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Problem:

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

$dY_t=X_tdt-Z_tdt$ 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 $\int_r^s a\exp(B_t\sigma)dt.$

saz
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user10248
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  • What do you call "solve the time integral of exponential of Brownian motion"? – Did May 09 '14 at 21:24
  • @Did Could you please explain how may I evaluate $\int_r^s a\exp(B_t\sigma)dt?$ Thanks – user10248 May 10 '14 at 00:48
  • 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. – Kai Sikorski May 10 '14 at 01:22
  • 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? – Did May 10 '14 at 06:18
  • 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. – Kai Sikorski May 10 '14 at 06:29
  • @Did I mean is there an expression to this integral so that I can remove the integral sign and just write an equation? Just like we evaluate an exact integral $\int_r^s x^2 dx=\frac{1}{3}(s^3-r^3)$. Thanks – user10248 May 11 '14 at 00:25
  • Then no, there is not. – Did May 19 '14 at 07:15
  • This is a case of "multiplicative" noise. Your equation is a special case of eq (1.5) at pag 2 of https://userswww.pd.infn.it/~orlandin/fisica_sis_comp/multiplicative_noise.pdf – Quillo Feb 14 '22 at 13:18
  • See also this, it should provide a full answer to your case: https://math.stackexchange.com/a/3709327/532409 – Quillo Feb 14 '22 at 13:22

1 Answers1

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As mentioned in the comments the expression $\int e^{B_s}ds$ is as closed form as it can get.

But if you interested to know more about its law, there is a large literature on "integrated Brownian motion" eg. "The Integral of Geometric Brownian Motion" by Daniel Dufresne.

Thomas Kojar
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