2318 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATION

This is easy to see from the observation that

∂x=

∂ (−x)(−1)

Thus if it involves n derivatives, you end up multiplying by (−1)n.Finally is the claim that

∂λHn (x,λ ) =−

12

∂ 2

∂x2 Hn (x,λ )

It is certainly true for n = 0,1,2. So suppose it is true for all k≤ n. Then from earlier claimsand induction,

(n+1)H(n+1)λ (x,λ ) = xHnλ (x,λ )−H(n−1) (x,λ )−λH(n−1)λ (x,λ )

= x(−12

)Hnxx−Hn−1 +λ

12

H(n−1)xx = x(−12

)Hn−2−Hn−1 +λ

12

H(n−3)

=−12(xHn−2−λHn−3 +2Hn−1) =−

12((n−1)Hn−1 +2Hn−1) =−

12((n+1)Hn−1)

comparing the ends,

H(n+1)λ =−12

Hn−1 =−12

H(n+1)xx

This proves the following theorem.

Theorem 68.1.2 Let Hn (x,λ ) be defined by

Hn (x,λ )≡(−λ )n

n!e

12λ

x2 ∂ n

∂xn

(e−

12λ

x2)

for λ > 0. Then the following properties are valid.

∂xHn (x,λ ) = Hn−1 (x,λ ) (68.1.4)

(n+1)Hn+1 (x,λ ) = xHn (x,λ )−λHn−1 (x,λ ) (68.1.5)

Hn (−x,λ ) = (−1)n Hn (x,λ ) (68.1.6)

∂λHn (x,λ ) =−

12

∂ 2

∂x2 Hn (x,λ ) (68.1.7)

With this theorem, one can also prove the following.

Theorem 68.1.3 The Hermite polynomials are the coefficients of a certain power series.Specifically,

exp(

tx− 12

t2λ

)=

∑n=0

Hn (x,λ ) tn