6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

5
$\begingroup$

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:

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

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 $t\in [0,T_1)$.

$\endgroup$
3
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks! Indeed this is the partiicular case I'm looking for. Cheers. $\endgroup$
    – CA-Quant
    Jul 20, 2020 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.