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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, Ito´s fórmula is: $$df(t,X_t)=(\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x}a_t)dt + (f(t, X_t+J_t)-f(t,X_t))dN_t$$

I don´t know how to demonstrate it.

Note: I understand that when there is a jump $N_t$ jumps 1 unit, so $X_t$ jumps $J_t$ and hence $f$ jumps $f(t, X_t+J_t)-f(t,X_t)$ when there is a ump at time t.

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    $\begingroup$ Would an answer regarding the second term on the RHS be sufficient? $\endgroup$ – muaddib May 26 '15 at 19:49
  • $\begingroup$ Hi muaddib. I don´t neither undertand the other part. However, any help will be very usefull for me. Thank you very much. $\endgroup$ – Edin_91 May 26 '15 at 19:53
  • $\begingroup$ Do you know how to demonstrate it ? $\endgroup$ – Edin_91 May 26 '15 at 20:52
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    $\begingroup$ I am but not making much progress. $\endgroup$ – muaddib May 26 '15 at 21:23
  • $\begingroup$ Thank you. What did you know about the second therm of RHS? $\endgroup$ – Edin_91 May 26 '15 at 21:31
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As you mentioned in your note, the outcome of the jump process part of the sde is a step functions (of times) that jumps up a distance $J_{t_i}$ at a sequence of random times $t_i$ which are an outcome of the Poisson process $N_t$.

Let's just consider the SDE: $dX_t = J_tdN_t$. Then given one of the outcomes I mentioned above, $f(X_t)$ is simply $f$ evaluated at where $X_t$ is at that moment of time. At an exact jump moment $t_i$, this will be $f(X_{t_i} + J_{t_i})$. Since before the jump we were at the value $f(X_{t_i})$ we need to add the difference to our current location to get our next location.
$$f(X_t) = f(X_{t_i}) + (f(X_{t_i} + J_{t_i}) - f(X_{t_i}))$$ for all $t > {t_i}$ but before the next jump $t_{i+1}$. If right this in terms of all the jumps that have occurred we have: $$f(X_t) = f(X_0) + \sum_{k=0}^i (f(X_{t_i} + J_{t_i}) - f(X_{t_i}))$$ We can rewrite each summand as time integral by integral a dirac mass at time $t_i$: $$f(X_t) = f(X_0) + \sum_{k=0}^i \int_0^t (f(X_s + J_s) - f(X_s)) \delta_{t_i}(s) ds$$ and this equals $$f(X_0) + \int_0^t (f(X_s + J_s) - f(X_s)) \sum_{k=0}^i \delta_{t_i}(s) ds$$

Now if we defined $dN_t$ to be a measure with dirac masses at time $t_i$ we get: $$f(X_t) = f(X_0) + \int_0^t (f(X_s + J_s) - f(X_s)) dN_s $$

In other words, this is just a very elaborate way of setting up a telescoping sum to form step functions.

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  • $\begingroup$ Oh! Thak you very much muaddib. So If we have that $f(X_t)= \int_{0}^t (f(X_s+J_s))-f(X_s)dNs$ we would have that $df(X_t)=(f(X_t+J_t)-f(X_t)dNt$ Correct? $\endgroup$ – Edin_91 May 26 '15 at 22:32
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    $\begingroup$ Yes, the latter is really shorthand (though it can be made well defined) for the former. $\endgroup$ – muaddib May 26 '15 at 22:33
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    $\begingroup$ Like you said, any help. That part though you can just look up in the standard derivation of ito's lemma. $\endgroup$ – muaddib May 26 '15 at 22:34
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    $\begingroup$ I mean the integrals have good definitions. It's hard but not impossible to come up with "derivatives" of processes whose outcomes are non-differentiable. $\endgroup$ – muaddib May 26 '15 at 22:38
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    $\begingroup$ Okey! Thank you very much :) :) $\endgroup$ – Edin_91 May 26 '15 at 22:41

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