I'm reading the Duffie, Pan, and Singleton (2000) paper now and I've stumbled upon something that seems to me as an inconsistency. Whenever I look up the SVJJ model, I see that its log-transform is formulated in the following way $$ \begin{align} dY_{t} &= \left(r - \frac{1}{2}V_{t} - \lambda\mathbb{E}[\mathrm{e}^{J^{Y}} - 1]\right)dt + \sqrt{V_{t}}dB^{Y}_{t} + J^{Y}dN^{Y}_{t}\\ dV_{t} &= \kappa(\theta - V_{t})dt + \sigma\rho\sqrt{V_{t}}B^{Y}_{t} + \sigma\sqrt{(1-\rho^{2})V_{t}}B^{V}_{t} + J^{V}dN^{V}_{t}. \end{align} $$ However, Duffie, Pan, and Singleton (2000) seems to just drop the $J$, i.e. they formulate the SVJJ as $$ \begin{align} dY_{t} &= \left(r - \frac{1}{2}V_{t} - \lambda\mathbb{E}[\mathrm{e}^{J^{Y}} - 1]\right)dt + \sqrt{V_{t}}dB^{Y}_{t} + dN^{Y}_{t}\\ dV_{t} &= \kappa(\theta - V_{t})dt + \sigma\rho\sqrt{V_{t}}B^{Y}_{t} + \sigma\sqrt{(1-\rho^{2})V_{t}}B^{V}_{t} + dN^{V}_{t}. \end{align} $$ (I refer to equation 4.1 in their paper). Can someone tell me why this is done or how I should understand this. Papers that perform some type of derivation from the price process to the log-transform price process would also be appreciated.
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1$\begingroup$ Equation (4.1) in this paper ? If yes they don't write $dN$ which presumably is a Poisson process but $dZ$ which is more general than that. See their definitions. $\endgroup$– Kurt G.Feb 7 at 18:50
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$\begingroup$ Yes, that is the paper I'm referring to. What is the difference between using $dN$ and $dZ$. I mean if it is clear that it is a compensated Poisson process. $\endgroup$– CasMathFeb 7 at 22:35
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$\begingroup$ To add to my previous comment. Would the above two definitions be equivalent if the DPS(2000) equation I wrote had $dZ$ instead of $dN$? i.e. the first two equations have $dN$ but the second two have $dZ$. $\endgroup$– CasMathFeb 8 at 0:06
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1$\begingroup$ Strictly speaking I don't know what $N$ is because you just posted that equation without a reference. Duffie et al write on p. 1349 very clearly that $Z$ is a pure jump process whose jumps have a fixed probability distribution $\nu$ on $\mathbb R^n\,.$ As I said before this is more general than a Poisson process. $\endgroup$– Kurt G.Feb 8 at 8:24
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