I am trying to do the following stochastic integral

$$ \int_0^T \mu(B_s) dB_s - \frac{\int_0^T (\mu(B_s))^2 ds}{2} $$ where $ \{ B_t \}$ is a standard Brownian motion, and $\mu(x) = \frac{\beta}{2x} (x^2 - \frac{4\gamma}{\sigma^2} - \frac{1}{ \beta})$.

I currently only know how to integrate when the integrands are constant, i.e. $\mu$.is constant function. I was wondering how to integrate under more complicated case as above.

Thanks and regards!

  • $\begingroup$ Is that a $\sqrt{x}$ in the denominator? What do you want it to do when $x=B_s$ is negative? $\endgroup$ – Nate Eldredge Nov 20 '10 at 4:36
  • $\begingroup$ @Nate: Sorry, I made a mistake when deriving this expression. I just made the correction to the post. $\endgroup$ – Tim Nov 23 '10 at 2:28
  • $\begingroup$ How do you define to compute? Is it: to express as a function of $B_0$, $B_T$ and $T$? $\endgroup$ – Did Mar 17 '11 at 11:40
  • $\begingroup$ @Didier: Yes, I think so. It has been a while, and I am still learning towards solving similar kinds of problems. . $\endgroup$ – Tim Mar 17 '11 at 13:01
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    $\begingroup$ My point is that it is not always possible to get rid of every integral term--as answers by @Sivaram and @TheBridge basically show. So what you can get are several equivalent expressions for your Brownian functional, none of which actually computes it. $\endgroup$ – Did Mar 17 '11 at 13:22

You can do this by different methods. One way I like is to use Ito's formula.

For instance, consider $B_t^2$.

$d(B_t^2) = 2B_t dB_t + dt$.

Hence, $\int_{0}^{T} B_t dB_t = \frac{1}{2} \int_{0}^{T} (d(B_t^2) - dt) = \frac{1}{2} (B_T^2 - T)$.

Similarly, consider $B_t^3$.

Use Ito's formula,

$d(B_t^3) = 3B_t^2 dB_t + 3 B_t dt$.

Hence, $\int_{0}^{T} B_t^2 dB_t = \frac{1}{3} \int_{0}^{T} (d(B_t^3) - 3 B_t dt) = \frac{1}{3} (B_T^3 - 3 \int_{0}^{T} B_t dt)$.

Use these to get your desired answer. (Plug in for $\mu(B_s)$ and then simplify to get integrals of the form $\int_{0}^{T} B_t^2 dB_t$ and $\int_{0}^{T} B_t dB_t$)

Hope this helps.


To finish the job initiated by Sivaram Ambikasaran, first you can say that $\mu(x)=\alpha x+ \beta 1/x$

The stochastic integral term with $\beta=0$ case is answered by Sivaram Ambikasaran.

For the $\alpha=0$ case you have using Itô's lemma :

$dLnB_t= \frac{1}{B_t}dB_t-\frac{1}{2.B_t^2}dt$

So integrating this gives you :

$\beta.\int_0^T \frac{dB_s}{B_s}=\beta.(Ln(\frac{B_T}{B_0})+1/2\int_0^T \frac{1}{B_s^2}ds)$

If you start your Brownian Motion at 0, the problem is not well defined. If $B_0\not=0$ then returning to the intial problem (still with $\alpha=0$) you have :

$ \int_0^T \mu(B_s) dB_s - \frac{\int_0^T (\mu(B_s))^2 ds}{2} = Ln(\frac{B_T}{B_0})+\frac{\beta.(1-\beta)}{2}.\int_0^T \frac{1}{B_s^2}ds$ if I am not mistaken.

For the general case ($\alpha\not= 0$ and $\beta\not= 0$) just adds the solutions ( i.e. respectively for $\alpha=0$ and $\beta=0$) and also add the cross product term from $\int_0^T (\mu(B_s))^2 ds$ (which should be something like $-T.\alpha.\beta$).


  • $\begingroup$ Thanks for digging this up! It has been a while, and I have been learning things more preliminary before eventually come back to learn how to solve stochastic integral problems. $\endgroup$ – Tim Mar 17 '11 at 13:05

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