$dX_t = \sigma X_tdB_t$, $X_0=x$.

$dY_t=X_tdt-Z_tdt$ find $Y_t$, where $Z_t$ is a control and $B_t$ is standard Brownian motion.

My attempt:

From Ito's lemma,

$\partial_BX_t=\sigma X_t$, therefore $X_t=c(t)\exp(\sigma B_t)$.

$\partial_tX_t+\frac{1}{2}\partial^2_BX_t=0$, substitue in above expression for $X_t$ and

$c^\prime(t)+\sigma^2\frac{1}{2}c(t)=0$ which implies $c(t)=a\exp(-\sigma^2\frac{t}{2}).$ Putting all together $$X_t=a\exp(-\sigma^2\frac{t}{2}+B_t\sigma)$$ which can be solved for $X_t=x$ to fine $a$.

Then $Y_s-Y_r=\int_r^s a\exp(-\sigma^2\frac{t}{2}+B_t\sigma)dt+\int^s_rP(t)dt$. How can I solve the time integral of exponential of Brownian motion? A hint is sufficient. I would like to do it myself. Thank you all.

Edit: I need help to find a closed form expression for $\int_r^s a\exp(B_t\sigma)dt.$

  • $\begingroup$ What do you call "solve the time integral of exponential of Brownian motion"? $\endgroup$ – Did May 9 '14 at 21:24
  • $\begingroup$ @Did Could you please explain how may I evaluate $\int_r^s a\exp(B_t\sigma)dt?$ Thanks $\endgroup$ – user10248 May 10 '14 at 0:48
  • $\begingroup$ You can't get an explicit expression just in terms of $B_s$ etc if that's what you're looking for. You can compute moments relatively easily though and often that's sufficient. $\endgroup$ – Kai Sikorski May 10 '14 at 1:22
  • $\begingroup$ This integral is a random variable. What do you call "evaluate" a randoim variable? The result cannot be that the random variable is $3.14159\pm0.00001$, right? $\endgroup$ – Did May 10 '14 at 6:18
  • $\begingroup$ I think he was hoping that it could be expressed explicitly just in terms of $B_t$ like for example $\int B \mathrm{d}B$ etc. $\endgroup$ – Kai Sikorski May 10 '14 at 6:29

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