Maximal inequality and Large Deviations 6.1 Kolmogorov’s maximal inequality

Proof. To see L2-convergence, we will prove that (Sn)n>1 is Cauchy in L 2. It is enough to show that ||Sn − Sm||2 < ε for all n,m > N(ε). Suppose n > m, then we obtain ||Sn − Sm||2 = E(Sn − Sm) = E(Xm+1 +Xm+2 + · · ·+Xn) = Var(Xm+1 +Xm+2 + · · ·+Xn) = Var(Xm+1) + Var(Xm+2) + · · ·+ Var(Xn). For any ε > 0, there exists N = N(ε) with ∑∞ i=N Var(Xi) < ε 2, thus we have ||Sn−Sm||2 < ε for all n,m > N(ε) which implies that (Sn)n>1 is a Cauchy sequence in L2 and therefore converges in L2. Now, it remains to show a.s. convergence. In other words, we need to show P(Sn converges) = 1. Define MN = max{|Sn − Sm| : n,m > N}, then clearly MN is decreasing. Indeed, it is enough to