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I have a question regarding the definitions of quadratic variation for stochastic processes and p-variation for general functions.

Specifically, for a stochastic process $X_t$, the quadratic variation $[X]_T$ is defined using the limit as the mesh of the partition goes to zero:

$$ [X]_T = \lim_{\|P\| \to 0} \sum_{i} (X_{t_{i+1}} - X_{t_i})^2, $$

where $\|P\|$ is the mesh of the partition, and the limit ensures that the partition becomes finer and finer.

On the other hand, for a general function $f(t)$, the p-variation $V^p(f, [0, T])$ is defined using a supremum over all possible partitions:

$$ V^p(f, [0, T]) = \sup_{P} \left( \sum_{i} |f(t_{i+1}) - f(t_i)|^p \right)^{1/p}. $$

My question is: Why does the definition of quadratic variation rely on a limit as the partition's mesh goes to zero, whereas $p$-variation uses a supremum over all partitions?

Thank you in advance for your insights!

FromLaTeXBeginer
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    Quadratic variation concerns the local, second order in stochastic processes, while the $p$-variation of functions is a measure of the global roughness of functions. Limit vs. supremum in their respective definitions does match their purpose: the quadratic variation looks at intricacies in time, while the $p$-variation looks at variability from a larger perspective. – Furdzik Zbignew Nov 22 '24 at 02:30

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Because they are different things.

The quadratic variation (used in stochastic calculus) requires that it is 0 for a smooth function, let's pick $f(x)=x$ in $[0,1]$.

But look what is the sup of the $V^2$? It will be (at least) 1 with a partition of a single element from 0 to 1.

Read this, it explains a bit more.

https://en.wikipedia.org/wiki/P-variation

and this How to show the 2-variation of Brownian motion sample paths is infinite

Andrea
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