I am confused on how to apply Ito's formula on certain problems, especially where expectations are involved. For example, if $W_t$ is a Wiener process and $X_t$ satisfies a below SDE:
$ dX_t = (X_t-\mu)dt + \sigma\sqrt{X_t}dW_t,~~~~~~ X_0 = x_o$
How do I find $\partial_t \phi$ or $\partial_\xi \phi$ where $\phi(t,\xi)=E[e^{i\xi X_t}]$ is characteristic function of $X_t$?
I don't quite understand how to approach this problem. Should I first solve the SDE for $X_t$, then compute the expectation $E[e^{i\xi X_t}]$, and then apply Ito's Lemma to find $\partial_t\phi$?
Taking it a step further, how would I compute $\partial_t\psi$ where $\psi(t,\xi)=\ln\phi(t,\xi)$ and solve resulting SDE for $\psi(t,\xi)$?
Reference: Ito's lemma
Best Answer
Try this approach:
For the partial derivative with respect to $\xi$,
$$\partial_{\xi}\phi(t,\xi)=i\mathbb{E}\left[X_t e^{i\xi X_t}\right] \; .$$
For the partial derivative with respect to $t$, I'll take the differential but only varying t, so as to connect with the SDE:
$$d_t\phi(t,\xi)=\mathbb{E}\left[e^{i\xi (X_t+dX_t)} - e^{i\xi X_t}\right]=\mathbb{E}\left[(i\xi dX_t-\frac{\xi^2}{2}dX_t^2)e^{i\xi X_t}\right] \; .$$
Now
$$ dX_t = (X_t - \mu) dt + \sigma \sqrt{X_t}dW_t $$
and
$$ dX_t^2 = \sigma^2 X_t dt $$
This implies
$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[dX_t e^{i\xi X_t}\right]-\frac{\xi^2}{2}\mathbb{E}\left[dX_t^2 e^{i\xi X_t}\right] $$
and thus
$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[(X_t-\mu) e^{i\xi X_t}\right]dt+i\xi \mathbb{E}\left[\sqrt{X_t} dW_t e^{i\xi X_t}\right]-\frac{\sigma^2 \xi^2}{2}\mathbb{E}\left[X_t e^{i\xi X_t}\right]dt $$
Now, the middle term contains $dW_t$ as a consequence, taking its expectation gives zero. You can now recognize the other terms as containing $\phi$ or $\partial_{\xi}\phi$, so you've got yourself an ordinary PDE for $\phi$. That should be the aim of the computation.