$u(t,B_t)$ is a martingale if it satisfies a certain condition.

I want to show that for $$u(t,x)$$ which is a polynomial in $$t$$ and $$x$$ such that $$\frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}=0$$ we have $$u(t,B_t)$$ is a martingale where $$B_t$$ stands for the standard Brownian motion.

Durrett's "Probability theory and examples" shows that $$E_x u(t, B_t)$$ is constant in $$t$$ and concludes right away that $$u(t,B_t)$$ is a martingale. How is this possible?

Any help is appreciated.

2 Answers

Your self-answer is unsatisfactory.

Also: In OP, what do you mean by "a polynomial $$u(t,x)$$ in $$t,x$$ such that $$\frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}"\quad\dots\quad ?$$ The correct statement is that any function $$u(t,x)$$ (regardless if polynomial or not) that satisfies the Kolmogorov backward equation $$\frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}=0$$ ensures that $$u(t,B_t)$$ is a (local) martingale.

Proof. Use Ito's formula $$u(t,B_t)=u(0,0)+\underbrace{\int_0^tu_x(s,B_s)\,dB_s}_{\text{local martingale}}+ \underbrace{\int_0^tu_t(s,B_s)\,ds+\frac12u_{xx}(s,B_s)\,ds}_{0}\,.$$ It is fairly easy to see that that local martingale does in general not have independent increments.

• Yes. I agree that my initial answer is very unsatisfactory :< Feb 18 at 3:54
• @Focus please also fix the title. It is not the heat equation. Feb 18 at 4:53

A mean-constant process with Markov and independent increment properties is a martingale.

• How do you know that $u(t,B_t)$ has independent increments? Feb 17 at 5:53
• @KurtG. I think because $u$ is a polynomial. Feb 17 at 5:57
• I'd like to see a non trivial polynomial in $t,x$ that satisfies the heat equation. Also: does $B_t^2$ have independent increments? Unlikely: $B_t^2-B_s^2=(B_t-B_s)(B_t+B_s)\,.$ Feb 17 at 6:20
• Ok. Some examples of polynomials that satisfy the heat equation are $$u(t,x)=x\,,\quad u(t,x)=t+x^2\,,\quad u(t,x)=tx+\tfrac13x^3\,.$$ Only for the first one $u(t,B_t)$ has independent increments. Feb 17 at 7:02
• Finally: do you know about the difference between heat equation and Kolmogorov BW equation? Hint: change the sign of $t\,.$ How will you have to modify the polynomials in the previous comment such that all of them lead to martingales? Feb 17 at 7:24