Relation between Hermite polynomials and Brownian motion (on martingale property)

This question already has an answer here:

  • Hermite Polynomials and Brownian motion

    1 answer

Solutions Collecting From Web of "Relation between Hermite polynomials and Brownian motion (on martingale property)"

It follows from the identity

$$e^{\theta x- \frac{1}{2} \theta^2} = \sum_{n \in \mathbb{N}_0} \frac{\theta^n}{n!} H_n(x), \qquad \theta \in \mathbb{R} \tag{1}$$

that

$$M_t^{\lambda} := f(\lambda,t,B_t) := e^{\lambda B_t- \frac{t}{2} \lambda^2} \stackrel{(1)}{=} \sum_{n \in \mathbb{N}_0} t^{n/2} \frac{\lambda^n}{n!} H_n \left( \frac{B_t}{\sqrt{t}} \right). \tag{2}$$

Moreover, it a straight-forward application of Itô’s formula shows that $(M_t^{\lambda})_{t \geq 0}$ is a martingale for each fixed $\lambda \in \mathbb{R}$, i.e.

$$\mathbb{E}(M_t^{\lambda} \mid \mathcal{F}_s) = M_s^{\lambda}. \tag{3}$$

Since, by $(2)$,

$$\frac{\partial^n}{\partial \lambda^n} f(\lambda,t,y) \bigg|_{\lambda=0} = t^{n/2} H_n \left( \frac{y}{\sqrt{t}} \right) \tag{4}$$

we obtain by differentiating $(3)$ $n$ times (with respect to $\lambda$) and evaluating at $\lambda=0$ that

$$\mathbb{E}\left[ t^{n/2} H_n \left( \frac{B_t}{\sqrt{t}} \right) \mid \mathcal{F}_s \right] = s^{n/2} H_n \left( \frac{B_s}{\sqrt{s}} \right).$$

This finishes the proof.


Edit: Note that $(3)$ is equivalent to

$$\int_F M_t^{\lambda} \, d\mathbb{P} = \int_F M_s^{\lambda} \, d\mathbb{P} \tag{5}$$

for all $F \in \mathcal{F}_s$. From the definition of $M_t^{\lambda}$ we have

$$\frac{\partial}{\partial \lambda} M_t^{\lambda} = M_t^{\lambda} (B_t-\lambda t).$$

In particular, we get

$$\sup_{|\lambda| \leq 1} \left| \frac{\partial}{\partial \lambda} M_t^{\lambda} \right| \leq (|B_t|+t) e^{|B_t|}.$$

Since $B_t$ has exponential moments, the right-hand side is a uniform (with respect to $\lambda$) integrable bound for the derivative. Consequently, we obtain from the differentiation lemma for parametrized integrals that we may interchange integration and differentiation, i.e.

$$\frac{\partial}{\partial \lambda} \int_F M_t^{\lambda} \, d\mathbb{P} = \int_F \frac{\partial}{\partial \lambda} M_t^{\lambda} \, d \mathbb{P}$$

for all $t$ and $|\lambda| \leq 1$. If we calculate derivatives of higher order with respect to $\lambda$, it is not difficult to see that there exists a polynomial $p$ such that

$$\sup_{|\lambda| \leq 1} \left| \frac{\partial^n}{\partial \lambda^n} M_t^{\lambda} \right| \leq p(|B_t|, |t|) e^{|B_t|}.$$

By iterating the above procedure (… use induction if you want to have a formal proof) we get

$$\frac{\partial^n}{\partial \lambda^n} \int_F M_t^{\lambda} , d\mathbb{P} = \int_F \frac{\partial^n}{\partial \lambda^n} M_t^{\lambda} \, d \mathbb{P}.$$

From $(5)$ we thus obtain

$$\int_F \frac{\partial^n}{\partial \lambda^n} M_t^{\lambda} \, d \mathbb{P} = \int_F \frac{\partial^n}{\partial \lambda^n} M_s^{\lambda} \, d \mathbb{P}$$

for all $F \in \mathcal{F}_s$. This is equivalent to

$$\mathbb{E} \left( \frac{\partial^n}{\partial \lambda^n} M_t^{\lambda} \mid \mathcal{F}_s \right) = \frac{\partial^n}{\partial \lambda^n} M_s^{\lambda}.$$

Now we evaluate this identity at $\lambda = 0$ and use $(4)$.