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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

1 Answers 1

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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.

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    Hi George, if you still read this, what about taking the expectation of the increments $dX$ conditional on knowing the value of $X$ and then taking the expectation of that. It should suffice to prove that the middle term is zero.2010-12-09