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Notation: $M_2^c$ are continuous square integrables martingales, $\langle X\rangle_t$ is the quadratic variation and $V_t^{(2)}(\Pi)$ the sum of square of increments of $X$ over the partition $\Pi$.

In the book Brownian motion and stochastic calculus by Karatzas and Shreves, 3 results are used to motivate the study of quadratic variation.

Theorem 5.8. Let $X$ be in $M_2^c$. For partition $\Pi$ of $[0,t]$, we have $\lim_{\|\Pi\|\to0} V_t^{(2)}(\Pi)=\langle X\rangle_t$ (in probability); i.e., for every $\varepsilon>0$,$\eta>0$ there exists $\delta>0$ such that $\|\Pi\|<\delta$ implies $$P[|V_t^{(2)}(\Pi)-\langle X\rangle_t|>\varepsilon]<\eta.$$

5.11 Problem. Let $\{X_t,\mathcal{F}_t;0\le t<\infty\}$ be a continuous process with the property that for each fixed $t>0$ and for some $p>0$, $$\lim\limits_{\|\Pi\|\to0}V_t^{(p)}(\Pi)=L_t\;\;\;\text{(in probability),}$$ where $L_t$ is a random variable taking values in $[0,\infty)$ a.s. Show that for $q>p$, $\lim_{\|\Pi\|\to0}V_t^{(q)}(\Pi)=0$ (in probability), and for $0<q<p, \lim_{\|\Pi\|\to0}V_t^{(p)}(\Pi)=\infty$ (in probability) on the event $\{L_t>0\}$.

5.12 Problem. Let $X$ be in $M_2^c$, and $T$ be a stopping time of $\{\mathcal{F}_t\}$. If $\langle X\rangle_T=0$, a.s. $P$, then we have $P[X_{T\wedge t}=0;\forall 0\le t<\infty]=1$.

The conclusion to be drawn from Theorem 5.8 and Problems 5.11 and 5.12 is that for continuous, square-integrable martingales, quadratic variation is the "right" variation to study. All variations of higher order are zero, and, except in trivial cases where the martingale is a.s. constant on an initial interval, all variations of lower order are infinite with positive probability. Thus, the sample paths of continuous, square-integrable martingales are quite different from "ordinary" continuous functions. Being of unbounded first variation, they cannot be differentiable, nor is it possible to define integrals of the form Ito $\int_0^t Y_s(\omega) dX_s(\omega)$ with respect to $X\in M_2^c$, in a pathwise (i.e., for every or P-almost every $\omega\in\Omega$), Lebesgue-Stieltjes sense.

First, I don't understand how problem 5.12 comes into play and how it is used to argue that the quadratic variation is "right" variation to study (see last quoted paragraph).

Second, I don't understand how they can deduce from this discussion that we cannot define a pathwise integral with respect to martingales of unbounded variations.

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First, I don't understand how problem 5.12 comes into play and how it is used to argue that the quadratic variation is "right" variation to study.

What this is saying is that a martingale is either constant or it has non-vanishing (and locally finite) quadratic variation. This highlights the fact that the quadratic order is the right order at which to study the variation of martingales.

Second, I don't understand how they can deduce from this discussion that we cannot define a pathwise integral with respect to martingales of unbounded variations.

The Lebesgue-Stieltjes theory allows you to define integrals where your integrators have bounded variation, as functions of bounded variation are the difference of two increasing functions (which in turn can be associated with measures). If you wish to define a stochastic integral with a sufficiently rich class of integrands, a pathwise integral will fail for the following reason:

Theorem: Suppose $G:[0,T] \to \mathbb{R}$ is a function for which the Riemann-Stieltjes integral $\int_0^T f dG$ exists for all continuous functions integrands $f$. Then $G$ is of bounded variation.

Proof. This is a consequence of the Banach-Steinhaus theorem. $\blacksquare$

A pathwise stochastic calculus does exist (e.g. see here for some references); it certainly is more elementary than Itô calculus, but the class of integrands is too restrictive for some applications.

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  • $\begingroup$ Thank you for the answer. I think I understand the second point. However, for the first point I am having a hard time relating problem 5.12 and what you wrote. $\endgroup$
    – edamondo
    Commented Jan 13, 2022 at 16:30
  • $\begingroup$ @edamondo To make it easier to grasp, let your stopping time be any deterministic time. It's just saying if you look at a martingale up to a time $T$ and you measure its quadratic variation to be zero, then if the martingale started at zero it stays at zero. Any other martingale will accumulate quadratic variation. $\endgroup$ Commented Jan 13, 2022 at 16:38
  • $\begingroup$ Okay. However, I still don't see from these result why the third variation for instance is not the "right" one to study. $\endgroup$
    – edamondo
    Commented Jan 13, 2022 at 18:06
  • $\begingroup$ @edamondo Because if the quadratic variation of a martingale is finite and nonvanishing then the third order variation (or any other variation $>2$) will be zero. Thus, they give you no information about the martingale. $\endgroup$ Commented Jan 13, 2022 at 18:09
  • $\begingroup$ But we could say the same thing of the third variation : if the third variation of a martingale is finite and nonvanishing then the third order variation (or any other variation > 3) will be zero, no? So what makes the second variation so special? $\endgroup$
    – edamondo
    Commented Jan 13, 2022 at 18:43

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