,the consecutive digits are not only not independent, but, after a finite number of them isgiven, all the infinite set of the other ones is determined with probability approachingcertainty. Indeed, the probability of any value is represented by a convergent infinite productwhose consecutive multipliers are all rapidly tending to unity. We see thus that somearithmetization of geometric totalities changes them into totalities of either the second or thefourth type. If both F 1 (x) <strong>and</strong> F 2 (x) differ from zero, we have a mixed or general case of a
totality of the second type (§3.1.5), which is easily reduced to a totality of the fourth type,<strong>and</strong> a simple totality of the second type.In Chapter 4 I shall return to the considerations that guide us when arithmetizing totalities.Here, it is appropriate to note that difficulties <strong>and</strong> contradictions appear, because, whenestablishing a certain arithmetizing function F(x), we keep at the same time to intuitivenotions incompatible with it. For example, recognizing that F(x) is continuous, we find itdifficult to imagine that the possibility of a definite proposition x = a is incompatible withour assumption; <strong>and</strong> that, when admitting that possibility, we ought to make a a point ofdiscontinuity of F(x). But it is hardly needed to say that such contradictions between intuitive<strong>and</strong> logical conclusions are rather usual in mathematics, <strong>and</strong> that they cannot be resolved bysome compromise such as “not any proposition of an infinite totality having probability zerois impossible”. Thus, in the theory of functions, we are not embarrassed by the contradictionbetween our intuitive notion of a curve <strong>and</strong> the existence of continuous functions lacking aderivative; <strong>and</strong> it will certainly never occur to anybody to assume, that a continuous functionis absolutely arbitrary, <strong>and</strong> to consider, at the same time, a tangent at some point of the curvedepicting that function.Having any absolutely arbitrary totality of the fourth type of any cardinal number as atotality of all the points of a segment, <strong>and</strong> of all of their possible joins, we will alwayspreserve, after its arithmetization, only a countable totality of elementary propositions, <strong>and</strong>we will be obliged to consider the other elementary propositions impossible. Indeed, therecannot be more than one elementary proposition with probability higher than 1/2; or morethan two of them having probabilities exceeding 1/3 etc.The choice of the elementary propositions which should be considered possible, is in manycases an unsolvable problem. Indeed, who, for example, will be able to indicate thatcountable totality of the points of a segment, which anybody at all had already indicated orchosen, or will indicate or choose (as, for example, 1/2, or 1/2, or ln2, etc)? Nevertheless, itis obvious that this totality is countable 32 whereas all the other numbers ought to beconsidered impossible, because they never were, <strong>and</strong> never will be actually realized, <strong>and</strong>,consequently, cannot be realized. This inability is proper; in accord with the dem<strong>and</strong>s ofexperience, practice in most cases compels us to ab<strong>and</strong>on the attempts to arithmetizetotalities of the fourth type, <strong>and</strong> to replace them by those of the second type, naturallywithout violating the principles of the theory.Here is the usual reasoning: When having two equal {congruent} finite segments, theprobabilities that a definite number is contained in either of them are equal. However, thisconsideration is not quite rigorous. The less is the length of the intervals, the moreconsiderable is the inaccuracy of that assumption which cannot be absolutely admitted sinceit would have led us to an arithmetizing function F(z) = z, incompatible, as I showed above,with the realization of definite equalities x = a. On the contrary, when considering ourarithmetization as only approximate; when assuming that the probabilities are not equal butdiffering one from another less than by some very small but not exactly known number , weshould remember that our arithmetization is relatively the less satisfactory the smaller are thesegments (so that, in particular, the probability of the equality x = a is not always equal tozero). We have thus solved the paradox consisting in that, for the arithmetizing function F(z)exactly equal to z, all the totality of the never realizable 33 (impossible) numbers withmeasure 1 would have probability 1 (equal to certainty).When arithmetizing a totality of the fourth type, the issue of determining the probabilitiesof the so-called non-measurable totalities of points should also be indicated. For us, thisproblem does not present difficulties, because, after choosing the arithmetizing function, –that is, after selecting the countable totality of elementary propositions, – any totality ofpoints, whether measurable or not, acquires a probability on the strength of the generalized
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of All Countries and to the Entire
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(Coll. Works), vol. 4. N.p., 1964,
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individuals of the third class, the
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From the theoretical point of view
- Page 13 and 14: Second case: Each crossing can repr
- Page 15 and 16: On the other hand, for four classes
- Page 17 and 18: f i = i S + i , i = 1, 2, 3, 4, (
- Page 19 and 20: f 1 = C 1 P(f 1 ; …; f n+1 ), C 1
- Page 21 and 22: ut in this case f = 2 , f 1 = 2 ,
- Page 23 and 24: I also note the essential differenc
- Page 25 and 26: A 1 23n1 + 1 A 1 A 1 … A 11A 2 A
- Page 27 and 28: coefficient of 2 in the right side
- Page 29 and 30: h(A r h - c h A r 0 ) = - A r0we tr
- Page 31 and 32: Notes1. Our formulas obviously pres
- Page 33 and 34: Bernstein’s standpoint regarding
- Page 35 and 36: Corollary 1.8. A true proposition c
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- Page 43 and 44: propositions (B i and C j ) can be
- Page 45 and 46: A ~ A 1 and B = B 1 , we will have
- Page 47 and 48: included in a given totality as equ
- Page 49 and 50: For unconnected totalities we would
- Page 51 and 52: proposition given that a second one
- Page 53 and 54: On the other hand, let x be a parti
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- Page 65 and 66: x = /2 + /(23) + … + /(23… p n
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- Page 79 and 80: |(x 1 ; t 0 ; t 1 ) - 1 t0tf(t)dt|
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second, it is not based on assumpti
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experimentation and connected with
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Russian, and especially of the Sovi
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station in England. This book, as h
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Uspekhi Matematich. Nauk, vol. 10,
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variety and detachment of those lat
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46. On the distribution of the regr
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119. On the Markov method of establ
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No lesser difficulties than those e
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Separate spheres of work considerab
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10. Anderson, O. Letters to Karl Pe
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Hier sind, im Allgemeinen, ganz ana
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Jedenfalls, glaube ich erwiesen zu
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werde ich das ganze Material in kur
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considered as the limiting case of
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and, inversely,] = m ...1 2 N[ ch h
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µ 2 2 = m 2 2 - 2m 2 m 1 2 + m 1 4
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(x k - x k+1 ) … (x k - x +) = E(
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the thus obtained relations as pert
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[1/S(S - 1)(S - 2)][(Si = 1Sx i ) 3
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( N −1)((S − N )(2NS− 3S− 3
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µ 5 + 2µ 2 µ 3 = U [S/S] 5 + 2U
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case, the same property is true wit
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It follows that the question about
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then expressed my doubts). And Gned
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For Problem 1, formula (7) shows th
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Let us calculate now, by means of f
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ϕ′1(x)1E(a|x 1 ; x 2 ; …; x n
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Theorem 3. If the prior density 3
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P( ≤ ≤ |, 1 , 2 , …, s )
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6. A Sensible Choice of Confidence
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0 = A 0 n, = B2, = B2, 0 = C 0 n
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Note also that (95),(96), (83),(85)
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Γ(n / 2)Γ [( n −1) / 2]k = (1/2
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f (x 1 , x 2 , …, x n ) = 1 if x
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and the probability of achieving no
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E = kEµ. (14)In many particular ca
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a = np, b = np 2 = a 2 /n, = a/nand
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with number (2k - 2), we commit an
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(67)which is suitable even without
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" = 1/[1 - e - ], = - ln [1 - (1/
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Such structures are entirely approp
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11. As a result of its historical d
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exaggeration towards a total denial