2314 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATION

Example 68.0.3 You could let H = R and let ξ be normally distributed density with mean0 and variance 1. Then let W (a)≡ ξ a. Does it work?

E(

abξ2)= abE

(ξ

2)= ab

Is W (a) normally distributed? E(

eiaξ t)=∫R eiaxtξ dx = e−(1/2)a2t2

which is the charac-teristic function of a normally distributed random variable having mean 0 and variancea2. It is clear that any linear combination of aiξ is normally distributed and so the vector(a1ξ , · · · ,anξ ) is normally distributed. This is by Theorem 59.16.4.

The above implies W is actually linear.

E((W ( f +g)− (W ( f )+W (g)))2

)= E

((W ( f +g))2 +

[W ( f )2 +W (g)2 +2W ( f )W (g)

]−2 [W ( f +g)W ( f )+W ( f +g)W (g)]

)

= E((W ( f +g))2

)+E

(W ( f )2

)+E

(W (g)2

)+E (W ( f )W (g))

−2(E (W ( f +g)W ( f ))+E (W ( f +g)W (g)))

which from the above equals

| f +g|2 + | f |2 + |g|2 +2( f ,g)−2 [( f +g, f )+( f +g) ,g]

= 2 | f |2 +2 |g|2 +4( f ,g)−2[| f |2 + |g|2 +2( f ,g)

]= 0

Thus W ( f +g)− (W ( f )+W (g)) = 0. Is it true that

(W (a f )) = aW ( f )?

This is easier to show.

E((W (a f )−aW ( f ))2

)= E

(W (a f )2−2W (a f )aW ( f )+a2W ( f )2

)

= |a f |2−2aE (W (a f )W ( f ))+a2E(

W ( f )2)

= a2 | f |2−2a2 | f |2 +a2 | f |2 = 0

Thus W is indeed linear.

68.1 Hermite PolynomialsConsider

exp(

tx− t2

2

)= exp

(x2

2− 1

2(x− t)2

)