Course in Probability Theory

5 .3 CONVERGENCE OF SERIES 1 1 2 5We can now prove the "three series theorem" of Kolmogorov (1929) .Theorem 5 .3 .3 . Let IX,,) be independent r .v .'s and define for a fixed constantA > 0 :n (w), ifY n IX,, (w) I < A(w) - X0, if IXn (w)I > A .Then the series >n X n converges a .e . if and only if the following three seriesall converge :(i) E„ -OP{IXnI> A} _ En 9~1{Xn :A Yn},(11) E17 cr(Yll ),(iii) En o 2 (Yn ) .PROOF . Suppose that the three series converge . Applying Theorem 5 .3 .1to the sequence {Y n - d (Y,, )}, we have for every m > 1 :If we denote the probability on the left by /3 (m, n, n'), it follows from theconvergence of (iii) that for each m :lim lim _?(m, n, n') = 1 .n-aoo n'-ooThis means that the tail of >n {Y n - 't(Y ,1 )} converges to zero a .e ., so that theseries converges a .e . Since (ii) converges, so does En Y n . Since (i) converges,{X„ } and {Y n } are equivalent sequences ; hence En X„ also converges a .e . byTheorem 5 .2 .1 . We have therefore proved the "if' part of the theorem .Conversely, suppose that En X n converges a .e. Then for each A > 0 :° l'(IXn I> A i .o .) = 0 .It follows from the Borel-Cantelli lemma that the series (i) must converge .Hence as before E n Y,, also converges a .e . But since I Y n - (r (Y n ) I< 2A, wehave by Theorem 5 .3 .2j=nmaxn