forward contract on a defaultable zero-coupon bond

Question

I'am trying to calculate the price of a forward on a defaultable zero-coupon bond. It is also true that the price will be given by Price a forward contract on a zero-coupon bond ? I guess the defautability of the bond shoud result in higher price for the forward than in the standard case.

Thanks.

Answer 1

Based on notations in this question, assuming the market value recovery mechanism, the pre-default value at time $T_1$ of a zero-coupon bond with maturity $T_2$, where $T_1 < T_2$, is given by \begin{align*} P(T_1, T_2) = E\Big(e^{-\int_{T_1}^{T_2}(r_s +(1-R)h_s)ds}\,\big|\, \mathscr{F}_{T_1}\Big). \end{align*} Let $B_t=e^{\int_0^t r_s ds}$ be the credit risk free money-market account value at time $t$. The pre-default forward price $K$ determined at time $t$, for $0\le t \le T_1$, is a value such that \begin{align*} 0 &= E\Big(\pmb{1}_{\tau>T_1}\frac{B_t}{B_{T_1}}(K-P(T_1, T_2)) \,|\,\mathscr{G}_t\Big)\\ &=\pmb{1}_{\tau>t}E\left(\Big(K e^{-\int_t^{T_1}(r_s+h_s) ds} - e^{-\int_t^{T_1}(r_s+h_s) ds-\int_{T_1}^{T_2}(r_s +(1-R)h_s)ds} \Big) \,|\,\mathscr{F}_t\right)\\ &=\pmb{1}_{\tau>t}E\left(\Big(K e^{-\int_t^{T_1}(r_s+h_s) ds} - e^{-\int_t^{T_2}(r_s+h_s) ds+\int_{T_1}^{T_2}Rh_sds} \Big) \,|\,\mathscr{F}_t\right). \end{align*} That is, \begin{align*} K = \frac{E\Big(e^{-\int_t^{T_2}(r_s+h_s) ds+\int_{T_1}^{T_2}Rh_sds} \,|\,\mathscr{F}_t\Big)}{E\Big(e^{-\int_t^{T_1}(r_s+h_s) ds} \,|\,\mathscr{F}_t\Big)}. \end{align*}

Your observation appears correct if you assume that the interest rate is defined by $r_t+h_t$ in the standard case.