1

Let $dX_t=a(t,X_t)dt+\sigma (t,X_t)dB_t$.

Let $\sigma (t,X_t)$ satisfies Lipschitz condition and linear growth condition.

Prove that $$\int_0^t \sigma (s,X_s)dB_s$$ is a martingale ($B_t$ is a brownian motion).

Do I need additional conditions here? How to approach this problem?

Snoop
  • 18,347
Paul R
  • 2,413

1 Answers1

1

We can deduce that $E[\int_0^T|\sigma(t,X_t)|^2dt]<\infty$ from the existence theorem (see e.g. Schilling Th.21.13 p. 396). For simplicity here the process is $\mathbb{R}$-valued over $[0,T]$, starts at $X_0=x_0\in \mathbb{R}$ and $a,\sigma :[0,T]\times \mathbb{R}\to \mathbb{R}$. From the linear growth condition we have $|\sigma(t,x)|^2\leq M_T^2(1+|x|)^2\leq 2M_T^2(1+|x|^2)$ for some $M_T>0$ for all $x \in \mathbb{R}$ and the existence theorem states that the unique solution satisfies $E[\sup_{s\leq T}|X_s|^2]\leq k_T(1+|x_0|)^2$ for some $k_T>0$. So $$\begin{aligned}E\bigg[\int_0^T|\sigma(t,X_t)|^2dt\bigg]&\leq 2M_T^2\bigg(T+E\bigg[\int_0^T|X_t|^2dt\bigg]\bigg)\\ &\leq 2TM_T^2\bigg(1+E\bigg[\sup_{t\leq T}|X_t|^2\bigg]\bigg)<\infty \end{aligned}$$ as we wanted to show.

Snoop
  • 18,347
  • how did you show that $2M_T^2\bigg(T+E\bigg[\int_0^T|X_t|^2dt\bigg]\bigg)\leq 2TM_T^2\bigg(1+E\bigg[\sup_{t\leq T}|X_t|^2\bigg]\bigg)$? – Paul R Oct 05 '24 at 16:47
  • 1
    @eMathHelp $|X_t|^2\leq \sup_{t\leq T}|X_t|^2$ for all $t\leq T$ so $\int_0^T|X_t|^2dt\leq \sup_{t\leq T}|X_t|^2\int_0^Tdt=T\cdot \sup_{t\leq T}|X_t|^2$ – Snoop Oct 05 '24 at 16:51
  • What about $E\bigg[\int_0^T|\sigma(t,X_t)|^2dt\bigg]\leq 2M_T^2\bigg(T+E\bigg[\int_0^T|X_t|^2dt\bigg]\bigg)$? You've bounded the expectation, but there are no bounds mentioned for the expectation. – Paul R Oct 05 '24 at 17:02
  • 1
    @eMathHelp From the linear growth condition on $\sigma$ we know that $\int_0^T|\sigma(t,X_t)|^2dt\leq 2M_T(T+\int_0^T|X_t|^2dt)$. Then, take expectation on both sides. – Snoop Oct 05 '24 at 17:23
  • Thank you very much! We didn't use Lipschitz condition, right? It was used just in the theorem. – Paul R Oct 05 '24 at 17:34
  • 2
    Indeed we just used the linear growth condition for the inequality, while the Lipschitz condition is there for the referenced theorem to hold. – Snoop Oct 05 '24 at 17:39
  • How does finiteness of second moment show that this is a martingale? – Paul R Oct 06 '24 at 17:23
  • Couldn't we just right that $E\left[\int_0^T|\sigma(t,X_t)|^2dt\right]\leq E\left[\int_0^T M_T^2(1+|x|^2)dt\right]=TM_T^2(1+|x|^2)$. Why do we need the $sup$ condition? – Paul R Oct 06 '24 at 17:33
  • 1
    @eMathHelp (1) please see the third paragraph of this answer. (2) No, you are fixing some $x$ on one side and you can't do it. The linear growth condition is on the second argument of $\sigma$. – Snoop Oct 06 '24 at 17:37
  • Is there a proof of this statement? – Paul R Oct 06 '24 at 17:44
  • 1
    Please look at Klenke's Th.25.18 p. 570 @eMathHelp – Snoop Oct 23 '24 at 15:12