I need help with the following problem. How can I derive $d\sqrt v$ using Ito's lemma for the following process:
$$d\sqrt v=(\alpha−\beta \sqrt v)dt+\delta dX$$
The parameters $\alpha, \beta, \delta$ are constant. Using Itô's lemma show that
$$dv = (\delta^2 + 2\alpha\sqrt v − 2\beta v)dt + 2\delta \sqrt vdX$$
any ideas?
It is really easy so I show you the solution, the important thing being to make some progress to try to understand it and to apply it in some other context.
Let's note $Y_t= \sqrt{V}_t$ then $Y^2_t=V_t$ (supposing $V$ is a positive process by the way).
So let's apply Itô's lemma with $f(Y)= Y^2$ we get : \begin{align} dV_t=df(Y_t)&= \frac{1}{2}.\frac{\partial^2 f(Y)}{ \partial Y^2} d⟨Y⟩_t+ \frac{\partial f(Y)}{ \partial Y}dY_t\\ = d\langle Y\rangle_t+ 2 Y_tdY_t &= (\delta^2 + 2.Y_t.(\alpha -\beta. Y_t))dt +2\delta.Y_t.dX_t\\ &=(\delta^2 + 2.\alpha .\sqrt{V}_t-2.\beta.V_t).dt +2\delta.\sqrt{V}_t.dX_t \end{align}
NB: $\frac{1}{2}.\frac{\partial^2 f(Y)}{ \partial Y^2} d⟨Y⟩_t=\frac{1}{2}.2.d⟨Y⟩_t$
As $d⟨Y⟩_t=⟨\delta.dX⟩_t=\delta^2.d⟨X⟩_t=\delta^2.dt$ giving the result.
Thank's for pointing out my mistake and also changed my notation on quadratic variation which was confusing. Best regards