3
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Yes. I agree that my initial answer is very unsatisfactory :< $\endgroup$
    – Focus
    Feb 18 at 3:54
  • $\begingroup$ @Focus please also fix the title. It is not the heat equation. $\endgroup$
    – Kurt G.
    Feb 18 at 4:53
1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ How do you know that $u(t,B_t)$ has independent increments? $\endgroup$
    – Kurt G.
    Feb 17 at 5:53
  • $\begingroup$ @KurtG. I think because $u$ is a polynomial. $\endgroup$
    – Focus
    Feb 17 at 5:57
  • $\begingroup$ 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: $B_t^2-B_s^2=(B_t-B_s)(B_t+B_s)\,.$ $\endgroup$
    – Kurt G.
    Feb 17 at 6:20
  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Feb 17 at 7:02
  • $\begingroup$ 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? $\endgroup$
    – Kurt G.
    Feb 17 at 7:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .