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I would like to solve the following equation explicitly using Ito's lemma: $$ x_t := a_t -b_t c_t , $$ where $x_t$ is an Ornstein-Uhlenbeck process (see here) $$ dx_t = \theta (\mu-x_t) dt+ \sigma dW_t $$

Can I simply use the solution provided here? So:

$$ f(x_t, t) = x_t e^{\theta t}$$ and then:

\begin{align} df(x_t,t) & = \theta x_t e^{\theta t}\, dt + e^{\theta t}\, dx_t \\[6pt] & = e^{\theta t}\theta \mu \, dt + \sigma e^{\theta t}\, dW_t. \end{align} $$ x_t = x_0 e^{-\theta t} + \mu(1-e^{-\theta t}) + e^{-\theta t}\int_0^t \sigma e^{\theta s}\, dW_s. $$

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  • $\begingroup$ are you trying to relate $a_t,b_t$ and $c_t$ to something? $\endgroup$ – Chinny84 Jun 9 '15 at 14:00
  • $\begingroup$ @Chinny84 Yes I would like to eventually have a separate equation for each. $\endgroup$ – student441 Jun 9 '15 at 14:03
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You have correctly found $x_t$ as

$$x_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})+\sigma e^{-\theta t}\int_0^te^{\theta s}dW_s$$

We can rewrite this as

$$x_t=a_t-b_tc_t$$

where

$$\begin{align} &a_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})\\\\ &b_t=-\sigma e^{-\theta t}\\\\ &c_t=\int_0^te^{\theta s}dW_s \end{align}$$


Note,

$$dx_t=da_t-c_tdb_t-b_tdc_t$$

with

$$\begin{align} &da_t=\theta(\mu- a_t)dt\\\\ &db_t=-\theta b_tdt\\\\ &dc_t=e^{\theta t}dW_t \end{align}$$

Then,

$$\begin{align} dx_t&=da_t-c_tdb_t-b_tdc_t\\\\ &=\theta(\mu-a_t+b_tc_t)dt-(-\sigma e^{-\theta t})(e^{\theta t}dW_t)\\\\ &=\theta(\mu-x_t)dt+\sigma dW_t \end{align}$$

as expected!!

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  • $\begingroup$ Why did the Wiener term disappear? $\endgroup$ – student441 Jun 9 '15 at 14:53
  • $\begingroup$ @student441 Good catch. A typo that has been edited. $\endgroup$ – Mark Viola Jun 9 '15 at 14:57
  • $\begingroup$ @student441 it shouldn't of disappeared. Because as it stands the function is purely deterministic .. Also $\theta$ is not a function of $t$ (or at least as stated in the problem) so $s$ should not be subscript. So OP don't follow this answer until he makes a few changes. $\endgroup$ – Chinny84 Jun 9 '15 at 14:57
  • $\begingroup$ @Chinny84 The typo has been edited. $\endgroup$ – Mark Viola Jun 9 '15 at 14:58
  • $\begingroup$ @Dr.MV What bothers me is that when $\sigma= 0 $, we get $b_t=0$ and $x_t = a_t$, while if we go back to $dx_t = \theta (\mu-x_t) dt+ \sigma dW_t$ for $\sigma=0$ we get: $d x_t= \theta (\mu-x_t) dt$ where $x_t=a_t+b_t$.. Can you explain? $\endgroup$ – student441 Jun 9 '15 at 16:52

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