Is it possible to construct a stochastic process $X_t$ where the limit
$\lim_{\Delta \rightarrow 0} \rm{Var}\left(\frac{X_{t_0+\Delta}-X_{t_0}}{\Delta}\right)$
does not exist but the sample paths are still differentiable?
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$\begingroup$ You want an example where the variance always exists, right? Otherwise, you could take a r.v. $\xi$ with nonexistent 2nd moment, and let $X_t=\xi t$. $\endgroup$– Serguei PopovCommented Nov 20, 2015 at 10:23
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$\begingroup$ Thanks @Serguei. Yes, for every $t$, the variance of the random variable $X_t$ should be defined. $\endgroup$– mcsanchezCommented Nov 21, 2015 at 4:49
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consider $\sum X_n e^{int}$ where the $X_n$ have the property that $X_n$ are independent and eventually 0. Then every sample path is a trig polynomial and infinitely differentiable, however by arranging that $\sum var(X_n) < \infty, \sum var(nX_n) = \infty $ you'll probably get what you want.
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$\begingroup$ Thanks @Michael. It seems like Gaussians with mean $1/n$ and variance $1/n^2$ would work. $\endgroup$ Commented Nov 21, 2015 at 5:09
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$\begingroup$ Gaussians will not work. If they are Gaussian they answer to your question is no, if the pointwise limit of the difference quotient exists then the variance must converge, You can see this via ptwise implies in distribution implies cnvgs of characteristic fctns implies cnvgs of variance. $\endgroup$– MichaelCommented Nov 23, 2015 at 9:47