$dX_t = \delta dt+ 2\sqrt{X_t} dW_t$, 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!


1 Answer 1


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

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


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