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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

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1 Answer

up vote 2 down vote accepted

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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@Raskolnikov, isn't $\partial_{\varepsilon} \phi(t,\varepsilon) = \partial_{\varepsilon} \int_{-\infty}^{\infty} p_{X_t}(u) e^{i \varepsilon X_t } = i \int_{-\infty}^{\infty} X_t p_{X_t} (u) e^{i \varepsilon X_t }$? So shouldn't $\partial_{\varepsilon} \phi(t, \varepsilon) = i \mathbb{E}[ X_t e^{i \varepsilon X_t } ]$? –  user4143 Dec 7 '10 at 19:15
    
You're right, I simply forgot the $i$, thank you! Besides, I've checked the rules of Itô calculus and I think the middle term in my final line does indeed cancel. Are you user957, the OP writer? –  Raskolnikov Dec 7 '10 at 19:17
    
@Raskolnikov, I am not the OP on this question. FWIW, I've usually seen $dX_t$ defined as $dX_t = (X_t - \mu)dt + \sigma X_t dW_t $ (without the $\sqrt{}$), though maybe that's how the OP wanted to define it. –  user4143 Dec 7 '10 at 20:07
    
@Raskolnikov: yes, it cancels. The argument isn't very rigorous (but it can be made rigorous). –  George Lowther Dec 7 '10 at 20:42
1  
I just meant integrability of X. $dX=\lambda(\mu-X)dt + \sigma X^2 dW$ is the kind of thing that would cause problems, as the $X^2dW$ term is not a martingale, and doesn't have zero mean. That is only an issue where ther coefficients grow faster than linearly, so not a problem with the question here. –  George Lowther Dec 7 '10 at 21:38
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