Solution for a SDE for a Bond found in Bugard & Kjaer

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 ${Z_i:igeq2}$ are irrelevant. You then notice that as soon as the process jumps once, the product of jump sizes becomes null. We therefore have: $$ P_t = P_0e^{int_0^tr_sds}mathbf{1}_{{N_t=0}} = P_0e^{int_0^tr_sds}mathbf{1}_{{t<T_1}} $$ where $T_1$ is the default time of the issuer.

I'll assume $$ J_t = sum_{i=1}^{N_t} Z_i$$ be a compound Poisson process, with $(T_n)_{ngeq 1}$ being the jump times for Poisson process $(N_t)_{tgeq 0}$ and $(Z_i)_{igeq 1}$ sequence of i.i.d. variables independent of $(N_t)_{tgeq 0}$.

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:

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

Adding drift

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

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 $tin [0,T_1)$.