4
$\begingroup$

If $B_t$ is a Brownian motion then using Hermite polynomials one can find that

$$1, B_t, B_t^2-t, B_t^3 - 3tB_t,...$$

are martingales.

If $X_t$ is a diffusion

$dX_t = \mu(X_t,t)dt + \sigma(X_t,t)dt$

Is it possible to create sequence of polynomials that would be martingales, but now with $X_t$ instead of Brownian motion?

I will be very grateful for any resources, articles that researched this question

Improve this question
$\endgroup$
5
  • $\begingroup$ If $H_{n}(x)$ is n'th degree (probabilists) Hermite polynomial then $v(x,t)=t^{n/2}H_{n}(x/\sqrt{t})$ satisfies backward heat equation $v_{t}+\frac{v_{xx}}{2}=0$ (with the boundary condition $v(x,0)=x^{n}$). This is the reason that the process $v(B_{t},t)$ is the martingale. Now for the stochastic process $dX_{t} = \mu(X_{t},t)dt+\sigma(X_{t},t)dt$, the new process $u(X_{t},t)$ is martingale if and only if $u(x,t)$ satisfies PDE $u_{t}+\mu u_{x}+\frac{\sigma^{2} u_{xx}}{2}$. $\endgroup$
    – Paata Ivanishvili
    Commented Feb 8, 2020 at 2:36
  • $\begingroup$ So you are asking for which $\mu$ and $\sigma$ the solutions of the PDE for $u$ with boundary condition $u(x,0)=x^{n}$ are polynomials in variables $(x,t)$. I doubt that there is a nice answer in such general setting, however, for some concrete $\mu$ and $\sigma$ they can be polynomials. $\endgroup$
    – Paata Ivanishvili
    Commented Feb 8, 2020 at 2:46
  • $\begingroup$ For example if $\mu$ and $\sigma$ are independent of $t$, then by letting $L:=\mu \partial_{x} + \frac{\sigma^{2}}{2} \partial_{xx}^{2}$, we want to solve PDE $u_{t}+Lu=0$ with boundary data $u(x,0)=x^{n}$. Informally you can write $u(x,t)=e^{-tL} x^{n}$. So this is polynomial if $L^{k} x^{n}=0$ starting for some $k \geq N(n)$. And this is true if $\mu$ and $\sigma$ are constants. $\endgroup$
    – Paata Ivanishvili
    Commented Feb 8, 2020 at 3:07
  • $\begingroup$ @PaataIvanishvili thank you very much for your reply! Regarding your last comment that this will be true if $\mu$ and $\sigma$ are constants, do I understand correctly that this will be true if some $m$th order derivatives of $\mu$ and $\sigma$ are zero? $\endgroup$
    – Kate
    Commented Feb 8, 2020 at 12:38
  • $\begingroup$ We need $\mu$ and $\sigma$ to be constant. If $\mu$ and $\sigma$ are polynomials (nonconstant) this will not help. If $\mu$ and $\sigma$ are constants then it is quite easy to recover the polynomials $u(x,t) = (\sum_{k=0}^{\infty} \frac{(-t)^{k}L^{k}}{k!})x^{n}$. For example, if $n=2$, then $u(x,t) =e^{-tL}x^{2}=(1-tL+\frac{t^{2}L^{2}}{2})x^{2} = x^{2}-2t\mu x - t\sigma^{2}+t^{2}\mu^{2}$, i.e., $X^{2}_{t}-2t\mu X_{t}-t\sigma^{2}+t^{2}\mu^{2}$ is martingale. $\endgroup$
    – Paata Ivanishvili
    Commented Feb 9, 2020 at 2:58

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .