# Computing infinitesimal generator for a jump process with continuous component

I am trying to show that the infinitesimal generator of the following process

$$dS_t = (\alpha S_{t-}+\beta )dt + (\gamma S_{t-}+\delta)dX_t,$$

where $$X_t$$ is a $$(\lambda,G)$$-compound Poisson process (that is $$X_t=\sum_{k=1}^{N_t}Z_k$$, $$N_t$$ is a Poisson process with intensity $$\lambda$$, and $$Z_k\sim G, supp(G)=\mathbb{R}$$) is given by

$$\mathcal{L}(f)(x)=(\alpha x+\beta)f'(x)+\lambda \int_{-\infty}^{\infty}[f(x+\gamma xz+\delta z)-f(x)]dG(z)$$

This is Proposition 8.6.7.1 from Jeanblanc, Chesney and Yor (2009) "Mathematical Methods in Financial Markets."

I was hoping if someone could clarify to me one of the steps in the proof of this result. The details are below.

The approach in the book is to apply path-by-path Stieltjes integration to write,

$$f(S_t)-f(x) = \int_0^tf'(S_{s-})(\alpha S_{s-}+\beta)ds+\sum_{0\geq s\geq t} \Delta(f(S_t))$$ and then take the expectations of this expression to obtain

$$\mathbb{E}^x[f(S_t)-f(x)]= \mathbb{E}^x\left[\int_0^tf'(S_{s-})(\alpha S_{s-}+\beta)ds+\sum_{0\geq s\geq t} \Delta(f(S_t))\right]\\ =\mathbb{E}^x\left[\int_0^tf'(S_s)(\alpha S_{s-}+\beta)ds\right]+\mathbb{E}^x\left[\sum_{0\geq s\geq t} f(S_{s-}+ \Delta S_s)-f(S_{s-})\right]$$

Up to this point, everything is clear. It is easy to show that the first term will converge to $$(\alpha x+\beta)f'(x)$$ when we divide by $$t$$ and take the limit as $$t\downarrow 0$$.

The step I do not understand is the following. The authors write:

$$\mathbb{E}^x\left[\sum_{0\geq s\geq t} f(S_{s-}+\Delta S_s)- f(S_{s-})\right] \\ =\mathbb{E}^x\left[ \int_0^t\int_{\mathbb{R}} f(S_{s-}+(\gamma S_{s-}+\delta)z) -f(S_{s-})dG(z)\lambda dt\right] (*)$$

Why this equality follows? How to justify it? Or is there a way to proceed alternatively?

I tried writing this sum as $$\mathbb{E}^x\left[\sum_{0\geq s\geq t} f(S_{s-}+\Delta S_s)-f(S_{s-})\right]=\mathbb{E}^x\left[ \int_0^t f(S_{s-}+(\gamma S_{s-}+\delta)Z) -f(S_{s-})dN_t\right]$$ I can then rewrite this expectations as $$\mathbb{E}^x\left[ \int_0^t f(S_{s-}+(\gamma S_{s-}+\delta)Z) -f(S_{s-})dN_t\right] = \mathbb{E}^x\left[ \int_0^t f(S_{s-}+(\gamma S_{s-}+\delta)Y) -f(S_{s-})d(N_t-\lambda t)\right] \\ + \mathbb{E}^x\left[ \int_0^t f(S_{s-}+(\gamma S_{s-}+\delta)Z) -f(S_{s-})d(\lambda t)\right] = \mathbb{E}^x\left[ \int_0^t f(S_{s-}+(\gamma S_{s-}+\delta)Z) -f(S_{s-})d(N_t-\lambda t)\right] + \\\lambda \int_0^t\int_{\mathbb{R}} f(S_{s-}+(\gamma S_{s-}+\delta)z) -f(S_{s-}) dG(z)ds$$ So if I could argue that the expectations of the first term is 0 I would be done. But the integrand is right continuous, and so I do not think it is a martingale.

I went very carefully through the book today, and it is hinted that they use a random measure, of the form $$\mu=\sum_n \delta_{T_n,Y_n}$$ and claim that $$\int_0^t\int_{\mathbb{R}} f(S_{s-}+(\gamma S_{s-}+\delta)z) -f(S_{s-})d(\mu(dz,ds)-\lambda G(ds))$$ is a local martingale. But I am not sure how to use that fact.

• "[...] does not seem to be a martingale" Why not? The process $N_t-\lambda t$ is a martingale and therefore the stochastic integral $\int_0^t H(s) \, d(N_s-\lambda s)$ is a martingale for any nicely integrable and measurable $H$. – saz Apr 25 at 9:21
• @saz Thank you so much for your response. Shouldn't the integrand be left continuous for the integral wrt $(N_t-\lambda t)$ to be a martingale? Also, I made a mistake, I should have arrived at the following integral $\int_{0}^t f(S_{s-}+(\gamma S_{s-}+\delta)Z)-f(S_{s-})d(N_t-\lambda t)$ rather than $\int_{0}^t\int_{\mathbb{R}} f(S_{s-}+(\gamma S_{s-}+\delta)z)-f(S_{s-})dG(z)d(N_t-\lambda t)$. I updated the question accordingly with some additional context. Do you have any idea how to deal with the updated integral? I am most likely missing something trivial... – Mdoc Apr 26 at 5:29
• Why do you think that the integrand is right-continuous? Since $t \mapsto S_t$ is càdlàg (=right continuous and left-hand limits) it follows that $t \mapsto S_{t-}$ is càglàd (=left-continuous and right-hand limits). – saz Apr 26 at 5:41
• @saz But the integrand is of the form $f(S_t)−f(S_{t−})$ since $(\gamma S_{s−}+\delta )Y$ is the jump of the process $S_t$. Isn't then the difference right continuous? – Mdoc Apr 26 at 6:01
• I just had once more a look at your computations, and I think that the very first equation (when you rewrite the sums) is actually wrong... the problem is that you are using the same $Z$ over the whole time which is not what is happening in reality (because the $k$-th jump height $Z_k$ will, in general, depend on $k$)). I suppose the only way is indeed to use the random measure which you are mentioning at the very end of your answer. – saz Apr 26 at 6:30