37 — Martingales for Brownian Motion
1. Martingales in Continuous Time
Consider a filtered probability space
\((\Omega,\mathcal F, \{\mathcal F_t\}_{t\ge 0}, P).\)
A process ${X_t,\mathcal F_t}_{t\ge 0}$ is a martingale if:
-
$E X_t < \infty$ for all $t \ge 0$. - $X_t$ is $\mathcal F_t$-measurable.
- $E[X_{t+h} \mid \mathcal F_t] = X_t$ for all $t\ge0$, $h>0$.
We want to study martingales associated with Brownian motion, using its natural (canonical) filtration ${\mathcal F_t}$.
Let \(M:\mathbb R \times [0,\infty) \to \mathbb R,\qquad (x,t)\mapsto M(x,t).\)
Main question:
When is the process
\(M(B_t^x, t),\qquad t\ge0,\)
a martingale?
This requires \(E_x\!\left[\,M(B_{t+h}, t+h)\mid \mathcal F_t \right] = M(B_t, t) \quad\text{a.s. for all } x\in\mathbb R,\ t\ge 0,\ h>0.\)
By the Markov property of Brownian motion, \(E_x[M(B_{t+h},t+h)\mid \mathcal F_t] = E_{B_t}[M(B_h,t+h)].\)
Define \(\varphi(y)=E_y[M(B_h,t+h)].\) The condition becomes \(\varphi(B_t)= M(B_t,t).\)
2. Example: The Famous Exponential Martingale
Define, for $\theta\in\mathbb R$, \(M(x,t)=\exp(\theta x - \tfrac12\theta^2 t).\)
Then, with $B_{t+h}\sim N(x,h)$, \(\begin{aligned} E_x[M(B_{t+h},t+h)] &= E_x\left[\exp\left(\theta B_{t+h}-\frac12\theta^2(t+h)\right)\right] \\ &= e^{-\frac12\theta^2(t+h)} E_x(e^{\theta B_{t+h}}). \end{aligned}\)
Write \(B_{t+h}=x+\sqrt{h}Z,\qquad Z\sim N(0,1).\)
Then \(\begin{aligned} E_x(e^{\theta B_{t+h}}) &= e^{\theta x} E(e^{\theta\sqrt h\, Z}) = e^{\theta x} e^{\frac12 \theta^2 h}. \end{aligned}\)
Putting it together, \(E_x[M(B_{t+h},t+h)] = e^{-\frac12\theta^2 t} e^{\theta x} = M(x,t).\)
Thus the process \(\boxed{\,e^{\theta B_t - \frac12\theta^2 t}\,}\) is a martingale under $P_x$ for every $x\in\mathbb R$.
3. Differentiating the Exponential Martingale
Because \(E_x\Big[ M(B_{t+h},t+h;\theta)\Big] = M(x,t;\theta)\) for all $\theta$, we may differentiate under the expectation.
First derivative w.r.t. $\theta$:
\[\frac{\partial}{\partial\theta}M(x,t;\theta) = (x-\theta t)\, M(x,t;\theta).\]Evaluate at $\theta=0$: \(\left.\frac{\partial}{\partial\theta}M(x,t;\theta)\right|_{\theta=0} = x.\)
Thus the process \(\boxed{\,B_t\,}\) is a martingale.
Second derivative:
\[\frac{\partial^2}{\partial\theta^2} M(x,t;\theta) = (x-\theta t)^2 M(x,t;\theta).\]Evaluate at $\theta=0$: \(\left.\frac{\partial^2}{\partial\theta^2}M(x,t;\theta)\right|_{\theta=0} = x^2.\)
Thus \(\boxed{\,B_t^2 - t\,}\) is a martingale.
Third derivative:
Written out (from notes on page 3), \(\frac{\partial^3}{\partial\theta^3}M = M(\cdot)(x-\theta t)^3 + M(\cdot)\cdot 2(x-\theta t)(-t) -tM(\cdot)(x-\theta t).\)
At $\theta=0$, this produces the martingale \(\boxed{\,B_t^3 - 3t B_t\,}.\)
Continuing similarly gives the whole Hermite–polynomial martingale family: \(1,\quad B_t,\quad B_t^2 - t,\quad B_t^3 - 3 t B_t,\quad B_t^4 - 6 t B_t^2 + 3t^2,\ \ldots\)
4. PDE Characterization of Martingales
A smooth function $M(x,t)$ satisfies $$ \boxed{\qquad \frac{\partial M}{\partial t}
- \frac12 \frac{\partial^2 M}{\partial x^2} = 0, \qquad x\in\mathbb R,\ t>0, } \(if and only if\) M(B_t,t) $$ is a martingale.
This is the backward heat equation.
The book proves: \(\frac{d}{dt} E_x[M(B_t,t)] = 0,\) so \(\boxed{\quad E_x[M(B_t,t)] = M(x,0),\quad t\ge 0.\quad}\)
5. Connection with Brownian Transition Density
The transition density of Brownian motion is \(P_t(x,y) = \frac{1}{\sqrt{2\pi t}} \exp\!\left(-\frac{(y-x)^2}{2t}\right) = p_{N(x,t)}(y).\)
It satisfies the forward heat equation: \(\frac{\partial P_t}{\partial t} = \frac12 \frac{\partial^2 P_t}{\partial y^2}.\)
This PDE duality explains why solving the backward heat equation produces martingales.
6. Optional Stopping for These Martingales
For a stopping time $\tau$ with $\tau \le t$ a.s., the optional stopping theorem gives \(E_x\!\left[M(B_t,t)\mid\mathcal F_\tau\right] = M(B_\tau,\tau).\)
This holds for all martingales $M(B_t,t)$ generated by the backward heat equation.
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