Relation between Hermite polynomials and Brownian motion (on martingale property) Let us define Hermite polynomials as $H_n(x)=(-1)^n e^{x^2/2}\frac{d^n}{dx^n} e^{-x^2/2}.$ One can prove that $e^{\theta x-\frac{1}{2} \theta^2}=\sum_{n=0}^{\infty} \frac{1}{n!}H_n(x)\theta^n \quad (*).$
Now denote $H_n(t,x):=t^{n/2}H_n(x/\sqrt t)$.
We want to prove that $H_n(t,B(t))$ is a martingale. Now, I've thought to use the identity $H_{n+2}(t,x)=xH_{n+1}(t,x)-(n+1)tH_{n}(t,x)$, so $$\mathbb E(H_{n+2}(t,B(t))|\mathcal F_s)=\mathbb E(B(t)H_{n+1}(t,B(t))|\mathcal F_s)-(n+1)t \mathbb E(H_{n}(t,B(t))|\mathcal F_s),$$
but that seems not to be the right way to solve the exercise. So I guess one shall use $(*)$, but I don't know how to do it. Can someone help me?
If it is possible, I'de like to prove it without using the fact that $dH_{n+1}(t,B(t))= H_{n}(t,B(t))$, because the exercise I proposed here is from a sheet that contains also the exercise $e^{\theta x-\frac{1}{2} \theta^2}=\sum_{n=0}^{\infty} \frac{1}{n!}H_n(x)\theta^n$, and I think that I should use that fact, but I'm not sure.
 A: 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 $(4)$.
