59.7. KOLMOGOROV’S INEQUALITY 1879

Now, summing these yields

ε2P(A)≤

∫Ω

XA

∣∣∣∣∣ n

∑j=1

X j

∣∣∣∣∣2

dP≤∫

Ω

∣∣∣∣∣ n

∑j=1

X j

∣∣∣∣∣2

dP

= ∑i, j

∫Ω

(Xi,X j)dP

By independence of the random vectors the mixed terms of the above sum equal zero andso it reduces to

n

∑i=1

∫Ω

|Xi|2 dP

This theorem implies the following amazing result.

Theorem 59.7.3 Let {Xk}∞

k=1 be independent random vectors having values in a separablereal Hilbert space and suppose E (|Xk|)< ∞ for each k and E (Xk) = 0. Suppose also that

∞

∑j=1

E(∣∣X j

∣∣2)< ∞.

Then∞

∑j=1

X j

converges a.e.

Proof: Let ε > 0 be given. By Kolmogorov’s inequality, Theorem 59.7.2, it followsthat for p≤ m < n

P

([max

m≤k≤n

∣∣∣∣∣ k

∑j=m

X j

∣∣∣∣∣≥ ε

])≤ 1

ε2

n

∑j=p

E(∣∣X j

∣∣2)≤ 1

ε2

∞

∑j=p

E(∣∣X j

∣∣2) .Therefore, letting n→ ∞ it follows that for all m,n such that p≤ m≤ n

P

([max

p≤m≤n

∣∣∣∣∣ n

∑j=m

X j

∣∣∣∣∣≥ ε

])≤ 1

ε2

∞

∑j=p

E(∣∣X j

∣∣2) .It follows from the assumption

∞

∑j=1

E(∣∣X j

∣∣2)< ∞

there exists a sequence, {pn} such that if m≥ pn

P

([max

k≥m≥pn

∣∣∣∣∣ k

∑j=m

X j

∣∣∣∣∣≥ 2−n

])≤ 2−n.