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For Ito's integral: \begin{equation*} \int_0^T f(t, X_t) dX_t, \end{equation*} where $X_t$ is Brownian motion. Because, the Ito integral is martingale, which means that \begin{equation} \mathbb{E}\left\{\int_0^T f(t, X_t) dX_t\right\}=\lim_{N\rightarrow\infty} \sum_{i=0}^{N-1} f(t_i, X_i) \mathbb{E} \left\{X_{i+1} - X_i\right\} =0 \end{equation} By using the Fubini's Theorem, we have \begin{equation*} \mathbb{E}\left\{\int_0^T f(t, X_t) dX_t\right\}=\int_0^T \mathbb{E}\left\{f(t, X_t)\right\}dX_t=0 \end{equation*} Can anyone suggest how to prove that: \begin{equation*} \int_0^T \mathbb{E}\left\{f(t, X_t)\right\}dX_t=0? \end{equation*}

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  • $\begingroup$ This proof is not correct, you always have to leave random variables within the expectation. @Antinous What are you saying? These are Itô integrals. $\endgroup$
    – shalop
    Aug 7 '17 at 9:22
  • $\begingroup$ @Shalop thx, do you mean the Fubini's theorem is not applicable to Ito integral? $\endgroup$
    – Stephen Ge
    Aug 7 '17 at 9:24
  • $\begingroup$ Definitely not the way that you are using it. $\endgroup$
    – shalop
    Aug 7 '17 at 9:24
  • $\begingroup$ @Shalop Sorry I wasn't sure what an Itô integral is - I should have checked. But I thought $E$ was expected value, so maybe we could use the definition of expected value somehow. A fleeting thought as I originally said. $\endgroup$
    – Pixel
    Aug 7 '17 at 9:25
  • $\begingroup$ $E$ is the expectation $\endgroup$
    – Stephen Ge
    Aug 7 '17 at 9:31
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Fubini's theorem requires you to have two (not necessarily distinct) measure spaces. I think that if you spit this out explicitly, your life will be much easier. Here, they are $([0,T],\lambda)$, where $\lambda$ is the 1-dimensional Lebesgue measure and, $(\Omega, \mathscr F, \mathbb P)$ the probabiity space on which your BM is defined.

The stochastic process $X$ is a function of two variables $X:[0,T]\times\Omega\to\mathbb R,\;\;$ $X:(t,\omega)\mapsto X_t(\omega)$. $\mathbb E$ denotes integration with respect to the $\omega\in \Omega$ variable. So, \begin{align} \mathbb E\int_0^Tf(t,X_t(\omega))dX_t(\omega)& =\int_\Omega\int_0^Tf(t,X_t(\omega))dX_t(\omega)d\mathbb P(\omega)\\ &=\int_0^T\int_\Omega f(t,X_t(\omega))dX_t(\omega)d\mathbb P(\omega)\\ &=\int_0^T\mathbb E\left[f(t,X_t(\omega))dX_t(\omega)\right]. \end{align}

This is all provided the hypotheses of Fubini's theorem hold, of course.


This answer is formal as $dX_t$ is not actually a measure... but you'll get into those details if you decide to pursue stochastic analysis further.

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  • $\begingroup$ This answer is quite misleading for two reasons. First of all, you use the notation $\int_0^T f(t,X_t(\omega))dX_t(\omega)$ which would suggest that there is a well-defined pathwise notion of integration against each individual sample path $X(\omega)$. This is not at all the case, as stochastic integrals are defined via an isometry of appropriate Banach spaces, and heavily rely on the probabilistic structure of the process. One has to be very careful with the specific types of integrands he puts into the integral, otherwise bad things can happen. $\endgroup$
    – shalop
    Aug 10 '17 at 8:57
  • $\begingroup$ The other issue is the way you have switched the two integrands here... like you say it's only formal, so this offense is a bit more understandable, but still Fubini's theorem does not apply in any way here. $\endgroup$
    – shalop
    Aug 10 '17 at 9:00
  • $\begingroup$ @Shalop $\omega\mapsto \int_0^Tf(t,X_t(\omega))dX_t(\omega)$ is a random variable which is $a.s.$ well-defined, if you like, and here we don't care how it is defined outside this set of measure 0. $\endgroup$
    – Bowditch
    Aug 10 '17 at 11:31
  • $\begingroup$ As for the other issue, I agree with you. One ought write $$E\int f(t,X_t)dX_t = \lim_{|D|\to 0}E\sum_{t_i\in D}f(t_i, X_{t_i})(X_{t_{i+1}}-X_{t_i})]$$ etc and justify pulling the limit out. $\endgroup$
    – Bowditch
    Aug 10 '17 at 11:33
  • $\begingroup$ You are missing my point, which is that the Riemann approximations $$\sum_{t_i \in D} f(t_i,X_{t_i})(X_{t_{i+1}}-X_{t_i})$$ do not converge a.s. to $\int f(t,X)dX_t$ (see this discussion). In general, the convergence only holds in $L^2(\Omega)$, which is what I mean when I say there is no "pathwise" notion of stochastic integration. Perhaps you are aware of this. Therefore, the better notation is just to write $\bigg[ \int f(t,X_t)dX_t \bigg] (\omega) $ instead of the other thing. $\endgroup$
    – shalop
    Aug 10 '17 at 11:49

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