Does there exist a continuous time martingale $X_t$ not a.s. constant in $t$ that is almost surely everywhere differentiable?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
5
The answer is no.
Indeed, if a martingale is a.s. everywhere differentiable, then its quadratic variation is a.s $0$. So, by the Burkholder--Davis--Gundy inequality, the martingale is a.s. constant.
Details: Suppose that $X:=(X_t)_{t\in[0,1]}$ is an almost surely (a.s.) everywhere differentiable martingale. Replacing $X_t$ by $X_t-X_0$, without loss of generality let us assume that $X_0=0$. Take any real $a>0$ and consider the bounded martingale $X^a:=(X_t^a)_{t\ge0}$, where $X_t^a:=X_{\min(t,T_a)}$ and $T_a:=\inf\{t\in[0,1]\colon|X_t|=a\}$, with $\inf\emptyset:=\infty$.
For any real-valued function $f$ on $[0,1]$, define its quadratic variation by the formula $$[f]:=\limsup\sum_{j=1}^n(f(t_j)-f(t_{j-1}))^2,$$ where the $\limsup$ is taken over all "partitions" $0=t_0<\cdots<t_n=1$ of $[0,1]$ as $\max_{1\le j\le n}(t_j-t_{j-1})\to0$.
By the Burkholder--Davis--Gundy inequality, \begin{equation} c\,E([X^a]^{1/2})\le EM^a\le C\,E([X^a]^{1/2}), \tag{BDG} \end{equation} where $c$ and $C$ are universal positive real constants and $M^a:=\max_{t\in[0,1]}|X_t^a|$.
By the first one of inequalities (BDG), $[X^a]<\infty$ a.s. So, using (i) Corollary 23 in this paper, (ii) the remark on line 3 of page 4228 of the same paper that $\mu_f=0$ iff $f\in V_2^0$, (iii) the obvious identity $\mu_f([0,1])=[f]$, and (iv) the definition of $V_2$ as the set of all functions $f$ with $[f]<\infty$, we conclude (as in this answer) that $[X^a]=0$ a.s. Therefore, by the second one of inequalities (BDG), $M^a=0$ a.s. for each $a$ and hence $X=0$ a.s., as desired.
answered Jul 16, 2021 at 1:06
-
1$\begingroup$ Ah, I had picked up that tge quadratic variation would be $0$, but didn’t think to use the BDG inequality to conclude the a.s. constantness. Thanks! $\endgroup$ Commented Jul 16, 2021 at 3:34
-
1$\begingroup$ I believe the quadratic variation process that appears in the BDG inequality is defined in a different way, is it not? E.g. the quadratic variation process of the Brownian motion $B_t$ is just $[B]_t=t$, despite the fact that the quadratic variation of almost every Brownian path is infinite (on every time interval). $\endgroup$ Commented Jul 22, 2021 at 21:46
-
$\begingroup$ @MateuszKwaśnicki : Thank you for this comment. Do you have an idea how to fix it? I will leave this answer for now, as a partial one, which works at least when the differentiability condition is strengthened to the continuous differentiability. $\endgroup$ Commented Jul 23, 2021 at 13:06
-
$\begingroup$ I guess if the true "quadratic variation" is zero, then the "quadratic variation process" is zero, too — by the very definition of these two notions. But this is just what my intuition tells me, I did not think about that. $\endgroup$ Commented Jul 23, 2021 at 21:17
-
$\begingroup$ @MateuszKwaśnicki : Yes, the problem here is whether a differentiable martingale or, in general, a differentiable process can have a finite nonzero quadratic variation. For a non-random differentiable function, this cannot happen. $\endgroup$ Commented Jul 23, 2021 at 21:41
$\begingroup$
$\endgroup$
This is also a direct consequence of Ito's formula. Let $T_n:=\inf\{t:|X_t|>n\}$ and define $Y^{(n)}_t:=X_{t\wedge T_n}$. As noted, $X$ has $0$ quadratic variation, hence so does $Y^{(n)}$. By Ito, the process $(Y^{(n)}_t-Y^{(n)}_0)^2$ is a bounded non-negative martingale with initial value $0$. Therefore $\Bbb E[(X_{t\wedge T_n}-X_0)^2]=\Bbb E[(Y^{(n)}_t-Y^{(n)}_0)^2]=0$ for each $t\ge 0$. By Fatou, $\Bbb E[(X_t-X_0)^2]=0$ for each $t\ge 0$.
$\begingroup$
$\endgroup$
1
Others have answered that there does not exist an a.s. differentiable martingale process that's not constant. One related fact (not asked for but interesting) is that an a.s. differentiable Markov process satisfies a deterministic ODE with random initial conditions. See https://www.ams.org/journals/notices/196808/196808FullIssue.pdf page 748
Somehow martingale/Markov don't work well with differentiability.
-
$\begingroup$ Interesting indeed. I wonder if differentiability a.e. runs into the same problem. $\endgroup$ Commented Jul 18, 2021 at 23:49