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

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.

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

• 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: $$P_t=P_0e^{\int_0^tr_sds}1_{\{N_t=0\}}=P_0e^{\int_0^tr_sds}1_{\{t<T_1\}}.$$ – Daneel Olivaw Jul 16 '20 at 19:52
• @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 :). – ir7 Jul 16 '20 at 22:00
• Thank you very much. Indeed your answer, together with @DaneelOlivaw, helped me solve this. – CA-Quant Jul 20 '20 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: $$P_t = P_0e^{\int_0^tr_sds}\mathbf{1}_{\{N_t=0\}} = P_0e^{\int_0^tr_sds}\mathbf{1}_{\{t where $$T_1$$ is the default time of the issuer.

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