I'm following a solution of an SDE from here


<http://www.math.ethz.ch/~delbaen/ftp/preprints/CEV.pdf>


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?



Change of time variable in Wiener process
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 $Y\_t$.


