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(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
    
I tried to upload an image of the question which is written in a pdf document but as I am a new member the website wouldn't allow me to do so. Instead I have typed the question above as best as I could. –  Mike Feb 1 '13 at 9:48
    
I have posted the question as an image. Please click on the link to view the question. –  Mike Feb 1 '13 at 10:01
    
@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

1 Answer 1

(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}

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Hi Ju'x thanks for explaining the calculation to reach the solution –  Mike Feb 1 '13 at 14:02

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