1878 CHAPTER 59. BASIC PROBABILITY

Written more succinctly, ∣∣∣∣∣ n

∑j=1

X j

∣∣∣∣∣2

=

∣∣∣∣∣ k

∑j=1

X j

∣∣∣∣∣2

+ ∑j>k or i>k

(Xi,X j)

Now multiply both sides by XAk and integrate. Suppose i ≤ k for one of the terms in thesecond sum. Then by Lemma 59.3.4 and Ak ∈ σ (X1, · · · ,Xk), the two random vectorsXAk Xi,X j are independent,∫

Ω

XAk (Xi,X j)dP =

(∫Ω

XAk XidP,∫

Ω

X jdP)= 0

the last equality holding because by assumption E (X j) = 0. Therefore, it can be assumedboth i, j are larger than k and

∫Ω

XAk

∣∣∣∣∣ n

∑j=1

X j

∣∣∣∣∣2

dP =∫

Ω

XAk

∣∣∣∣∣ k

∑j=1

X j

∣∣∣∣∣2

dP

+ ∑j>k,i>k

∫Ω

XAk (Xi,X j)dP (59.7.10)

The last term on the right is interesting. Suppose i > j. The integral inside the sum is of theform ∫

Ω

(Xi,XAk X j

)dP (59.7.11)

The second factor in the inner product is in

σ (X1, · · · ,Xk,X j)

and Xi is not included in the list of random vectors. Thus by Lemma 59.3.4, the two randomvectors Xi,XAk X j are independent and so 59.7.11 reduces to(∫

Ω

XidP,∫

Ω

XAk X jdP)=

(0,∫

Ω

XAk X jdP)= 0.

A similar result holds if j > i. Thus the mixed terms in the last term of 59.7.10 are all equalto 0. Hence 59.7.10 reduces to

∫Ω

XAk

∣∣∣∣∣ n

∑j=1

X j

∣∣∣∣∣2

dP =∫

Ω

XAk

∣∣∣∣∣ k

∑j=1

X j

∣∣∣∣∣2

dP

+∑i>k

∫Ω

XAk |Xi|2 dP

and so ∫Ω

XAk

∣∣∣∣∣ n

∑j=1

X j

∣∣∣∣∣2

dP≥∫

Ω

XAk

∣∣∣∣∣ k

∑j=1

X j

∣∣∣∣∣2

dP≥ ε2P(Ak) .