# Bugard＆Kjaer中的Bond SDE解決方案

I'll assume $$J_t = \sum_{i=1}^{N_t} Z_i$$ be a compound Poisson process, with $$(T_n)_{n\geq 1}$$ being the jump times for Poisson process $$(N_t)_{t\geq 0}$$ and $$(Z_i)_{i\geq 1}$$ sequence of i.i.d. variables independent of $$(N_t)_{t\geq 0}$$.

For SDE

$$dP_t = P_{t^-} dJ_t$$

we notice that at jump times we have

$$dP_{T_i} = P_{T_i} - P_{T_i^-} = Z_{i} P_{T_i^-}$$

so

$$P_{T_i} = (1+Z_i) P_{T_i^-}$$

From here we can conclude that:

$$P_t = P_0 \prod _{i=1}^{N_t} (1+Z_i)$$

$$dP_t = r_t P_t dt + P_{t^-} dJ_t$$

gives

$$P_t = P_0 \mathrm{e}^{\int_0^t r_s ds}\prod _{i=1}^{N_t} (1+Z_i)$$

as between jump times $$P_t$$ evolves as $$r_t P_t dt$$ and gets multiplied by $$1+Z_{i}$$ at $$T_{i}$$, starting with

$$P_t = P_0 \mathrm{e}^{\int_0^t r_s ds}$$

for $$t\in [0,T_1)$$.

As a complement to @ir7’s comprehensive derivation, in the case of Burgard and Kjaer’s the jump process $$J_t$$ models the default of the issuer. You specialize the process by setting $$Z_1=-1$$, while the values of $$\{Z_i:i\geq2\}$$ are irrelevant. You then notice that as soon as the process jumps once, the product of jump sizes becomes null. We therefore have: $$P_t = P_0e^{\int_0^tr_sds}\mathbf{1}_{\{N_t=0\}} = P_0e^{\int_0^tr_sds}\mathbf{1}_{\{t where $$T_1$$ is the default time of the issuer.