kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

µ 2 = U [S/S] 2 = [(S – 1)/S] 2 , µ 3 = U [S/S] 3 = [(S – 1)(S – 2)/S 2 ] 3 ,µ 4 = U [S/S] ( S −1)(S − 2)( S − 3) ( S −1)(2S− 3)4 = 34 + (34 + 3 22 ),SS2µ 2 = U [S/S] 2( S −1)(S − 2)( S − 3) ( S −1)22 = 322 + (3 4 + 3 22 )SSso that 2 = [S/(S – 1)]µ 2 , 3 = 4 = 22 =3Sµ 4 –( S −1)(S − 2)( S − 3)3Sµ 2 2 –( S −1)(S − 2)( S − 3)2Sµ 3 ,( S −1)(S − 2)S(2S− 3)(µ 4 + 3µ 2 2 ),( S −1)(S − 2)( S − 3)S(µ 4 + 3µ 2 2 ).( S − 2)( S − 3)We thus derive all the already known to us formulas for the transition from a uniformsystem to its special case, the layout of the unreturned ticket. Making use of the relation (13),we obtain without difficulties the sought expressions for U 2 [N/S] , U 3 [N/S] , U 4 [N/S] and U 22 [N/S] .Another example. In the scheme of the returned ticket we have (cf. [3, p. 186, formula (7)]U ]N[ 2( N −1)(N − 2)( N − 3)( N − 4) 5( N −1)(N − 2)5 = µ45 +4NN(µ 5 + 2µ 2 µ 3 )and, as it follows, for a uniform system,U [N] 2( N −1)(N − 2)( N − 3)( N − 4) 5( N −1)(N − 2)5 = 45 +4NN( 5 + 2 23 ).When applying the expectation of U 23 (N) we obtain for the totality of trials (cf. Note 11)U ]N[ 5 + 2 U ]N[ ( N −1)(N − 2)23 = (µ25 + 2µ 2 µ 3 )Nand, consequently, for a uniform system,U [N] 5 + 2 U [N] ( N −1)(N − 2)23 = (25 + 2 23 ).NNoting that, on the other hand, when N = S,U 5 [S/S] = µ 5 , U 23 [S/S] = µ 2 µ 3 ,we obtain, analogous to the above,µ 5 = U [S/S] 2( S −1)(S − 2)( S − 3)( S − 4) ( S −1)(S − 2)5 = 45 + 54SS( 5 + 2 23 ),