Hermite Polynomials and Brownian motion I am asked to prove the following :
Let $B_t$ be a standard brownian motion. 
The $n$th Hermite polynomial is $\displaystyle H_n(t,x)=\frac{(−t)^n}{n!} e^{x^2/(2t)} \frac{d^n}{dx^n}e^{-x^2/(2t)}$. 
Show that the $H_n$ play the role that the monomials $\dfrac{x_n}{n!}$ play in ordinary calculus. 
In other words show : $dH_{n+1}(t, B_t) = H_n(t, B_t)$
Now my guess is that we can take the integral of both sides and use ito integration ? But I am not sure how to proceed from there. 
 A: First of all, the statement
$$dH_{n+1}(t,B_t) = H_n(t,B_t)$$
doesn't make sense. Or can you explain what you mean by it? I guess, it should read
$$dH_{n+1}(t,B_t) = H_n(t,B_t) \, dB_t;$$
at least that's what I prove in this answer.


Lemma 1 $$H_{n+1}(t,x) = \frac{x}{n+1} H_n(t,x) - \frac{t}{n+1} H_{n-1}(t,x). \tag{0}$$

Proof: Denote by $$G_n(y) := (-1)^n \exp \left( \frac{y^2}{2} \right) \frac{d^n}{dy^n} \exp \left(- \frac{y^2}{2} \right)$$ the "standard" Hermite polynomial. Then it follows from the chain rule that $$H_n(t,x) = t^{n/2} \frac{1}{n!} G_n \left( \frac{x}{\sqrt{t}} \right). \tag{1}$$ Using the recursion relation $$G_{n+1}(y) = y G_n(y)- n G_{n-1}(y) \tag{2}$$ we find $$\begin{align*} H_{n+1}(t,x) &\stackrel{(1)}{=} t^{(n+1)/2} \frac{1}{(n+1)!} G_{n+1} \left( \frac{x}{\sqrt{t}} \right) \\ &\stackrel{(2)}{=} t^{(n+1)/2} \frac{1}{(n+1)!} \left[ \frac{x}{\sqrt{t}} G_n \left( \frac{x}{\sqrt{t}} \right) - n G_{n-1} \left( \frac{x}{\sqrt{t}} \right) \right] \\ &\stackrel{(1)}{=} \frac{x}{n+1} H_n(t,x) - \frac{t}{n+1}H_{n-1}(t,x). \end{align*}$$


Lemma 2 $$\frac{\partial}{\partial t} H_n(t,x) = - \frac{1}{2} H_{n-2}(t,x).$$

Proof: It follows straight from the product rule that 
$$\begin{align*}
\frac{\partial}{\partial t} H_n(t,x) &= - \frac{(-t)^{n-1}}{(n-1)!} \exp \left(\frac{x^2}{2t}\right) \frac{d^n}{dx^n} \exp \left(-\frac{x^2}{2t} \right)\\ 
&\quad - \frac{(-t)^n}{n!} \frac{x^2}{2t^2} \exp \left(\frac{x^2}{2t}\right) \frac{d^n}{dx^n} \exp \left(-\frac{x^2}{2t} \right) \\ 
&\quad + \frac{(-t)^n}{n!} \frac{1}{2t^2} \exp \left(\frac{x^2}{2t}\right) \frac{d^n}{dx^n} \bigg( x^2 \exp \left(-\frac{x^2}{2t} \right)\bigg). \tag{3}
\end{align*}$$ 
Applying Leibniz' product rule 
$$\frac{d^n}{dx^n} (f \cdot g) = \sum_{k=0}^n {n \choose k} f^{(k)}(x) g^{(n-k)}(x)$$ 
we get 
$$\begin{align*}
\frac{d^n}{dx^n} \bigg( x^2 \exp \left(-\frac{x^2}{2t} \right)\bigg) &= x^2 \frac{d^n}{dx^n} \exp \left( - \frac{x^2}{2t} \right) + 2nx  \frac{d^{n-1}}{dx^{n-1}} \exp \left( - \frac{x^2}{2t} \right) \\ &\quad + (n-1) n \frac{d^{n-2}}{dx^{n-2}} \exp \left( - \frac{x^2}{2t} \right) \tag{4}
\end{align*}$$
Plugging $(4)$ into $(3)$ yields$$\begin{align*} \frac{\partial}{\partial t} H_n(t,x) &= - \frac{(-t)^{n-1}}{(n-1)!} \exp \left( \frac{x^2}{2t} \right) \frac{d^n}{dx^n} \exp \left(- \frac{x^2}{2t} \right) \\ &\quad + \frac{1}{2} \frac{(-t)^{n-2}}{(n-2)!}\exp \left( \frac{x^2}{2t} \right) \frac{d^{n-2}}{dx^{n-2}} \exp \left(- \frac{x^2}{2t} \right) \\ &= \frac{n}{t} H_n(t,x) + \frac{1}{2} H_{n-2}(t,x) \\ &\stackrel{(0)}{=} \frac{n}{t} \left( \frac{x}{n} H_{n-1}(t,x) - \frac{t}{n} H_{n-2}(t,x) \right)+ \frac{1}{2} H_{n-2}(t,x) = - \frac{1}{2} H_{n-2}(t,x). \end{align*}$$


Lemma 3 $$\frac{\partial}{\partial x} H_{n}(t,x) = H_{n-1}(t,x).$$

Proof: $$\begin{align*} \frac{\partial}{\partial x} H_n(t,x) &= \frac{(-t)^n}{n!} \exp \left( \frac{x^2}{2t} \right) \bigg[ \frac{x}{t} \frac{d^n}{dx^n} \exp \left(- \frac{x^2}{2t} \right) + \frac{d^{n+1}}{dx^{n+1}} \exp \left(- \frac{x^2}{2t} \right) \bigg] \\ &= \frac{x}{t} H_n(t,x) + \frac{n+1}{-t} H_{n+1}(t,x) \\ &\stackrel{(0)}{=} H_{n-1}(t,x). \end{align*}$$


Theorem $$dH_{n+1}(t,B_t) = H_n(t,B_t) \, dB_t$$

Proof: The claim follows directly from Itô's formula and Lemma 2+3. Note that Lemma 3 implies $$\frac{\partial^2}{\partial^2 x} H_{n+1}(t,x) = \frac{\partial}{\partial x} H_n(t,x) =  H_{n-1}(t,x). $$
Remark Note that this shows in particular that $(H_n(t,B_t))_{t \geq 0}$ is a martingale (see also this question).
