I'm following a solution of an SDE from here


Start with the SDE $$ dX_t = \delta dt + 2\sqrt{X_t} dW_t $$ consider a deterministic time change $$ \tau = \frac{\sigma^2}{2\nu(2-\delta)}\left(1-\exp\left(-\frac{2\nu t}{2-\delta}\right)\right) $$ the process $Y_t$ is defined as $$ Y_t = exp(\nu t) X_{\tau}^{1-\delta/2} $$ then, using Ito's lemma, we get $$ dY_t = \nu Y_t dt + \sigma Y_t^{\frac{1-\delta}{2-\delta}} dW_t $$

I'd like to understand how the Ito's lemma is applied here. The problem that I have is that when you write it, the $Y$ is derivated with respect to the $X$, but in the definition of $Y_t$ is used $\tau$, not $t$, in $X$, so there's a mixture between different times. How is this handled?


1 Answer 1


The time change described in the question may be handled as follows. Recall that if $W(t)$ is a standard Brownian motion then $$ W(\tau(b))-W(\tau(a)) $$ has the same distribution as $$ \int_a^b \sqrt{\tau'(t)} dW(t) $$ where $a \le b$. Therefore the time-changed process $\tilde X_t = X_{\tau(t)}$ satisfies: $$ d \tilde X_{t} = \delta \tau'(t) dt + 2 \sqrt{ \tilde X_{t} } \sqrt{\tau'(t)} dW(t) \;. $$ Now apply Ito's chain rule to $Y_t = \exp(\nu t) \tilde X_t^{1 - \delta/2}$ to obtain $$ d Y_t = \nu Y_t dt + \exp(\nu t) (1-\frac{\delta}{2}) \tilde X_{t}^{-\delta/2} d \tilde X_{t} +\underbrace{\exp(\nu t) \left( 2 ( -\frac{\delta}{2} ) (1-\frac{\delta}{2}) \tau'(t) \tilde X_{t}^{-\delta/2} \right) dt}_{\text{Ito correction to standard chain rule}} \;. $$ Eliminate $\tilde X(t)$ to obtain the desired SDE for $Y_t$.

  • $\begingroup$ great, I never saw a change in the time coordinates until now actually. Do you have a reference where this is considered explicitly? $\endgroup$
    – chuse
    Apr 15, 2015 at 13:07
  • $\begingroup$ Check Oksendal's book, SDE: An introduction with applications. $\endgroup$
    – nabla
    Apr 17, 2016 at 11:46

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