<s>[INST] Stochastic Calc (a) Consider the process
$$
\mathrm d\sqrt{v} = (\alpha - \beta\sqrt{v})\mathrm dt + \delta \mathrm dW
$$
Here $\alpha, \beta,$ and $\delta$ are constants. Using Ito's Lemma show that 
$$
\mathrm dv = (\delta^2 + 2\alpha\sqrt{v} - 2\beta v)\mathrm dt + 2\delta\sqrt{v}\mathrm dW.
$$
(b) Using Ito's Lemma to find the SDE satisfied by $U$ given that $U =\ln(Y)$ and $Y$ satisfies
$$
\mathrm dY = \frac{1}{2Y}\mathrm dt + \mathrm dW \\
Y(0) = Y_0.
$$
 [/INST] (a)  Notice that $v = f(\sqrt{v})$ for $f(x)=x^2$. We have $f'(x)=2x$ and $f''(x)=2$.
Ito's lemma yields:
$$
dv = f'(\sqrt{v})\,d\sqrt{v} + \frac{1}{2}f''(\sqrt{v})\,d\langle\sqrt{v}\rangle.
$$
Hence
\begin{align}
dv &= 2\sqrt{v}\times\left((\alpha-\beta\sqrt{v})\,dt + \delta \,dW\right) + \frac{1}{2}\times 2\times\delta^2\,dt\\
&=(\delta^2 + 2\alpha\sqrt{v} - 2\beta v)\,dt + 2\delta \sqrt{v}\, dW.
\end{align}

(b)  Take $f(y)=\ln y$. We have $f'(y)=\dfrac{1}{y}$ and $f''(y)=-\dfrac{1}{y^2}$. Hence $U(0)=\ln Y(0)$ and
\begin{align}
dU &= df(Y) = \frac{1}{Y}\left(\frac{1}{2Y}dt+dW\right) - \frac{1}{2}\frac{1}{Y^2}dt\\
& = \frac{1}{Y}dW\\
& = e^{-U}dW
\end{align}
</s>