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Ito's lemma: $$\text{if } dX_t = \sigma_tdW_t+\mu_t dt, \text{ where $W_t$ is a Wiener process, then } df(X_t)=\sigma_tf'(X_t)dW_t + (\mu_tf'(X_t)+ \frac{1}{2}\sigma_t^2f''(X_t))dt$$

now I want to apply this to $$Y_t=e^{\sigma W_t+\mu t}$$

Where $W$ is a Wiener process.

There are two ways I can think of to apply Ito's lemma to $Y_t$ but they give different results:

(1). We can of course set $X_t = \sigma W_t + \mu t$, and set $f(x)=e^x$. Then Ito's Lemma gives $df= f(X_t)(\sigma dW_t+(\mu + \frac{1}{2} \sigma ^2 )dt)$

(2). However, I am wondering why I get a different result if instead, I set $X_t=W_t$, and $f(x)=e^{\sigma x + \mu t}$.

Then Ito's lemma gives: $df=f(X_t)(\sigma dW_t + \frac{1}{2}\sigma^2dt)$.

Which of these applications is correct, and more importantly, why is the other incorrect?

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  • $\begingroup$ Because in the second case it's $f(t,x)$, not $f(x)$. $\endgroup$ – zhoraster Mar 29 '17 at 14:12
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The first one is correct.

In the second one, $f$ depends explicitly on $t$. This is not a problem per se, but it means that the correct form of Ito's lemma should include an additional $\frac{\partial f}{\partial t} dt$ term. This is an advantage of the otherwise misleading notation $\frac{dX_t}{dt}=\mu_t + \sigma_t \frac{dW_t}{dt}$: it reminds us what we are differentiating with respect to, so that we know not to forget about any explicit dependence on that variable.

To see the point, consider a more extreme example: $f(t,x)=t$. If I then define $Y_t=f(t,W_t)$, $dY_t$ can't possibly be $0$. Yet this is what your form of Ito's lemma without the $\frac{\partial f}{\partial t}$ term would imply.

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  • $\begingroup$ you mean with respect to $t$? Are we differentiating with respect to $t$? I find this very confusing. If we are really differentiating with respect to $t$ then why don't we just write it in the form you just wrote? Why is this notation "otherwise misleading" as you say? $\endgroup$ – user56834 Mar 29 '17 at 14:22
  • $\begingroup$ @Programmer2134 Roughly speaking, we are differentiating with respect to $t$, yes. Put another way, you might think of $df$ as $\frac{df}{dt} dt$. The notation I suggested is only misleading because $\frac{dW_t}{dt}$ does not exist in the classical sense. $\endgroup$ – Ian Mar 29 '17 at 14:27

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