I am new to stochastic calculus but the statement below confuses me:

Beside the issue of the impossible consensus on a probability measure, the representation of the gain from trading lacks a pathwise meaning: while being a limit in probability of approximating Riemann sums, the stochastic integral does not have a well-defined value on a given ‘state of the world’. This causes a gap in the use of probabilistic models, in the sense that it is not possible to compute the gain of a trading portfolio given the realized trajectory of the underlying asset price, which constitutes a drawback in terms of interpretation.

Source: Riga C., (2015), Pathwise functional calculus and applications to continuous-time finance, Page 3

since the computation of a stochastic integral entails essentially computing a distribution, why can we then not define it pathwise? What am I missing? e.g. $\int_{0}^{t} X_{s} dB_{s} \sim \mathcal{N}(0, \sigma^{2})$ surely it is then simply pathwise defined on any state $\omega \in \Omega$ as $\mathcal{N}(0,\sigma^{2})(\omega)$.

Can anyone describe my misunderstanding, and help me?

  • $\begingroup$ What is the source of your quotation? $\endgroup$ Mar 16, 2020 at 19:52

1 Answer 1


Breezing through the referenced paper, the point of it seems to be to develop Ito calculus for non-anticipative functionals $$ F(t, X_t),$$ where $X_t := \left\{X(u)\mid 0\leq u\leq t \right\}$ and $\left(X(u)\right)_{u\geq 0}$ is a stochastic process. For example, for $$ F(t,X_t)=t^{-1}\int_0^t X(u)du.$$ I think that, in that introductive paragraph, the author simply mentions that classical Ito calculus (formula) didn't have 'pathwise interpretability' until the machinery behind functional Ito calculus arrived, which essentially generalizes it, not that classical Ito calculus had any shortcomings.


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