kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

f i = i S + i , i = 1, 2, 3, 4, (36)all the stationary solutions are determined by the common solution of equations (27), weconclude that all these solutions are determined by formulas (35) with the parameter k = 4 / 3taking all possible values from 0 to .There thus exist such positive values 4 / 3 that the other coordinates 1 / 3 and 2 / 3 determined by the equations (35) are also positive. Therefore, by continuously varyingthe parameter we can make at least one coordinate (for example, 4 / 3 ) vanish with the otherones being non-negative. Then, with 1 , 2 , 3 taking the respective positive values andreplacing 4 by zero, we note that (36.4) vanishes which is impossible because all thecoefficients there are positive.Let us now pass on to the general case and show by the same method that if the Theoremis valid for some n it holds for (n + 1). Indeed, if it is valid for n, the equations (36) with i =1, 2, …, n cannot include dependent equations 1 = 0, 2 = 0, …, n – 1 = 0when all the coefficients in f i are positive. Therefore, the similar equations 1 = 0, 2 = 0, …, n = 0,where f i are the same as in (36) but with i = 1, 2, …, (n + 1), cannot be connected by morethan one dependence; i.e., the stationarity condition for (n + 1) classes cannot depend onmore than one parameter.Consequently, the requirement thatkf n – f n+1 = 0,if only it does not hold identically for some k, leads to a restricted number of possible valuesfor f 1 , f 2 , …, f n , f n+1 . Therefore, uniting the n-th and the (n + 1)-th classes into one, andassuming that in the initial distribution n = /(1 + k), n+1 = k/(1 + k),the functions i = f i [ 1 ; 2 ; …; /(1 + k); k/(1 + k)], i = 1, 2, …, n – 1, n = f n [ 1 ; 2 ; …; /(1 + k); k/(1 + k)] +f n+1 [ 1 ; 2 ; …; /(1 + k); k/(1 + k)]if onlyF k = kf n [ 1 ; 2 ; …; /(1 + k); k/(1 + k)] –f n+1 [ 1 ; 2 ; …; /(1 + k); k/(1 + k)] = 0 (37)can take only a restricted number of values, and, owing to their continuity, have only onedefinite system of values 15 . It follows that if{the left side of} equation (37) is not an exactsquare, then i = i ( 1 + … + n–1 + ) 2 + µ i F k , i = 1, 2, …, n. (38)