Hermite polynomial with brownian motion is martingale

Question

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!

Answer 1

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 $n\ge 1$.