Consider a basket with $K=10$ names. Default times of the names, $\tau_k$, are i.i.d. random variables with distribution $P(\tau_k \leq t) = 1 - e^{-\lambda t}$. Suppose that each name in the basket has a notional $N_k = N$, the basket notional, $N=KN$, the discount factors, $D(t) = exp(-rt)$. Denote by $s$ the payment rate (also called the spread). Denote by $T$ the maturity of the $m$-to-default BDS written on this basket. Assume that the number of payment dates in $M$.

Find the value of the default and premium legs for $m=1$ and $m=2$.

My attempt: For $m=1$: Let the random variable $\tau^* = \min_{i\in \{1,...,K \}}\tau_i$ denote the first order statistic. The distribution of $\tau^*$ is easily calculated: $$ f_{\tau_1^*}(t) = {K \choose 1} f_{\tau}(t) F_{\tau}(t)(1-F_{\tau}(t))^{k-1}$$

Let $T_1,...T_M$ denote the term structure of payment dates.

The event {$\tau^*<T_i$} implies that the first default occurred before $T_i$.

For the default leg, I think: $$V_0^{Def} = E_0^{Q}[\sum_{i=1}^Me^{-rT_i} s *N*\mathbb{1}_{{0<\tau ^* <T_i}}]$$

But I unsure about this logic. I have little background with these financial terms.


Let $\tau_{(1)} = \min(\tau_1, \ldots, \tau_K)$ be the first-to-default time. Moreover, for $1< m \le K$, let \begin{align*} \tau_{(m)} = \min\left(\tau_k: k=1, \ldots, K, \tau_{k} > \tau_{(m-1)}\right). \end{align*} be the $m^{\rm th}$-to-default time. In particular, $\tau_{(K)} = \max(\tau_1, \ldots, \tau_K)$. Note that, for $t \ge 0$, \begin{align*} P\left(\tau_{(1)} > t\right) &= \prod_{i=1}^K P(\tau_i > t)\\ &=e^{-K \lambda t}. \end{align*} Then, the density function is of the form \begin{align*} \frac{d P\left(\tau_{(1)} \le t\right)}{dt} &= K \lambda\, e^{-K \lambda t}. \end{align*} Generally, for $1 \le m \le K$, the event $\left(\tau_{(m)} > t\right)$ happens as long as there are $K-m+1$ defaults happen later than time $t$, while the remaining $m-1$ defaults happens earlier than $t$. That is, \begin{align*} P\left(\tau_{(m)} > t\right) &= \sum_{j=K-m+1}^K \sum_{\pi \in \Pi_j}\prod_{i_k \in \pi}P(\tau_{i_k} > t) \prod_{i_l \not\in \pi}P(\tau_{i_l} \le t)\\ &=\sum_{j=K-m+1}^K {K \choose j} e^{-j\lambda t} \Big(1-e^{-(K-j)\lambda t} \Big), \end{align*} where $\Pi_j$ denotes the family of subsets of $(1, \ldots, K)$ consisting of $j$ elements. Here, $ \prod_{i_l \not\in \pi}P(\tau_{i_l} \le t)=1$, if $j=K$. The density function is then given by \begin{align*} \sum_{j=K-m+1}^K {K \choose j} \lambda\, e^{-j\lambda t}\Big(j-Ke^{-(K-j)\lambda t} \Big). \end{align*} We consider the first-to-default case below, that is, $m=1$. The general $m^{\rm th}$-to-default case is similar based on the density function above.

Let $R$ be the recovery rate (e.g., R = 40 %). Note that the default leg is also called the protection leg.

Default Leg. The value of the default leg, if we assume that the default payment is made at the default time, is given by \begin{align*} (1-R)N \, E\Big( D\big(\tau_{(1)}\big) 1_{0 < \tau_{(1)} \le T} \Big) &= (1-R)N K \lambda \int_0^T e^{-(r+K \lambda)t}dt\\ &=\frac{(1-R)N K \lambda}{r+K \lambda}\left(1- e^{-(r+K \lambda)T}\right). \end{align*} However, if we assume that the default payment is made at the next premium payment date, then the value of the default leg is given by \begin{align*} (1-R)N \, E\left(\sum_{j=1}^M D(T_j) 1_{T_{j-1} < \tau_{(1)} \le T_j} \right) &= (1-R)N K \lambda\sum_{j=1}^M e^{-rT_j} \int_{T_{j-1}}^{T_j} e^{-K \lambda t}dt\\ &=(1-R)N \sum_{j=1}^M e^{-rT_j} \left(e^{-K \lambda T_{j-1}} - e^{-K \lambda T_j}\right). \end{align*}

Premium Leg. For $j=1, \ldots, M$, let $\Delta T_j = T_j - T_{j-1}$. We assume that the premium $s$, for the payment period $(T_{j-1}, \, T_j]$ $(j=1, \ldots, M)$, is paid at the end date $T_j$. Moreover, the accrued interest to default, $s(\tau_{(1)}-T_{j-1}) 1_{T_{j-1} < \tau_{(1)} \le T_j}$, is also paid at $T_j$. Then value of the premium leg is then given by \begin{align*} &\ N s E\left(\sum_{j=1}^M D(T_j) \Big[\Delta T_j 1_{\tau_{(1)} > T_j} + \big(\tau_{(1)}-T_{j-1}\big) 1_{T_{j-1} < \tau^* \le T_j} \Big] \right) \\ =& \ N s\sum_{j=1}^M e^{-r T_j} \bigg[\Delta T_j e^{-K \lambda T_j} + K \lambda \int_{T_{j-1}}^{T_j}(t-T_{j-1})e^{-K \lambda t} dt \bigg]\\ \approx& \ N s\sum_{j=1}^M e^{-r T_j} \bigg[\Delta T_j e^{-K \lambda T_j} +\frac{\Delta T_j}{2} \left(e^{-K \lambda T_{j-1}} - e^{-K \lambda T_j}\right) \bigg]. \end{align*} For the last step, we basically assume that, if the default happens, it happens in the middle of the payment period.

| improve this answer | |
  • $\begingroup$ Changed $\sum_{(i_1, \ldots, i_{K-m+1}) \in \pi_m}$ to $\sum_{(i_1, \ldots, i_{K-m+1}) \in \pi_{K-m+1}}$ $\endgroup$ – Gordon Jan 13 '16 at 21:13
  • $\begingroup$ Some updates for the density function. $\endgroup$ – Gordon Jan 23 '16 at 18:47

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