68.1. HERMITE POLYNOMIALS 2319

Proof: Replace Hn with Kn which really are the coefficients of the power series andthen show Kn = Hn. Thus

exp(

tx− 12

t2λ

)=

∞

∑n=0

Kn (x,λ ) tn

Then K0 = 1 = H0 (x) . Also K1 (x) = x = H1 (x) .

∂

∂ t

(exp(

tx− 12

t2λ

))= exp

(tx− 1

2t2

λ

)(x− tλ )

=∞

∑n=0

xKn (x,λ ) tn−∞

∑n=0

λKn (x,λ ) tn+1 =∞

∑n=0

xKn (x,λ ) tn−∞

∑n=1

λKn−1 (x,λ ) tn

Also,

∂

∂ t

(exp(

tx− 12

t2λ

))=

∞

∑n=1

nKn (x,λ ) tn−1 =∞

∑n=0

(n+1)Kn+1 (x,λ ) tn

It follows that for n≥ 1,

(n+1)Kn+1 (x,λ ) = xKn (x,λ )−λKn−1 (x,λ )

Thus the first two K0,K1 coincide with H0 and H1 respectively. Then since both Kn and Hnsatisfy the recursion relation 68.1.5, it follows that Kn = Hn for all n.

The first version is just letting λ = 1 in the second version.There is something very interesting about these Hermite polynomials Hn (x,λ ) . Let W

be the real Wiener process. Consider the stochastic process Hn (W (t) , t) ,n≥ 1. This endsup being a martingale. Using Ito’s formula, the easy to remember version of it presentedabove, and the above properties of the Hermite polynomials,

dHn = Hnx (W (t) , t)dW +Hnt (W (t) , t)dt +12

Hnxx (W (t) , t)dW 2

= Hn−1 (W (t) , t)dW − 12

Hnxx (W (t) , t)dt +12

Hnxx (W (t) , t)dt

Note that if n < 2, both of the last two terms are 0. In general, they cancel and so

dHn = Hn−1 (W (t) , t)dW

and soHn (W (t) , t) = Hn (W (0) ,0)+

∫ t

0Hn−1 (W (t) , t)dW

now the constant term in the above equation is F0 measurable and the stochastic integral isa martingale. Thus this is indeed a martingale assuming everything is suitably integrable.However, this is not hard to see because these Hn are just polynomials. It was shown inTheorem 64.1.3 that W (t) ∈ Lq (Ω) for all q. Hence there is no integrability issue in doing