kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

To compare, we adduce the elementary calculation {omitted} by means of formula (22):R 3 = 1 – (P 0 + P 1 + P 2 ) = 0.59503.Let us return now to justifying the formulas (27) and (30). They are derived from theexpansion of the probabilities into Charlier seriesP m = ∞k =0whereA k = e t mA k ∇ k (m; t) (42)∇ k (m; t) P m , ∇ 0 (m; t) = (m; t), (43; 44a)∇ k+1 (m; t) = ∇ k (m; t) – ∇ k–1 (m; t).(44b)The coefficients A k can also be represented askA k =i=0(– 1) i (t i /i!)[F k–1 /(k – 1)!] (45)where F s are the factorial moments of the random variable µ, i.e.,F 0 =mP m = 1, F 1 =mmP m = a, F s =mm(m – 1) … (m – s + 1)P m . (46)Formulas (42) – (46) are applicable not only to our special case in which the probabilitiesP m are represented by (22) but to any probabilities P m = P(µ = m) for an arbitrary randomvariable µ only taking a finite number of integer non-negative values m = 0, 1, 2, …, n 12 .The parameter t is here arbitrary. It is usually assumed that t = a, and then the formulas forthe first coefficients become somewhat more simple. Namely, if t = a, we will haveA 0 = 1, A 1 = 0, A 2 = (F 2 /2) – aF 1 + (a 2 /2),A 3 = (F 3 /6) – aF 2 /2 + (a 2 /2)F 1 – (a 3 /6).For our particular case, assuming thata = p 1 + p 2 + + … + p n , b = p 1 2 + p 2 2 + + … + p n 2 ,c = p 1 3 + p 2 3 + + … + p n 3 , d = p 1 4 + p 2 4 + + … + p n 4 ,we have, again for t = a,A 0 = 1, A 1 = 0, A 2 = – b/2, A 3 = – c/3, A 4 = – d/4 + (b 2 /8). (50)Since b, c, d, … are magnitudes of the order not higher than , 2 , 3 , … respectively, A 2 =O( ), A 3 = O( 2 ), A 4 = O( 2 ). It can be shown that, for any k ≥ 1,A 2k–1 = O( k ), A 2k = O( k ).A more subtle analysis shows that not only the terms of the series (42) having numbers (2k– 1) and 2k are, in our case 13 , of an order not higher thank ; the same is true with regard tothe sum of all the following terms. Thus, in our case, when curtailing our series by the term