2016 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

are independent ifE((

hi ◦X ′)(

g j ◦Y ′))

= 0

for all i, j. This is what I will show next.

E((

hi ◦X ′)(

g j ◦Y ′))

=14

E ((hi (X)+hi (Y ))(g j (X)−g j (Y )))

=14

E (hi (X)g j (X))− 14

E (hi (X)g j (Y ))

+14

E (hi (Y )g j (X))− 14

E (hi (Y )g j (Y )) (61.7.25)

Now from the above observation after the definition of Gaussian measure hi (X)g j (X)and hi (Y )g j (Y ) are both in L1 because each term in each product is normally distributed.Therefore, by Lemma 59.15.2,

E (hi (X)g j (X)) =∫

Ω

hi (Y )g j (Y )dP

=∫

Ehi (y)g j (y)dµ

=∫

Ω

hi (X)g j (X)dP

= E (hi (Y )g j (Y ))

and so 61.7.25 reduces to

14(E (hi (Y )g j (X)−hi (X)g j (Y ))) = 0

because hi (X) and g j (Y ) are independent due to the assumption that X and Y are indepen-dent. Thus

E (hi (X)g j (Y )) = E (hi (X))E (g j (Y )) = 0

due to the assumption that µ is symmetric which implies the mean of these random vari-ables equals 0. The other term works out similarly. This has proved the independence ofthe random variables, X ′ and Y ′.

Next consider the claim they have the same law and it equals µ . To do this, I will useTheorem 59.12.9 on Page 1897. Thus I need to show

E(

eih(X ′))= E

(eih(Y ′)

)= E

(eih(X)

)(61.7.26)

for all h∈ E ′. Pick such an h. Then h◦X is normally distributed and has mean 0. Therefore,for some σ ,

E(

eith◦X)= e−

12 t2σ2

.