# 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 $$\mathrm dY = \frac{1}{2Y}\mathrm dt + \mathrm dW \\ Y(0) = Y_0.$$

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Welcome to MSE! Looks like the image got blown away. It helps to format your questions using MathJax. Regards –  Amzoti Jan 31 '13 at 22:57
Did you delete the last line of your post that had a number and a link? That was necessary syntax for the image to show up. –  Sam DeHority Jan 31 '13 at 23:21
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@Mike: Please for future questions do not just post a link to a picture containing the question. Please type up the question and format it using MathJax (LaTeX). You can see more about how to do that here: meta.math.stackexchange.com/questions/5020/… I typed up this question for you. –  Thomas Feb 2 '13 at 13:56

(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: $$dv = f'(\sqrt{v})\,d\sqrt{v} + \frac{1}{2}f''(\sqrt{v})\,d\langle\sqrt{v}\rangle.$$
(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}