59.13. CHARACTERISTIC FUNCTIONS IN BANACH SPACE 1899

It suffices to consider only the case where each Ek = E. This is because you can con-sider each X j to have values in ∏

nk=1 Ek by letting X j take its values in the jth component

of the product and 0 in the other components. Can you draw the conclusion the randomvariables are independent? By Theorem 59.5.1, it suffices to show the random variables{

gmk ◦Xk}n

k=1 are independent where gmk =(x∗1, · · · ,x∗mk

)∈ (E ′)mk . This happens if when-

ever tmk ∈ Rmk and

P =n

∑k=1

tmk ·(gmk ◦Xk

),

it follows

E(eiP)= n

∏k=1

E(

eitmk ·(gmk◦Xk)). (59.13.21)

However, the expression on the right in 59.13.21 equals

n

∏k=1

E(

ei(tmk ·gmk)◦Xk)

and tmk · gmk ≡ ∑mkj=1 t jx∗j ∈ E ′. Also the expression on the left equals E

(ei∑

nk=1 tmk ·gmk◦Xk

)Therefore, by assumption, 59.13.21 holds.

There is an obvious corollary which is useful.

Corollary 59.13.4 Let {Xk}nk=1be random variables such that Xk has values in Ek, a real

separable Banach space. Then the random variables are independent if and only if

E(eiP)= n

∏j=1

E(

eit∗j (X j))

where P≡ ∑nj=1 t∗j (X j) for t∗j ∈M j where M j is a dense subset of E ′j.

Proof: The easy direction follows from Theorem 59.13.3. Suppose then the aboveequation holds for all t∗j ∈M j. Then let t∗j ∈ E ′ and let

{t∗n j

}be a sequence in M j such that

limn→∞

t∗n j = t∗j in E ′

Then define

P≡n

∑j=1

t∗j X j, Pn ≡n

∑j=1

t∗n jX j.

It follows

E(eiP) = lim

n→∞E(eiPn)

= limn→∞

n

∏j=1

E(

eit∗n j(X j))

=n

∏j=1

E(

eit∗j (X j))