I'm going over the paper -Partial Differential Equation Representation of Derivatives with Bilateral Counterparty Risk and Funding Costs- from Burgard and Kjaer. There the following SDE is given for a defaultable bond: $$ dP(t) = r(t)P(t)dt - P(t)dJ(t), $$ where $r(t)$ is an adapted process, and $J(t)$ is a jump process that changes from zero to one on default of the bond issuer.

I'm trying to solve this SDE by finding a closed form formula for $P(t)$, where I'm following the theory given in Steven Shreve's book: -Stochastic Calculus for Finance, Continuous-Time Models- (Chapter 11). I'm attempting to use Ito's formula for jumps, but I'm stuck. Any hints on how to proceed to formally get $P(t)$ from the SDE? Thanks in advance.

  • $\begingroup$ As an aside, while the authors claim these are the dynamics for a defaultable bond, if we assume a stochastic rate $r(t)$, they are probably inconsistent as there should be contribution from the rate's volatility, vanishing as we approach the bond's maturity. This is probably best viewed as continuously rolled-over commercial paper from the issuer. $\endgroup$ Feb 18, 2022 at 13:44

2 Answers 2


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 $(N_t)_{t\geq 0}$.


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

we notice that at jump times we have

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


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

From here we can conclude that:

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

Adding drift

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


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

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 $t\in [0,T_1)$.

  • $\begingroup$ Note that in Burgard & Kjaer’s paper, $J_t$ is used to model counterparty default, hence we set $Z_1=-1$. Then, as soon as $J_t$ jumps for the first time, the product becomes null, so that we can write: $$P_t=P_0e^{\int_0^tr_sds}1_{\{N_t=0\}}=P_0e^{\int_0^tr_sds}1_{\{t<T_1\}}.$$ $\endgroup$ Jul 16, 2020 at 19:52
  • $\begingroup$ @DaneelOlivaw Thank you. Could you please add a separate, complementary answer addressing default risk? It's well worth it to clearly make that distinction. I recited the general case, but you are are really answering OP's question :). $\endgroup$
    – ir7
    Jul 16, 2020 at 22:00
  • $\begingroup$ Thank you very much. Indeed your answer, together with @DaneelOlivaw, helped me solve this. $\endgroup$
    – CA-Quant
    Jul 20, 2020 at 7:08

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: $$ P_t = P_0e^{\int_0^tr_sds}\mathbf{1}_{\{N_t=0\}} = P_0e^{\int_0^tr_sds}\mathbf{1}_{\{t<T_1\}} $$ where $T_1$ is the default time of the issuer.

  • $\begingroup$ Thanks! Indeed this is the partiicular case I'm looking for. Cheers. $\endgroup$
    – CA-Quant
    Jul 20, 2020 at 7:09

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