I have this seemingly simple problem which I haven't been able to solve. I have two standard Brownian motions, $B$ and $W$, on the same probability space and under the same filtration (I am not so sure whether this last part is sensible). What is $\int_0^tB_s\,dW_s$?

The product rule gives $\int_0^tB_s\,dW_s + \int_0^tW_s\,dB_s = B_tW_t$. Obviously the expressions for $\int_0^tB_s\,dW_s$ and $\int_0^tW_s\,dB_s$ should be symmetric but that doesn't really narrow down the possible solutions that much. I played with a few two-variable functions using Ito's lemma but no result so far.

I also gave Riemann sum a shot but I get the same result as above (because of the way I manipulate the terms in the sum). Maybe there is another way to express the terms in the sum that leads to what I want but I doubt that for some reason.

I feel like this has a very simple solution which I am not seeing but I wanted to post it here in any case.

  • $\begingroup$ I know this doesn't answer the question but I think you can use Ito Isometry to show that $\mathbb{Var}[\int_0^t B_s dW_s] = \frac{t^2}{2}$ and obviously it has expectation $0$. $\endgroup$
    – Stanley
    May 25, 2015 at 22:00
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    $\begingroup$ Correction: $\mathbb{Var}[\int_0^tB_sdW_s] = \int_0^t B_s^2 ds$ $\endgroup$
    – Stanley
    May 25, 2015 at 23:41
  • $\begingroup$ Also relevant: link $\endgroup$
    – Stanley
    May 25, 2015 at 23:42
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    $\begingroup$ "What is $\int_0^tB_s\,dW_s$?" What kind of answer are you expecting? If you are after some function $G$ such that $\int_0^tB_s\,dW_s=G(B_t,W_t,t)$, this cannot exist. $\endgroup$
    – Did
    May 26, 2015 at 6:04
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    $\begingroup$ @Did I was wondering if such a closed form answer exists and if it does, what it is. The two Brownian motions have nothing to do with each other so what I did was a bit like expecting to get a closed form expression for a Lebesgue-Stieltjes integral of two arbitrary functions, which is plain dumb. But on the other hand, $B$ and $W$ are not that arbitrary. They belong to a very specific class of processes. So I couldn't convince myself that a closed form expression cannot exist. That is why I posted the question here. $\endgroup$
    – Calculon
    May 26, 2015 at 6:11


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