Squared Bessel Process and Ito Lemma $dX_t = \delta dt+ 2\sqrt{X_t} dW_t$, where $W_t$ is a standard Wiener process, 
Define $\tau =\frac{\sigma ^2}{2\nu(2 − \delta)}\left(1 − \exp \left(−\frac{2\nu t}{2−\delta}\right)\right)$
If $Y_t=\exp(\nu t) (X_\tau)^{1-(δ/2)}$, By Ito's Lemma, 
$dY_t = \nu Y_t dt + \sigma Y_t^{(1−δ)/(2−δ)} dW_t $
But I cannot understand how Ito's Lemma can be applied here. 
My concern is that I have only learnt Ito's Lemma of $f(t,X_t)$, but now, $Y_t$ is a function of $t$ and $X_\tau$, but not $X_t$. 
Thank you! 
 A: Since
$$
e^{-\nu t}Y_t=X_{\tau}^{1-\frac{\delta}{2}}\,,\tag{1}
$$
It$ô$'s lemma implies
$$
-\nu e^{-\nu t}Y_t\text dt + e^{-\nu t}\text dY_t = \text d(\color{red}{e^{-\nu t}Y_t}) = \text d(\color{red}{X_{\tau}^{1-\frac{\delta}{2}}}) = (1-\frac{\delta}{2})X_{\tau}^{-\frac{\delta}{2}}\text dX_{\tau}-\frac{1}{2}\frac{\delta(2-\delta)}{4}X_{\tau}^{-\frac{\delta+2}{2}}\text d\langle X\rangle_{\tau}\,.
$$
Using the definition of $\ \text dX_{\tau}\ $simplifies this to
$$
\text dY_t = \nu Y_t\text dt + e^{\nu t}(2-\delta)X_{\tau}^{\frac{1-\delta}{2}}\text dW_{\tau}\tag{2}
$$
Finally, noting that 
$$
\int_{0}^{t}\sqrt{\frac{\text d\tau}{\text du}}\text dW_u \stackrel{\text{law}}{=} W_{\tau}\implies\text dW_{\tau}=\sqrt{\frac{\text d\tau}{\text dt}}\text dW_t = \frac{\sigma}{(2-\delta)}e^{-\frac{\nu t}{2-\delta}}\text dW_t\,,
$$
together with (1), allows us to simplify (2) further. Thereby, we obtain
$$
\text dY_t = \nu Y_t\text dt + \sigma e^{\nu t}e^{-\frac{\nu t}{2-\delta}}e^{-\frac{\nu t(1-\delta)}{2-\delta}}Y_t^{\frac{1-\delta}{2-\delta}}\text dW_{t}
$$ 
This shows that

$$
\text dY_t = \nu Y_t\text dt + \sigma Y_t^{\frac{1-\delta}{2-\delta}}\text dW_{t}
$$ 

