Hi all,this question is a bit silly but I'm somewhat lost.Let z be an OU process driven by jumps: [$]dz = -c z dt - q dJ[$]where c>0, J is Poisson with intensity h and q is a r.v. on R+ (ie gamma, exponential, IG,...).Now let A=exp(a z) and B=exp(b z).I'm interested into compute expectations of the form [$] E[ dA/A dB/B ] [$]but it is not totally clear to me how does it work. I guess it's something like [$] h E[(exp(a z) - 1) (exp(b z) - 1)] [$]Any advice? thanks a lot.

I think you've almost got it. I'd say [$]h \, dt E[(e^{-a q}-1)(e^{-b q}-1)][$], where the expectation is taken over the jump-size distribution (distribution of q, given a jump event) As for justifying it, first, I will use slightly different notation: [$]A \rightarrow f[$], [$]B \rightarrow g[$], and [$]dJ_t \rightarrow dN_t[$]In general, Ito for jumps for [$]f(z_t)[$] is [$]df_t = df^c_t + \Delta f_t[$], where the first part isthe familiar (no-jump, continuous) part of Ito's rule and the second is the jump part. If [$]f_t = f(z_t) = e^{a z_t}[$], then [$]\Delta f_t = (e^{a z_t} -e^{a z_{t-}}) dN_t[$], where [$](t-)[$] is infinitesimally prior to [$]t[$].Similarly [$]g(z_t) = e^{b z_t}[$] has [$]\Delta g_t = (e^{b z_t} -e^{b z_{t-}}) dN_t[$]. Conditional on a jump, your SDE says [$]z_t = z_{t-} - q_t[$].Also, at any time, [$]dN_t \times dN_t = dN_t[$] since [$]dN_t = 0,1[$]Convince yourself the continuous drift term can be ignored here. That leaves[$]E_{t-}[\Delta f_t \Delta g_t] = E_{t-}[(e^{-a q_t } - 1) e^{a z_{t-}} (e^{-b q_t } - 1) e^{b z_{t-}} dN_t] = h \, dt \, E_{t-}[(e^{-a q_t } - 1) e^{a z_{t-}} (e^{-b q_t } - 1) e^{b z_{t-}}] = h \, dt \, (f_{t-}) \, (g_{t-}) E_t[(e^{-a q_t } - 1) (e^{-b q_t } - 1) ][$]

Thank you Alan I forgot dt and I used z but I was meaning q. However, now you've fixed my doubts.