64.6. WIENER PROCESSES, ANOTHER APPROACH 2219

Let λ k ∈ H. Consider t0 < t1 < · · ·< tn.

E

(exp i

n

∑k=1

(λ k,W (tk)−W (tk−1))

)=

n

∏k=1

E (exp(i(λ k,W (tk)−W (tk−1))))? (64.6.38)

Start with the left. There are finitely many increments concerned and so it can be assumedthat for each k one can have m→ ∞ such that the partial sums up to m in the definition ofW (tk)−W (tk−1) converge pointwise a.e. Thus

E

(exp i

n

∑k=1

(λ k,W (tk)−W (tk−1))

)

= limm→∞

E

(exp i

n

∑k=1

(λ k,

m

∑j=1

(ψ j (tk)−ψ j (tk−1)

)Jg j

))

= limm→∞

E

(exp

n

∑k=1

m

∑j=1

i(

λ k,(

ψ j (tk)−ψ j (tk−1))

Jg j

))

= limm→∞

E

(m

∏j=1

exp

(n

∑k=1

i(

λ k,(

ψ j (tk)−ψ j (tk−1))

Jg j

)))

Now from 64.6.36,{

∑nk=1 i

(λ k,(

ψ j (tk)−ψ j (tk−1))

Jg j

)}m

j=1are independent. Hence

the above equals

= limm→∞

m

∏j=1

E

(exp

(n

∑k=1

i(

λ k,(

ψ j (tk)−ψ j (tk−1))

Jg j

)))

= limm→∞

m

∏j=1

E

(n

∏k=1

exp(

i(

λ k,(

ψ j (tk)−ψ j (tk−1))

Jg j

)))Now from independence of the increments for the ψ j, this equals

= limm→∞

m

∏j=1

n

∏k=1

E(

exp(

i(

λ k,(

ψ j (tk)−ψ j (tk−1))

Jg j

)))

= limm→∞

m

∏j=1

n

∏k=1

E(

exp(

i(λ k,Jg j)(

ψ j (tk)−ψ j (tk−1))))

= limm→∞

m

∏j=1

n

∏k=1

e−12 (λ k,Jg j)

2(tk−tk−1) = lim

m→∞

m

∏j=1

e−12 ∑

nk=1(λ k,Jg j)

2(tk−tk−1)

= limm→∞

exp

(−1

2

m

∑j=1

n

∑k=1

(λ k,Jg j)2 (tk− tk−1)

)

= exp

(−1

2

n

∑k=1

∑j=1

(J∗λ k,g j)2 (tk− tk−1)

)(64.6.39)

64.6. WIENER PROCESSES, ANOTHER APPROACH 2219Let A; € H. Consider to < t) < +++ < ty.k=1£ (oS eww) (t,1)) n) = TLE exp (i(Ax, W (t,) — W (th-1))))? (64.6.38)Start with the left. There are finitely many increments concerned and so it can be assumedthat for each k one can have m — © such that the partial sums up to m in the definition ofW (t.) — W (t,_1) converge pointwise a.e. ThusE (es y: (Ax, W(t.) —W a)k=1= limE (eid is (v;( tk) — Wj (th »)10)k=1= lim E (Ef di i (Aas (vj) — vi (te »)»))= lim (Teo (3) (24, (Wyte) = Wy (4 »)4«3)))j=!Now from 64.6.36, {yn i (Ax, (v, (tk) — W; J8i) be are independent. Hencethe above equals= Jim 1 Gaps (A (v( tk) )~vj(t-1))48))))~ gin FL (Teo (#(2e (vs (0) vst) 9)Now from independence of the increments for the y;, this equals= = Jim) nC (exp (i (Ax, (v, (th) — Wj (n-1)) Jei)))m sn= Jin TTT? (exe (4 (Audi) (vi (a) — v;(4-1))))= lim Ie }(Aglgj) (ete) — lim lle Syn (Agdej) (te—te-1)me k=l m—r001 m n= lim exp | —> y y (Ax,Jaj)° (te —te-1)m—yoo 2 iIO= exp & y L (J*AK gj)” oo) (64.6.39)k=1 j=l