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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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    @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 $\partial_{\varepsilon} \phi(t, \varepsilon) = i \mathbb{E}[ X_t e^{i \varepsilon X_t } ]$?2010-12-07
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    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?2010-12-07
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    @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.2010-12-07
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    @Raskolnikov: yes, it cancels. The argument isn't very rigorous (but it can be made rigorous).2010-12-07
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    @user4143: Indeed, I've also seen it that way, the resulting partial differential equation is parabolic. In the case of the OP, it becomes a first order equation in the derivatives. @George: I don't know how rigorous the OP wants the answer, but if he assumes Itô calculus rules, it's fine. I wonder if he'll see the answer though. He posted it 4 Nov. I just unearthed it cause there was no answer and I wanted to try my skills at Itô calculus again.2010-12-07
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    @Raskolnikov: I meant that it isn't very rigouros because really you need to show that everything is integrable before taking expectations. There are cases where this approach can give the wrong answer if you aren't careful about integrability. Ito calculus itself is perfectly rigorous.2010-12-07
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    I see, but the functions in this case seem integrable to me, I didn't do anything fancy (although maybe the complex exponential?). Could you point me to an example of a case where it fails with a function of $X_t$. Is Riemann integrability of the function sufficient to guarantee Itô integrability?2010-12-07
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    Woops, that should be Lebesgue integrability since we're talking about the expectation value here.2010-12-07
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    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.2010-12-07
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    Also, the square root in this expression is quite common in finance in eg, interest rate models And stochastic volatility models. Mainly because it leads to reasonable models which are analytically tractable.2010-12-07
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    I think, based on other posts by the same person, that user957 probably is studying financial mathematics, so that would explain it.2010-12-07
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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