750 CHAPTER 27. ANALYTICAL CONSIDERATIONS

Definition 27.5.8 Let µ be a probability measure on B (E) where E is a real sep-arable Banach space. Then for x∗ ∈ E ′,

φ µ (x∗)≡

∫E

eix∗(x)dµ (x) .

φ µ is called the characteristic function for the measure µ .

Note this is a little different than earlier when the symbol φ X (t) was used and X wasa random variable. Here the focus is more on the measure than a random variable X suchthat its distribution measure is µ . It might appear this is a more general concept but in factthis is not the case. You could just consider the separable Banach space or Polish spacewith the Borel σ algebra as your probabililty space and then consider the identity map asa random variable having the given measure as a distribution measure. Of course a majorresult is the one which says that the characteristic function determines the measures.

Theorem 27.5.9 Let µ and ν be two probability measures on B (E) where E is aseparable real Banach space. Suppose

φ µ (x∗) = φ ν (x

∗)

for all x∗ ∈ E ′. Then µ = ν .

Proof: It suffices to verify that µ (A) = ν (A) for all A ∈ K where K is the set ofcylindrical sets. Fix gn ∈ (E ′)n . Thus the two measures are equal if for all such gn, n ∈ N,

µ(g−1

n (B))= ν

(g−1

n (B))

for B a Borel set in Rn. Of course, for such a choice of gn ∈ (E ′)n , there are measuresdefined on the Borel sets of Rn µn and νn which are given by

µn (B)≡ µ(g−1

n (B)), νn (B)≡ ν

(g−1

n (B))

and so it suffices to verify that these two measures are equal. So what are their character-istic functions? Note that gn is a random variable taking E to Rn and µn, νn are just theprobability distribution measures of this random variable. Therefore,

φ µn(t)≡

∫Rn

eit·sdµn =∫

Eeit·gn(x)dµ

Similarly,

φ νn(t)≡

∫Rn

eit·sdνn =∫

Eeit·gn(x)dν

Now t ·gn ∈E ′ and so by assumption, the two ends of the above are equal. Hence φ µn(t) =

φ νn(t) and so by Theorem 27.1.6, µn = νn which, as shown above, implies µ = ν . ■

27.6 Independence in Banach SpaceI will consider the relation between the characteristic function and independence of randomvariables having values in a Banach space. Recall an earlier proposition which relatesindependence of random vectors with characteristic functions. It is proved starting on Page744.

750 CHAPTER 27. ANALYTICAL CONSIDERATIONSDefinition 27.5.8 Le: Lt be a probability measure on &(E) where E is a real sep-arable Banach space. Then for x* € E',oy (0) = [le Madu (a).Eis called the characteristic function for the measure [.On uNote this is a little different than earlier when the symbol ¢, (£) was used and X wasa random variable. Here the focus is more on the measure than a random variable X suchthat its distribution measure is LW. It might appear this is a more general concept but in factthis is not the case. You could just consider the separable Banach space or Polish spacewith the Borel o algebra as your probabililty space and then consider the identity map asa random variable having the given measure as a distribution measure. Of course a majorresult is the one which says that the characteristic function determines the measures.Theorem 27.5.9 Let u and v be two probability measures on B(E) where E is aseparable real Banach space. SupposePu (x") =O, (x")for allx* € E'. Then b= Vv.Proof: It suffices to verify that u (A) = v(A) for all A € “ where -% is the set ofcylindrical sets. Fix g,, € (E’)”. Thus the two measures are equal if for all such g,,n € N,u (gn! (B)) =v (gn! (B))for B a Borel set in R". Of course, for such a choice of g, € (E’)", there are measuresdefined on the Borel sets of R” 1, and Vv, which are given byH, (B) =U (gn! (B)), Vn(B) =v (ga (B))and so it suffices to verify that these two measures are equal. So what are their character-istic functions? Note that g,, is a random variable taking E to R” and U,, Vn are just theprobability distribution measures of this random variable. Therefore,Ou, (t) = [etean, _ [etoauSimilarly,by, (t)= / et Sdy, = | elt nO ayJR" ENow t-g,, € E’ and so by assumption, the two ends of the above are equal. Hence U, (t) =oy, (£) and so by Theorem 27.1.6, H,, = Vn which, as shown above, implies u = v.27.6 Independence in Banach SpaceI will consider the relation between the characteristic function and independence of randomvariables having values in a Banach space. Recall an earlier proposition which relatesindependence of random vectors with characteristic functions. It is proved starting on Page744.