Timeline for Martingale polynomial functions
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2020 at 2:58 | comment | added | Paata Ivanishvili | 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. | |
| Feb 8, 2020 at 12:38 | comment | added | Kate | @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? | |
| Feb 8, 2020 at 3:07 | comment | added | Paata Ivanishvili | 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. | |
| Feb 8, 2020 at 2:46 | comment | added | Paata Ivanishvili | 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. | |
| Feb 8, 2020 at 2:36 | comment | added | Paata Ivanishvili | 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 $u_{t}+\mu u_{x}+\frac{\sigma^{2} u_{xx}}{2}$. | |
| Feb 7, 2020 at 20:45 | review | First posts | |||
| Feb 7, 2020 at 22:35 | |||||
| Feb 7, 2020 at 20:42 | history | asked | Kate | CC BY-SA 4.0 |