Hermite polynomial with brownian motion is martingale Let $(B_t)_{t\ge 0}$ be a standard brownian motion. I want to show that $(H_n(B_t,t))_{t\ge 0}$ is a martingale, where
$$H_n(x,t)=\frac{d^n}{du^n}e^{ux-\frac{u^2}{2}t}\Big|_{u=0},$$
such that the Taylor-formula is
$$e^{ux-\frac{u^2}{2}t}=\sum_{n\ge 0}H_n(x,t)\frac{u^n}{n!}$$
How do I calculate $E[|H_n(B_t,t)|]$? Is it possible to interchange integration and differentiation? I tried to calculate $H_n(x,t)$ first to find a formula, but I do not get a nice solution. Can someone give me a hint on this? Thanks in advance!
 A: You may do it inductively.

Lemma 1. We have for $n\ge 1$,
  $$
\frac{{\rm d}^n}{{\rm d}u^n}\left(uf(u)\right)=uf^{(n)}(u)+nf^{(n-1)}(u).
$$

This Lemma 1 can be easily shown by mathematical induction. Hence I skip its proof here.

Lemma 2. We have for $n\ge 2$,
  $$
H_{n+1}(x,t)=xH_n(x,t)-ntH_{n-1}(x,t).
$$

Proof. For $n\ge 2$, we have
\begin{align}
H_{n+1}(x,t)&=\frac{{\rm d}^{n+1}}{{\rm d}u^{n+1}}e^{ux-\frac{t}{2}u^2}\bigg|_{u=0}\\
&=\frac{{\rm d}^n}{{\rm d}u^n}\left(\frac{\rm d}{{\rm d}u}e^{ux-\frac{t}{2}u^2}\right)\bigg|_{u=0}\\
&=\frac{{\rm d}^n}{{\rm d}u^n}\left(e^{ux-\frac{t}{2}u^2}\left(x-tu\right)\right)\bigg|_{u=0}\\
&=x\frac{{\rm d}^n}{{\rm d}u^n}e^{ux-\frac{t}{2}u^2}\bigg|_{u=0}-t\frac{{\rm d}^n}{{\rm d}u^n}\left(ue^{ux-\frac{t}{2}u^2}\right)\bigg|_{u=0}\\
&=xH_n(x,t)-t\frac{{\rm d}^n}{{\rm d}u^n}\left(ue^{ux-\frac{t}{2}u^2}\right)\bigg|_{u=0}.
\end{align}
Now, apply Lemma 1 to the last term, and we obtain
$$
\frac{{\rm d}^n}{{\rm d}u^n}\left(ue^{ux-\frac{t}{2}u^2}\right)=u\frac{{\rm d}^n}{{\rm d}u^n}e^{ux-\frac{t}{2}u^2}+n\frac{{\rm d}^{n-1}}{{\rm d}u^{n-1}}e^{ux-\frac{t}{2}u^2}.
$$
Thus by taking $u=0$, we have
$$
\frac{{\rm d}^n}{{\rm d}u^n}\left(ue^{ux-\frac{t}{2}u^2}\right)\bigg|_{u=0}=n\frac{{\rm d}^{n-1}}{{\rm d}u^{n-1}}e^{ux-\frac{t}{2}u^2}\bigg|_{u=0}=nH_{n-1}(x,t).
$$
Thanks to this result, Lemma 2 follows immediately.#
Note that the last equation also proves

Lemma 3. We have for $n\ge 2$,
  $$
\frac{\partial H_n}{\partial x}(x,t)=nH_{n-1}(x,t),
$$

because
\begin{align}
\frac{\partial H_n}{\partial x}(x,t)&=\frac{\partial}{\partial x}\left(\frac{\partial^n}{\partial u^n}e^{ux-\frac{t}{2}u^2}\right)\bigg|_{u=0}\\
&=\frac{\partial^n}{\partial u^n}\left(\frac{\partial}{\partial x}e^{ux-\frac{t}{2}u^2}\right)\bigg|_{u=0}\\
&=\frac{\partial^n}{\partial u^n}\left(ue^{ux-\frac{t}{2}u^2}\right)\bigg|_{u=0}.
\end{align}
Now, your question can be figured out inductively.
Initially, note that
\begin{align}
H_1(B_t,t)&=B_t\\
H_2(B_t,t)&=B_t^2-t
\end{align}
are both martingales.
Inductively, suppose $H_n(B_t,t)$ and $H_{n-1}(B_t,t)$ are martingales. We need to show that $H_{n+1}(B_t,t)$ is a martingale. By Lemma 2,
\begin{align}
{\rm d}H_{n+1}(B_t,t)&={\rm d}\left(B_tH_n(B_t,t)-ntH_{n-1}(B_t,t)\right)\\
&=H_n(B_t,t){\rm d}B_t+B_t{\rm d}H_n(B_t,t)+{\rm d}\left<B_t,H_n(B_t,t)\right>-nH_{n-1}(B_t,t){\rm d}t-nt{\rm d}H_{n-1}(B_t,t),
\end{align}
where the last line uses Ito's formula. Note that by Ito's formula again,
\begin{align}
{\rm d}H_n(B_t,t)&=\frac{\partial H_n}{\partial t}(B_t,t){\rm d}t+\frac{\partial H_n}{\partial x}(B_t,t){\rm d}B_t+\frac{1}{2}\frac{\partial^2H_n}{\partial x^2}(B_t,t){\rm d}\left<B_t\right>\\
&=\frac{\partial H_n}{\partial t}(B_t,t){\rm d}t+\frac{\partial H_n}{\partial x}(B_t,t){\rm d}B_t+\frac{1}{2}\frac{\partial^2H_n}{\partial x^2}(B_t,t){\rm d}t\\
&=\left(\frac{\partial H_n}{\partial t}(B_t,t)+\frac{1}{2}\frac{\partial^2H_n}{\partial x^2}(B_t,t)\right){\rm d}t+\frac{\partial H_n}{\partial x}(B_t,t){\rm d}B_t\\
&=\frac{\partial H_n}{\partial x}(B_t,t){\rm d}B_t,
\end{align}
where the last line is due to the inductive assumption that $H_n(B_t,t)$ is a martingale, for which the coefficients in front of ${\rm d}t$ must vanish. Similarly, we have
$$
{\rm d}H_{n-1}(B_t,t)=\frac{\partial H_{n-1}}{\partial x}(B_t,t){\rm d}B_t.
$$
In addition, we also have
\begin{align}
{\rm d}\left<B_t,H_n(B_t,t)\right>&={\rm d}B_t{\rm d}H_n(B_t,t)\\
&=\frac{\partial H_n}{\partial x}(B_t,t){\rm d}B_t{\rm d}B_t\\
&=\frac{\partial H_n}{\partial x}(B_t,t){\rm d}t.
\end{align}
Thanks to all these results, we eventually have
\begin{align}
{\rm d}H_{n+1}(B_t,t)&=\left(\frac{\partial H_n}{\partial x}(B_t,t)-nH_{n-1}(B_t,t)\right){\rm d}t\\
&+\left(H_n(B_t,t)+B_t\frac{\partial H_n}{\partial x}(B_t,t)-nt\frac{\partial H_{n-1}}{\partial x}(B_t,t)\right){\rm d}B_t\\
&=\left(H_n(B_t,t)+B_t\frac{\partial H_n}{\partial x}(B_t,t)-nt\frac{\partial H_{n-1}}{\partial x}(B_t,t)\right){\rm d}B_t,
\end{align}
where the last line results from Lemma 3. This result shows that $H_{n+1}(B_t,t)$ is a stochastic integral of $B_t$, which obviously indicates that it is a martingale.
To sum up, we may conclude that $H_n(B_t,t)$ is a martingale for all $n\ge 1$.
