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Let, $N_t$ be a Poisson process and let $X_t$ solve the SDE $d{X_t}=a_t dt +J_t dN_t$. Then, what is the correct Ito´s fórmula:

i)$df(t,X_t)=(\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x}a_t)dt + (f(t, X_t+J_t)-f(t,X_t))dN_t$

or

ii)$df(t,X_t)=(\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} a_t)dt + (\frac{\partial f}{\partial x}J_t )dN_t$

I have seen both of the in literature, but I do not know if one of them is incorrect or they are equivelent.

Thank you :)

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    $\begingroup$ There is something wrong with your small $a$ and capital $A$ $\endgroup$ – wiskundeliefhebber May 26 '15 at 16:33
  • $\begingroup$ Thanks! I ve fixed it :) Do you know if they seem to bee the same formula or something is wrong? $\endgroup$ – Edin_91 May 26 '15 at 16:34
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I think that (i) is correct, I have never seen (ii) in the literature.

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  • $\begingroup$ Does someone know the demostration of i? I`m quite sure they must be equivalent or there must be a small error, $\endgroup$ – Edin_91 May 26 '15 at 16:41
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    $\begingroup$ @Edin_91 The following is deliberately phrased sketchily for purposes of intuition. When $N$ changes, it changes by a finite amount instantaneously. It does not change by an infinitesimal amount from one moment to the next. That means that although the $dN_t$ term arises through the dependence of $f$ on $x$, you have to add up all the changes in $f$ as the $x$ argument changes from $X_t$ to $X_t+J_t$, not as $x$ changes from one number to an infinitely close number. $\endgroup$ – Ian May 26 '15 at 16:47
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    $\begingroup$ @Edin_91 For illustration, consider an $f$ which depends only on $x$ and is zero on $[-1/2,1/2]$ and positive elsewhere. Suppose $X_0=0,a_t \equiv 0,J_t \equiv 1$. Then your second form would not change $f$ at the first jump, but $f$ clearly does change at the first jump (it should go to $f(1)$). $\endgroup$ – Ian May 26 '15 at 16:53
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    $\begingroup$ @Edin_91 That's right. $\endgroup$ – Ian May 26 '15 at 16:56
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    $\begingroup$ It is just the definition of the derivative except at the points where $N$ jumps. Where you have a jump, suppose $N$ jumps by some amount $\Delta N$ at time $t$. Then $X$ jumps by $J_t \Delta N$ at time $t$. Hence $f$ jumps by $f(t,X_t+J_t\Delta N)-f(t,X_t)$ at time $t$ (assuming here that $f \in C^1$). There is only a minor technical issue, which is that you need to be sure that the jumps are isolated from one another with probability $1$ in order to do the differentiation at the remaining points. But this is routine. $\endgroup$ – Ian May 26 '15 at 16:58

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