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EdisonCruise
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### What’s Ito’s lemma for Poisson process in this function?

If $N_t$ is a Poisson process with intensity $\lambda$,and $dX_t=\delta dN_t$, $q_t=-N_t$, then the Ito's lemma for function $H(X_t)$ should be
$$dH(X_t)=[H(X_t+\delta)-H(X_t)]dN_t$$
For the function  $H(X_t,q_t)$, why it is not something like this?
$$dH(X_t,q_t)=[H(X_t+\delta,q_t)-H(X_t,q_t)]dN_t+[H(X_t,q_t-1)-H(X_t,q_t)]dN_t$$
As I read from some book, it should be
$$dH(X_t,q_t)=[H(X_t+\delta,q_t-1)-H(X_t,q_t)]dN_t$$
Can anyone give me a formal derivation of the last equation? I cannot find in a book. Thank you.

Alan
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### Re: What’s Ito’s lemma for Poisson process in this function?

I don't know about formal. If a process $H_t$ jumps by $\Delta H_t$ when $N_t$ increases by one, then Ito's differential is just $dH_t = \Delta H_t \, dN_t$, which is the last one.

If the equivalence between $dH_t = \Delta H_t \, dN_t$ and your last one is not clear, take $N_{t^-}=6$ jumping to $N_t=7$, say. What's $H$ before and after the jump?

p.s.  Sec 19.4 here has a proof that the integrated form equivalent to what I wrote is correct for general $f(N_t)$ -- and so this applies to your case with $f(N_t) = H(\delta \times N_t, -N_t)$.

EdisonCruise
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### Re: What’s Ito’s lemma for Poisson process in this function?

Thank you Alan. May I ask one more question? In page 645 of your given file, there is an Ito formula for the compensated compund Poisson process:
$$f(Y_t)=f(0)+\int_0^t \!(f(Y_s)-f(Y_s-))(dN_s-\lambda ds)+\lambda \int_0^t \!(f(Y_s)-f(Y_s-))ds$$
where $dY_t=Z_{N_t} dN_t$ is a compound Poisson process with random jump size $Z_{N_t}$.
Is this equivalent to its differential form as below?
$$df(Y_t)=(f(Y_t)-f(Y_t-))(dN_t-\lambda dt)+\lambda E[f(Y_t)-f(Y_t-)]dt$$
where $E[]$ means expectation.

Since I read in page 26 of http://people.ucalgary.ca/~aswish/JumpProcesses.pdf , there is somehow an expectation there after conversion to its compensated version.
Last edited by EdisonCruise on November 19th, 2019, 12:32 am, edited 1 time in total.

Alan
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### Re: What’s Ito’s lemma for Poisson process in this function?

The second line looks wrong to me. The first line, which is correct, just adds and subtracts the same expression to an integral w.r.t. $dN_t$. The differential form of the first line would not contain an $E[\cdots]$.

Also, probably a typo on your part, the r.h.s. of the 2nd line should have t's not s's, in addition to dropping the expectation.

EdisonCruise
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Joined: September 15th, 2012, 4:22 am

### Re: What’s Ito’s lemma for Poisson process in this function?

Thank you. I have corrected the typo.
However, in $dY_t=Z_{N_t} dN_t$, $Z_{N_t}$ is a random variable, indicating the random jump size. I think it makes sense to include an expectation w.r.t.  $Z_{N_t}$ in the drift term, as in page 26 of http://people.ucalgary.ca/~aswish/JumpProcesses.pdf . It gives a different interpretation with expectation term (maybe equivalent actually) of ito's lemma for compensated compound Poisson process, which is not quite clear to me.
Actually I am thinking  in N. Privault's formula : .
$$f(Y_t)=f(0)+\int_0^t \!(f(Y_s)-f(Y_s-))(dN_s-\lambda ds)+\lambda \int_0^t \!(f(Y_s)-f(Y_s-))ds$$
Since $\int_0^t \!(f(Y_s)-f(Y_s-))ds$ is a random variable, maybe it should be replaced by $E[\int_0^t \!(f(Y_s)-f(Y_s-))ds]$ so that the drift term becomes deterministic.

Alan
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### Re: What’s Ito’s lemma for Poisson process in this function?

Take the trivial case where $f(Y)=Y$. Making your replacement seems to require that

$Y_t - Y_{t-} = E[Y_t - Y_{t-}]$

But that's generally false.

EdisonCruise
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Joined: September 15th, 2012, 4:22 am

### Re: What’s Ito’s lemma for Poisson process in this function?

Thank you Alan. Maybe I confuse it with something. I have made a new thread for this problem.