Bugard&Kjaer中的Bond SDE解決方案


6

我正在研究Burgard和Kjaer的論文-帶有雙邊交易對手風險和資金成本的衍生產品的偏微分方程表示-。此處為默認債券提供了以下SDE: $$ dP(t)= r(t)P(t)dt-P(t)dJ(t),$$ 其中 $ r(t)$ 是經過調整的過程,而 $ J(t)$ 是跳轉該過程在債券發行人違約時從零變為一。

我正在嘗試通過為 $ P(t)$ 找到一個封閉式公式來解決此SDE,在這裡我遵循了史蒂文·史瑞夫(Steven Shreve)的理論書籍:-金融隨機演算,連續時間模型-(第11章)。我正在嘗試使用伊藤公式進行跳躍,但我遇到了麻煩。關於如何繼續從SDE正式獲取 $ P(t)$ 的任何提示?提前致謝。

5

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 $t\in [0,T_1)$.


2

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.