kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

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1in which K(x; y) is positive and K(x; y) dy = 10has only one solution (up to a constant factor). If, however, the equation11 1 K(x; y)(x) dx = K(x; x 1 )K(x 1 ; y)(x) dx dx 100 0is satisfied by any positive and integrable function (x), then K(x, y) is a function of y only.Theorem B. If the equation1 11 1 K(x, y; z)(x)(y) dx dy = K(x; y; z) 1 (x) 1 (y) dx dy0 00 0is satisfied by any positive function (x) obeying the condition10and(x) dx = 11 1 1 (u) = K(x; y; u)(x)(y) dx dy0 0with a positive function K(x; y, z) symmetric with respect to x and y and such that10K(x; y; z) dz = 1,then K(x ; y ; z) is a function of z only.Without dwelling in more detail on the case n = or on its connection with the theory ofintegral equations, we shall consider now the next important case of a finite number ofclasses.Chapter 310. Suppose that there are in all N = n + 2 classes with two of them being pure races. Torepeat (cf. Note 3), each of these two produces, under internal crossing, only its ownindividuals, and, when being mutually crossed, gives rise to individuals of all the other(hybrid) classes. In accord with §6 we would have had Mendelian heredity if the entiretotality of the hybrids represented a class. We shall see now that if these hybrids representseveral classes, two possibilities should be distinguished from each other:1) Under internal crossing each of the hybrid classes produces individuals of one of thetwo pure classes.2) There exists a hybrid class, which, under the same condition, cannot produceindividuals of those two classes.Denote the functions of reproduction for our N classes by f and f 1 for the pure races and by i , i = 1, 2, …, n for the hybrid races, and the respective probabilities by , and i .Then ourmain assumption means that all the quadratic forms i have terms containing but that theydo not include 2 or 2 . On the contrary, the form f contains 2 (with coefficient 1) and doesnot include either or 2 and f 1 contains 2 (with coefficient 1)but does not include either or 2 .It is not difficult to prove, first of all, that in this case f does not at all depend on , nordoes f 1 depend on ; in other words, that crossing with one of the parents belonging to a purerace never produces an individual of the other pure race. Indeed, let us assume that initially i= 0 for all values of i; then, because of the principle of stationarity,( + ) 2 f = f 2 + fA i i + A ik i k + f 1 D i i ,
ut in this case f = 2 , f 1 = 2 , and 1 = 2c 1 where c 1 > 0. Since is not included in the leftside in a degree higher than the second, D i = 0 for all values of i which confirms the above.11. Before going on to the proof of the general proposition, we dwell for the sake ofgreater clearness on the case N = 4. The general statement will be its direct generalizationdemanding some additional essential considerations.Theorem. For N = 4 the formulas of reproduction should have one of the two followingforms: eitherf =[ + (1/2) A 1 1 + (1/2)A 2 2 ] 2 , f 1 = [ + (1/2)B 1 1 + (1/ 2)B 2 2 ] 2 , i = 2c i [ + (1/2)A 1 1 + (1/2)A 2 2 ][ + (1/2)B 1 1 + (1/2)B 2 2 ], i = 1, 2, (45)where c 1 + c 2 = 1, A 1 + B 1 = A 2 + B 2 = 2, A 1 c 1 + A 2 c 2 = 1. Or,f = ( + 1 )( + 2 ), f 1 = ( + 1 )( + 2 ), (46) i = ( + i )( + i ), i = 1, 2.Indeed, let us assume at first that there exists an identical dependencec 2 1 = c 1 2 (47)between 1 and 2 . Then, supposing from the very beginning that c 2 1 = c 1 2 , we may uniteboth hybrid classes in one so as to obtain a biotype of three classes that must obey theMendelian law. Consequently,f(; ; c 1 ; c 2 ) = ( + /2) 2 , f 1 (; ; c 1 ; c 2 ) = ( + /2) 2 .Therefore, assuming thatf = 2 + A i i + A ik i k , f 1 = 2 + B i i + B ik i k ,we find thatA 1 c 1 + A 2 c 2 = B 1 c 1 + B 2 c 2 = c 1 + c 2 = 1,A ik c i c k = B ik c i c k = (1/4)(c 1 + c 2 ) 2 . (48)But, forming the equation of stationarity for f, we obtainff 1 = f[(A 1 – 1) 1 + (A 2 – 1)] + A ik i k , (49)and, applying equalities (48) and the relation (47), we conclude thatff 1 = (1/4)( 1 + 2 ) 2 .It follows that f and f 1 should be exact squares and we immediately arrive at formulas (45).Let us suppose now that, on the contrary, there is no identical proportionality between thefunctions 1 and 2 . Then any dependence between the functions of reproduction shouldcontain at least three of them. We have seen, however (§10), that there exists an infinite setof stationary conditions, under which relation (47) holds with 2c 1 and 2c 2 being thecoefficients of in 1 and 2 respectively, and satisfying the equation
Page 3: of All Countries and to the Entire
Page 6 and 7: (Coll. Works), vol. 4. N.p., 1964,
Page 8 and 9: individuals of the third class, the
Page 10: 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: f 1 = C 1 P(f 1 ; …; f n+1 ), C 1
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
Page 37 and 38: It is important to indicate that al
Page 39 and 40: ut for the simultaneous realization
Page 41 and 42: devoid of quadratic divisors and re
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
Page 55 and 56: totality is perfect, but that the j
Page 57 and 58: In this case, all the finite or inf
Page 59 and 60: probabilities p 1 , p 2 , … respe
Page 61 and 62: where x is determined by the inequa
Page 63 and 64: totality of the second type (§3.1.
Page 65 and 66: x = /2 + /(23) + … + /(23… p n
Page 67 and 68: that the fall of a given die on any
Page 69 and 70: infinitely many digits only dependi
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10. (§2.1.5). Such two proposition
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F(x + h) - F(x) = Mh, therefore F(x
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“confidence” probability is bas
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x1+ Lp n (x) x1− Lx1+ Lf(t)dt < x
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|(x 1 ; t 0 ; t 1 ) - 1 t0tf(t)dt|
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5. The distribution ofξ , the arit
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P(x 1i < x) = F(x; a i ) = C(a i )
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egards his promises. Markov shows t
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other solely and equally possible i
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notion of probability and of its re
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However, already in the beginning o
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the revolution. My main findings we
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Nevertheless, Slutsky is not suffic
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path that would completely answer h
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on political economy as well as wit
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scientific merit. Borel was indeed
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[3] Already in Kiev Slutsky had bee
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different foundation. The difficult
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5. On the criterion of goodness of
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--- (1999, in Russian), Slutsky: co
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Here also, the author considers the
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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
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kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

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