Pick a sequence $b_i\in C^\infty_c(\Bbb R)$ and define $S_t=\sum_{i=1}^N X_ib_i(t)$ where $N$ is a discrete r.v. and the $X_i$'s are i.i.d. standard normals. It is clear that $S_t$ is a.s. smooth regardless of the distribution of $N$. However, when I try to compute its quadratic variation I run into the formula $$\Bbb E[(S_t-S_r)^2]=\sum_n p_n\sum_{i=1}^n(b_i(t)-b_i(r))^2=(t-s)^2\sum_n p_n\sum_{i=1}^nb_i'(c_i)^2$$ Which looks like I would have to pick the $b_i$'s really carefully in order to make the above behave nicely. It appears this is a problem of balancing the upper bounds on the derivatives of realizations of a process. If these upper bounds have a supremum the variation vanishes, but if too many of them blow up, so does the variation. Is there a smooth stochastic process with nonzero, finite variation?
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Assuming "smooth" means at least "continuous and finite variation," the quadratic variation will be $0$. That is because if $0 = t_0 < t_1 < \cdots < t_n = T$ is a partition of $[0,T]$ then $$\sum_{i=1}^n (S_{t_i}-S_{t_{i-1}})^2 \le \sup_{i \le n}|S_{t_i}-S_{t_{i-1}}| \cdot \sum_{i=1}^n |S_{t_i}-S_{t_{i-1}}| \le \sup_{i \le n}|S_{t_i}-S_{t_{i-1}}| \cdot V_T$$
where $V_T$ is variation of $S$ over $[0,T]$. Taking the limit as the mesh goes to $0$ then gives $0$ by the continuity of $S$.
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