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This is more of a sanity check. Let $\Gamma(t)$ be the random trajectory of a 2D particle at time $t$ that has a prescribed velocity vector $\dot\Gamma(t)=[v(t)\cos(B_t), v(t)\sin(B_t)]$, where the speed $v(t)$ is deterministic and the heading $B_t$ is a standard Brownian motion (Wiener process). Is the following assessment true?

Almost surely, for every sample path $\omega$ of $B_t$, $\Gamma(t, \omega)\in \text{C}^1$ but not in $\text{C}^2$.

My reasoning is simple. A sample path of $B_t$ is almost surely continuous so $$\Gamma(t)=\left[\int_0^{t}v(s)\cos(B_s(\omega))\text{d}s, \int_0^{t}v(s)\sin(B_s(\omega))\text{d}s\right],$$ is well-defined as a Riemann or Lebesgue integral, with derivative $\dot\Gamma(t)$ as above. But since $B_t$ is almost surely not differentiable $\dot\Gamma(t, \omega)$ is not differentiable.

Finally, does anyone know a reference that deals with this type of processes where Brownian motion directly appear in the orientation?

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Your formulation has a small problem. Maybe the integral should read $\int_0^t v(s)\cos(B_s(\omega))ds$?

Suppose $v$ is nice, say bounded. The map $t\mapsto \int_0^t v(s)\cos(B_s(\omega))ds$ is $P$-almost surely Lipschitz. Therefore, this map has a derivative Lebesgue-almost everywhere $P$-a.s., which equals $v(t)\cos(B_t(\omega))$.

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  • $\begingroup$ I fixed the dummy variable. Also, you are right, Lipschitz is indeed the case. Thanks! $\endgroup$
    – RozaTh
    Oct 6, 2021 at 0:49

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