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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.

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  • $\begingroup$ 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
    Commented Dec 5, 2017 at 16:44
  • $\begingroup$ Of course you are right. What’s the idea if $g$ is bounded? $\endgroup$
    – user427574
    Commented Dec 5, 2017 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
    Commented Dec 5, 2017 at 17:02

1 Answer 1

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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.

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  • $\begingroup$ 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 $[0,t]$. $\endgroup$
    – user427574
    Commented Dec 5, 2017 at 18:41
  • $\begingroup$ By locally BV, I meant of bounded variation on $[0,t]$, for each $t>0$. $\endgroup$ Commented Dec 6, 2017 at 17:00

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