I'm working through Protter's book on stochastic integration; this is problem 16 from chapter 2. I can't seem to crack it--maybe someone here can give me a hint? Let B be standard Brownian and H be a continuous, adapted process. I'm trying to prove that $$ \lim_{h \to 0} \frac{1}{B_{t+h} - B_t}\int_t^{t+h} H_sdB_s = H_t, $$ where the limit is taken in probability.

This looks a lot like the classical mean-value theorem, but I can't see how to apply that theorem or emulate its proof. Ito's formula lets us rewrite the problem as $$ \lim_{h \to 0} \frac{1}{B_{t+h} - B_t}\int_t^{t+h} (H_s - H_t)dB_s = 0. $$ But I'm not sure that helps. The integral goes to zero: continuity of $H$ upgrades to ucp continuity on $[t,t+h]$ (its a compact set) and the stochastic integral preserves ucp. But the denominator $B_{t+h} - B_t$ also goes to zero. In a classical setting I'd try l'Hopital, but nothing here is differentiable.

Also, it's interesting to me that the limit is in probability--not ucp. That suggests to me that commuting limits is not the right approach. Do I want to be looking at the variance of this expression (show that it goes to zero)? It's not obvious to me how to do that either...

Any observations would be greatly appreciated!

  • $\begingroup$ See this question: math.stackexchange.com/q/573000 - there, the result is proved for $t=0$. Using the stationarity of the increments, the claim follows easily for any $t \geq 0$. $\endgroup$ – saz Apr 19 '14 at 16:12
  • $\begingroup$ Thanks saz. Hmm... the hypotheses on H are slightly different in that problem (adapted versus progressive and bounded). And Protter does not appear to have covered Ito's isometry (actually, it doesn't even appear in the index of his book!). I will study your answer and post back here if I can adapt it to my situation (or if I find another proof). Thanks again! $\endgroup$ – mathrat Apr 19 '14 at 22:01
  • $\begingroup$ Since the process $H$ is continuous, it is in particular progressively measurable. And the boundedness is used to ensure the existence of the stochastic integral; so probably you have to use some stopping argument. Don't hesitate to ask if you don't get along with it. $\endgroup$ – saz Apr 20 '14 at 7:23
  • $\begingroup$ Whoops, meant to respond sooner. So the two key insights are to see the denominator can be controlled by marginalizing (what you showed in the linked post) and then that the integral can be controlled by Markov's inequality and the Ito isometry (which you also alluded to) the trick here being that we must consider the square root of the isometry (I'm a bit slow--took me awhile to see that). Also, in the context of Protter's book, while Protter does not discuss the Ito isometry in full generality he does have a nice discussion of it on page 77 as it applies in particular to BM. Thanks again $\endgroup$ – mathrat May 14 '14 at 14:06

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