Stochastic Calc

Question

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

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}

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(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}