2320 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATION

these things. Actually, Hn (W (0) ,0) = 0 To see this, note that E(

W (0)2)= (0,0)H = 0

and so W (0) = 0. Now it is not hard to see that Hn (0,0) = 0. Indeed,

exp(

tx− 12

t2λ

)=

∞

∑n=0

Hn (x,λ ) tn

Thus Hn (x,0) = ∑∞n=0 Hn (x,0) tn = exp(tx) = ∑

∞n=0

(tx)n

n! = ∑∞n=0 xn tn

n! and so for all n ≥1,Hn (0,0) = 0. Thus in fact, for n ≥ 1, t → Hn (W (t) , t) is a martingale which equals 0when t = 0.

68.2 A Remarkable Theorem, Hermite PolynomialsLemma 68.2.1 Say (X ,Y ) is generalized normally distributed and

E (X) = E (Y ) = 0,E(X2)= E

(Y 2)= 1.

Then for m,n≥ 0,

E (Hn (X)Hm (Y )) ={

0 if n ̸= m1n! (E (XY ))n if n = m

Proof: By assumption, sX + tY is normal distributed with mean 0. This follows fromTheorem 59.16.4. Also

σ2 ≡ E

((sX + tY )2

)= s2 + t2 +2E (XY )st

and so its characteristic function is

E (exp(iλ (sX + tY ))) = φ sX+tY (λ ) = e−12 σ2λ

2= e−

12 (s2+t2)λ

2e−E(XY )stλ 2

So let λ =−i. You can do this because both sides are analytic in λ ∈ C and they are equalfor real λ , a set with a limit point. This leads to

E (exp(sX + tY )) = e12 (s2+t2)eE(XY )st

Hence, multiplying both sides by e−12 (s2+t2),

e−12 (s2+t2)E (exp(sX + tY )) = E

(exp(

sX− s2

2

)exp(

tY − t2

2

))= exp(stE (XY ))

Now take ∂ n+m

∂ ns∂ mt of both sides. Recall the description of the Hermite polynomials givenabove

n!Hn (x) =dn

dtn exp(

tx− t2

2

)|t=0

Thus

E (n!Hn (X)m!Hm (Y )) =∂ n+m

∂ ns∂ mtexp(stE (XY )) |s=t=0