Let $B_t,t\geq 0$ a brownian motion and $u(t,x)$ a function satisfying the following PDE $$\frac{\partial u}{\partial t}+\frac{1}{2}\frac{\partial^2 u}{\partial x^2}=0.$$ Then we prove that $\frac{\partial }{\partial t}\Bbb{E}(u(t,B_t))=0$, but I want to prove that $u(t,B_t)$ is a martingale (without using Itô formula). I think it suffices to show that $u(t,B_t)$ is a submartingale or supermartingale. Thanks.

  • 1
    $\begingroup$ Actually, you would have to use at some point that $u$ solves the PDE. With applying Ito formula, the only way you can use it, as I see, is to write the explicit expression of $u$ via the fundamental solution (=Gaussian density) and deal with integrals w.r.t. conditional distributions of Brownian motion. Perhaps, doable - but Ito's formula seems to be much easier. $\endgroup$
    – SBF
    Commented Jun 5, 2013 at 15:22

1 Answer 1


You can apply the following theorem to prove $u(t,B_t)$ a martingale. The main idea is -as Ilya wrote- to use the fact that the density $p(t,x)$ of a Gaussian distribution with mean 0, variance $t$ solves $$\frac{\partial}{\partial t} p(t,x) = \frac{1}{2} \frac{\partial^2}{\partial x^2} p(t,x) \qquad (x \in \mathbb{R},t>0) \tag{1}$$.

Theorem Let $(B_t)_{t \geq 0}$ a Brownian motion and $u \in C^{1,2} \cap C([0,\infty) \times \mathbb{R})$ such that $u$ and its partial derivatives are exponentially bounded. Then $$M_t^u := u(t,B_t) - u(0,B_0) - \int_0^t \underbrace{\left( \frac{\partial }{\partial t} + \frac{1}{2} \frac{\partial^2}{\partial x^2} \right) u(r,B_r)}_{=:Lu(r,B_r)} \, dr$$ is a martingale.

Sketch of the proof: By the Markov property of the Brownian motion, one can show that

$$\begin{align*} \mathbb{E}(M_t^u-M_s^u \mid \mathcal{F}_s) &= \mathbb{E} \left( u(t,B_t)-u(s,B_s) - \int_s^t Lu(r,B_r) \, dr \mid \mathcal{F}_s \right) \\ &= \mathbb{E} \left( u(t,B_{t-s}+z)-u(s,B_0+z) - \int_0^{t-s} Lu(r+s,B_r+z) \, dr \right) \bigg|_{z=B_s} \\ &= \mathbb{E} \bigg( \underbrace{\varphi(t-s,B_{t-s}) - \varphi(0,B_s) - \int_0^{t-s} L\varphi(r,B_{r}) \, dr}_{M_{t-s}^{\varphi}-M_0^{\varphi}} \bigg) \bigg|_{z=B_s} \end{align*}$$

where $\varphi(t,x) := u(t+s,x+z)$. To prove that this expression equals $0$, use Fubini's Theorem, partial integration and $(1)$ to show that

$$\mathbb{E}(M_t^{\varphi}-M_{\varepsilon}^{\varphi}) = \int (p(t,x) \cdot \varphi(t,x)-p(\varepsilon,x) \cdot \varphi(\varepsilon,x)) \, dx \\ - \int_{\varepsilon}^t p(r,x) \cdot \left( \frac{\partial}{\partial t} + \frac{1}{2} \frac{\partial^2}{\partial x^2} \right) \varphi(r,x) \, dx \, dr = 0$$

for $t > \varepsilon>u$. Finally, by Doob's maximal inequality and the exponential boundedness, one can apply dominated converence to conclude that $$\mathbb{E}(M_t^{\varphi}-M_0^{\varphi}) = \lim_{\varepsilon \to 0} \mathbb{E}(M_t^{\varphi}-M_{\varepsilon}^{\varphi}) = 0$$

This finishs the proof.

(For more details, see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Theorem 5.6.)

Actually, the mentioned theorem is a special case of the following theorem:

Theorem Let $(X_t)_{t \geq 0}$ a Feller process with generator $A$. Then $$f(X_t)-f(X_0) - \int_0^t Af(X_r) \, dr$$ is a martingale for any $f$ in the domain of $A$.

So, in your case the process is given by $X_t := (t,B_t)$. With some knowledge about generators, it's not that difficult to show that the generator $A$ is given by $$A = \frac{\partial}{\partial t} + \frac{1}{2} \frac{\partial^2}{\partial x^2}.$$ (This leads to the topic of transition semigroup and its generator. It's also contained in the book, chapter 7.)

  • $\begingroup$ Thank you for your answer. I have a question maybe stupid why we don't compute directly $\Bbb{E}(u(t,B_t)-u(s,B_s)|\mathcal{F_s})$ and use the same arguments used in your proof. My objectif is to prove that $u(t,B_t)$ is a martingale not $M_t^u$. $\endgroup$
    – bmo
    Commented Jun 6, 2013 at 8:32
  • $\begingroup$ @bmo Yes, of course. In your example, we have $Lu = 0$, i.e. $M_t^u = u(t,B_t)-u(0,B_0)$. So you can either apply the theorem or adapt the proof. $\endgroup$
    – saz
    Commented Jun 6, 2013 at 10:42
  • $\begingroup$ very nice proof that deserves to be more seen (y) $\endgroup$ Commented May 4, 2021 at 20:05

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