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

November 16th, 2019, 10:25 am

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

November 16th, 2019, 2:44 pm

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 [$]f(N_t) = H(\delta \times N_t, -N_t) [$].
 
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EdisonCruise
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Re: What’s Ito’s lemma for Poisson process in this function?

November 18th, 2019, 8:53 am

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

November 18th, 2019, 2:53 pm

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

November 19th, 2019, 1:17 am

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

November 20th, 2019, 12:10 am

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

November 20th, 2019, 7:41 am

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