Hermite Polynomial Martingales of Brownian Motion

Why Brownian Motion Produces Polynomial Martingales

Brownian motion has Gaussian increments, and Gaussian moments are governed by a useful family of polynomials. The key point is that these are not merely polynomials of the current value $B_t$: they also depend on time. That time dependence cancels the drift term created by Ito's formula. The resulting processes have zero drift and are martingales.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration. In particular, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has law $N(0,t-s)$ when $s<t$. For an Ito integral on $[0,T]$, the integrand will be progressively measurable and in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Definition: Define the time-dependent Hermite polynomials $H_n(t,x)$, for integers $n\ge0$, by the generating function $$ \exp\left(ux-\frac{u^2t}{2}\right)=\sum_{n=0}^{\infty}H_n(t,x)\frac{u^n}{n!}. $$ Equivalently, $H_n(t,x)=n!\sum_{k=0}^{\lfloor n/2\rfloor}\dfrac{(-1)^k t^k x^{n-2k}}{2^k k!(n-2k)!}$. This polynomial definition applies also at $t=0$.

The convention here makes the first polynomials $H_0(t,x)=1$, $H_1(t,x)=x$, $H_2(t,x)=x^2-t$, $H_3(t,x)=x^3-3tx$, and $H_4(t,x)=x^4-6tx^2+3t^2$. The polynomials $H_n(t,B_t)$ will be the martingales of interest. Their time arguments are essential; for example, $B_t^2$ itself has a nonzero Ito drift.

Derivative Identities and the Martingale Result

Theorem: For every integer $n\ge0$, the process $M_t^{(n)}=H_n(t,B_t)$ is a martingale on each finite time interval. For $n\ge1$, $$ dH_n(t,B_t)=nH_{n-1}(t,B_t)\,dB_t, \qquad \mathbb{E}[H_n(t,B_t)\mid\mathcal{F}_s]=H_n(s,B_s) \quad(0\le s\le t). $$ For $n=0$, the process is the constant martingale $1$.

Proof. Differentiate the generating function with respect to $x$. Its derivative is $u\exp(ux-u^2t/2)$; comparing the coefficients of $u^n/n!$ gives $\partial_xH_n=nH_{n-1}$. Differentiating with respect to $t$ gives $-\tfrac12u^2\exp(ux-u^2t/2)$, hence $\partial_tH_n=-\tfrac12 n(n-1)H_{n-2}$ for $n\ge2$, while $\partial_tH_0=\partial_tH_1=0$. Applying the spatial derivative identity twice gives $\partial_{xx}H_n=n(n-1)H_{n-2}$. Thus, for $n\ge2$, $\partial_tH_n+\tfrac12\partial_{xx}H_n=0$; the same equation holds for $n=0,1$, since both derivatives in it vanish.

Fix $n\ge1$ and apply Ito's formula to the function $f(t,x)=H_n(t,x)$ and the process $B_t$, whose drift is zero and whose diffusion coefficient is one. The partial derivatives and correction term are $\partial_tf=-\tfrac12n(n-1)H_{n-2}$, $\partial_xf=nH_{n-1}$, and $\tfrac12\partial_{xx}f=\tfrac12n(n-1)H_{n-2}$ for $n\ge2$. For $n=1$, the time derivative and second spatial derivative are both zero. In each case the time derivative cancels the Ito correction, so $$ H_n(t,B_t)=H_n(0,B_0)+\int_0^t nH_{n-1}(s,B_s)\,dB_s. $$ For $n=0$, $H_0(t,B_t)=1$.

On any fixed interval $[0,T]$, $nH_{n-1}(s,B_s)$ is progressively measurable, since it is a polynomial in the adapted continuous process $B_s$ and time. Its square is integrable over time and probability: the Gaussian variable $B_s$ has finite moments of every order, and the polynomial coefficients are bounded for $s\in[0,T]$. Therefore the integrand belongs to $\mathcal{L}^2$. The Ito integral is square-integrable, and its increment from $s$ to $t$ has conditional expectation zero given $\mathcal{F}_s$. This gives the stated conditional identity and verifies the martingale property; in particular, each $M_t^{(n)}$ is integrable. The case $n=0$ is immediate. $\square$

The same conditional identity can also be checked directly from Brownian increments. Given $\mathcal{F}_s$, write $B_t=B_s+Y$, where $Y$ is independent of $\mathcal{F}_s$ and has law $N(0,t-s)$. For each fixed $b$, the Gaussian moment-generating function gives $\mathbb{E}[\exp(u(b+Y)-u^2t/2)]=\exp(ub-u^2s/2)$. Comparing coefficients of $u^n/n!$ yields $\mathbb{E}[H_n(t,b+Y)]=H_n(s,b)$. The coefficient comparison is valid because the Gaussian moment-generating function is finite for every real $u$; each coefficient is a finite polynomial moment. Substituting $b=B_s$ proves the conditional identity. This also shows how the time parameter compensates for the variance $t-s$ of the new increment.

Worked Polynomial Martingales

Worked Example: The Second Hermite Polynomial

For $M_t=B_t^2-t=H_2(t,B_t)$, calculate its Ito differential and verify its conditional identity between times $1$ and $2$.

Set $f(t,x)=x^2-t$. Then $\partial_tf=-1$, $\partial_xf=2x$, and $\partial_{xx}f=2$, so the Ito correction is $\tfrac12\partial_{xx}f\,dt=dt$. Ito's formula gives $$ d(B_t^2-t)=(-1+1)\,dt+2B_t\,dB_t=2B_t\,dB_t. $$ The integrand $2B_s$ is in $\mathcal{L}^2$ on $[0,2]$, since $\mathbb{E}\int_0^2(2B_s)^2\,ds=4\int_0^2s\,ds=8$. Thus the integral is square-integrable and has zero conditional mean over $[1,2]$.

For a direct check, let $Y=B_2-B_1$. It is independent of $\mathcal{F}_1$, with $\mathbb{E}[Y]=0$ and $\mathbb{E}[Y^2]=1$. Therefore $\mathbb{E}[(B_1+Y)^2-2\mid\mathcal{F}_1] =B_1^2+2B_1\mathbb{E}[Y]+\mathbb{E}[Y^2]-2 =B_1^2-1=H_2(1,B_1)$. Also $\mathbb{E}[B_2^2-2]=2-2=0=H_2(0,0)$, checking the expectation at time $2$.

Worked Example: The Third Hermite Polynomial

For $M_t=B_t^3-3tB_t=H_3(t,B_t)$, derive the stochastic differential and check the conditional identity from time $1$ to time $2$.

Set $f(t,x)=x^3-3tx$. Its derivatives are $\partial_tf=-3x$, $\partial_xf=3x^2-3t$, and $\partial_{xx}f=6x$. The Ito correction is $\tfrac12\partial_{xx}f\,dt=3x\,dt$. Consequently, $$ d f(t,B_t)=\left(-3B_t+3B_t\right)dt+ (3B_t^2-3t)\,dB_t =3H_2(t,B_t)\,dB_t. $$ The integrand is square-integrable on every finite interval because it is a polynomial in a Gaussian variable with finite fourth moment.

For the conditional calculation put $Y=B_2-B_1$. Here $\mathbb{E}[Y]=0$, $\mathbb{E}[Y^2]=1$, and $\mathbb{E}[Y^3]=0$. Expanding each term, $\mathbb{E}[(B_1+Y)^3\mid\mathcal{F}_1]=B_1^3+3B_1$, while $\mathbb{E}[-6(B_1+Y)\mid\mathcal{F}_1]=-6B_1$. Their sum is $B_1^3-3B_1=H_3(1,B_1)$, as required. At time $2$, symmetry of the centered normal distribution gives $\mathbb{E}[B_2^3]=0$, and $\mathbb{E}[B_2]=0$; hence $\mathbb{E}[H_3(2,B_2)]=0=H_3(0,0)$.

Worked Example: The Fourth Hermite Polynomial

For $M_t=B_t^4-6tB_t^2+3t^2$, derive the Ito differential and verify the conditional identity from time $1$ to time $2$, then check its expectation at time $2$.

Set $f(t,x)=x^4-6tx^2+3t^2$. Then $\partial_tf=-6x^2+6t$, $\partial_xf=4x^3-12tx$, and $\partial_{xx}f=12x^2-12t$. The correction is $\tfrac12\partial_{xx}f\,dt=(6x^2-6t)dt$. The drift terms cancel, giving $$ d f(t,B_t)=(4B_t^3-12tB_t)\,dB_t =4H_3(t,B_t)\,dB_t. $$ Again the polynomial integrand is in $\mathcal{L}^2$ on finite intervals, since its square has finite expectation and is bounded in expectation over each such interval.

Write $Y=B_2-B_1$, so its first four moments are $0,1,0,3$, respectively. With $b=B_1$, $\mathbb{E}[(b+Y)^4\mid\mathcal{F}_1]=b^4+6b^2+3$ and $\mathbb{E}[-12(b+Y)^2\mid\mathcal{F}_1]=-12(b^2+1)$. Adding the constant $12$ from $3(2)^2$ gives $b^4-6b^2+3=H_4(1,b)$. Finally $B_2\sim N(0,2)$, so $\mathbb{E}[B_2^2]=2$ and $\mathbb{E}[B_2^4]=3(2)^2=12$. Therefore $\mathbb{E}[H_4(2,B_2)]=12-6(2)(2)+3(2)^2=12-24+12=0=H_4(0,0)$.

Expectations, Uses, and a Common Pitfall

For every $n\ge1$, the conditional identity at $s=0$ gives $\mathbb{E}[H_n(t,B_t)\mid\mathcal{F}_0]=H_n(0,0)=0$. Taking expectations yields $\mathbb{E}[H_n(t,B_t)]=0$. Meanwhile $H_0(t,B_t)=1$ has expectation $1$. Thus the Hermite polynomials encode cancellations among the ordinary Gaussian moments. Their conditional identities are stronger than the unconditional moment equalities: they retain the dependence on the observed value $B_s$.

A common mistake is to use the ordinary chain rule on a power of Brownian motion. The wrong calculation $d(B_t^3)=3B_t^2\,dB_t$ omits the second-derivative correction. For $f(x)=x^3$, the derivatives are $f'(x)=3x^2$ and $f''(x)=6x$, so Ito's formula includes $\tfrac12 f''(B_t)dt=3B_tdt$: $$ d(B_t^3)=3B_t^2\,dB_t+3B_t\,dt. $$ The time-dependent term in $B_t^3-3tB_t$ cancels this drift. In fact, applying Ito's formula to $g(t,x)=3tx$ gives $\partial_tg=3x$, $\partial_xg=3t$, $\partial_{xx}g=0$, and correction $\tfrac12\partial_{xx}g\,dt=0$; hence $d(3tB_t)=3B_tdt+3t\,dB_t$. Subtracting recovers $d(B_t^3-3tB_t)=3(B_t^2-t)\,dB_t$. This is exactly the cancellation that makes the third Hermite polynomial a martingale.

The polynomial martingales are useful for organizing Gaussian moments and constructing explicit martingales from Brownian motion. The distinction between drift and diffusion is visible in their differentials: the time derivative and the Ito correction determine the drift, while the remaining coefficient multiplies $dB_t$. A zero drift alone is not enough for a martingale claim; here the finite-time square-integrability of the stochastic integrands ensures that the resulting local martingales are true martingales.

Check Your Understanding

Use the generating function, Ito's formula, and the conditional Gaussian increment as appropriate.

Write $H_0,\ldots,H_4$ explicitly and identify which are nonconstant martingales when evaluated at $(t,B_t)$.
For $H_2(t,B_t)$, compute $\mathbb{E}[H_2(3,B_3)\mid\mathcal{F}_1]$ in terms of $B_1$.
Apply Ito's formula to $H_1(t,B_t)$, showing its time derivative and second-derivative correction.
Explain why the integrand $3H_2(t,B_t)$ in the differential for $H_3(t,B_t)$ is square-integrable on every finite time interval.
Find the missing drift term in the ordinary-chain-rule calculation $d(B_t^4)=4B_t^3\,dB_t$.

Hermite Polynomial Martingales of Brownian Motion

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