Let $(B_t)_{t\geq 0}$ be a Brownian motion. Is it true that if the generalized Ito process $$ X_t=\int_0^tf(B_s)\;dB_s+\int_0^tg(B_s)\;ds, $$ $f,g\in\mathcal{C}(\mathbb{R})$, is a martingale, then $g\equiv 0$? It is easy to see that $X_t\in L^2$ for any $t\geq 0$ and hence $(M_t)_{t\geq 0}:=(X^2_t-\langle X\rangle_t)_{t\geq 0}$, where $\langle X\rangle$, denotes the quadratic variation is a martingale. Hence, if I could manage to show that $(M_t)_t$ is of BV for $g\neq 0$, I would be done since any BV martingale is constant as. However, I am struggling a bit with the argument.

  • $\begingroup$ I don't think your argument will work. Note also that it is $X_t^2-\langle X\rangle X_t$ which is a martingale. My feeling is the answer is yes, $g\equiv0$. The proof is not difficult if you assume $g$ is either bounded or Lipschitz. The general case seems to be a little more difficult. $\endgroup$ – Jason Dec 5 '17 at 16:44
  • $\begingroup$ Of course you are right. What’s the idea if $g$ is bounded? $\endgroup$ – julian Dec 5 '17 at 16:52
  • $\begingroup$ Observe that $(X_t)$ is a martingale if and only if $E[\int_s^tg(B_u)du|\mathcal F_s]=0$ a.s. for all $s<t$. If $g(x_0)\neq0$, assume $g(x_0)=:\alpha>0$, and let $\delta>0$ be such that $g(x)>\alpha/2$ for $|x-x_0|<\delta$. Let $A_{s,t}=\{\max_{u\in[s,t]}|B_u-B_s|>\delta/2\}$. Note that $P(A_{s,t})\to0$ as $t\to s$. Bound the conditional expectation below on the event $\{|B_s-x_0|<\delta/2\}$ by separately considering $A_{s,t}$ and $A_{s,t}^c$, which have respective lower bounds $(t-s)P(A_{s,t})\min g$ and $(t-s)P(A_{s,t}^c)\alpha/2$. The sum of these is positive if $t-s$ is small enough. $\endgroup$ – Jason Dec 5 '17 at 17:02

If $X$ is a martingale, then $\int_0^t g(B_s)\,ds = X_t-\int_0^t f(B_s)\,dB_s$ is a local martingale with paths of (locally) bounded variation. As you note, the only BV martingales are the constant ones, and this remains true for BV local martingales by the obvious localization argument.

  • $\begingroup$ thanks for your answer. why are the paths only locally bounded? the left-hand side of your equation is continuous differentiable, ie. is of BV on any $[0,t]$. $\endgroup$ – julian Dec 5 '17 at 18:41
  • $\begingroup$ By locally BV, I meant of bounded variation on $[0,t]$, for each $t>0$. $\endgroup$ – John Dawkins Dec 6 '17 at 17:00

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