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Technically speaking, are Ito integrals of stochastic processes $S$ with respect to Brownian motion $B$

$$ \int_{0}^{t} S_s dB_s $$

concrete stochastic processes in $t$ or only equivalence classes of indistinguishable stochastic processes in $t$?

Joker123
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    These integrals do not exist in Riemann Stieltje's sense and they can only be defined measure theoretically up to indistinguishability. – Kavi Rama Murthy Apr 23 '19 at 23:18
  • Ok, thanks, Do all these integrals necessarily define almost-surely continuous processes? – Joker123 Apr 23 '19 at 23:30
  • Typically, one considers the stochastic integral as a stochastic process $X_t=\int_0^t S_s dB_s$ and then restricts the attention to its continuous version. – Mdoc Apr 24 '19 at 00:13
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    @Mdoc You can't say "its continuous version" like such a thing is unique. I can always change the value of a continuous version on a null set and get another continuous version. – Rhys Steele Apr 24 '19 at 13:29

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Because Brownian motion has infinite variation but finite quadratic variation we are forced to work away from Riemann Stieltjes and even away from Young-integrals. Meaning we can only make sense of the limit

$$\sum f(t_{i})(B_{t_{i+1}}-B_{t_{i}})$$

as an $L^{2}$-function. From here the issues are no different than in $L^{p}$-space where we define functions up to measure-zero

However, one should mention that if one is adds some more conditions on the partition and the functions involved, one can even get almost sure convergence (but one can also pick a different sequence see Exercise 1.15 in Môrters-Peres Brownian motion book and not get it, and that's why we resorted to L2-convergence)

Understand better stochastic integral through a.s. convergence

and

What is the explicit obstruction to almost sure convergence in stochastic integrals?

If the meshes are chosen regularly with meshwidth $h_n\to 0$ such that $$ \sum_{n=1}^{\infty}\mathbb{E}\left[ \int_{0}^{1}(f(\omega,x)-f(\omega,\lfloor x/h_n\rfloor h_n)^{2}\;dx\right]<\infty, $$ then the Itô sums converge almost surely to the Itô integral (under usual conditions on the integrand)

This is proven, for example, in Remark 7.7 of the textbook [Brownian motion][1] by Mörters and Peres. It is an immediate consequence of Itô's isometry.

Thomas Kojar
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