CDS Spread and Par Bond Yield Spread

Question

It is claimed that the credit default swap (CDS) spread should approximate the risky par bond yield or coupon rate spread from the riskless bond on the same entity. This comes about when we assume discount factor $B(t)=e^{-rt}$ with constant riskless interest rate $r$ together with infinitesimal coupon period. However, this is not true in general and it is dubious under what condition this holds even approximately.

Let us examine this claim mathematically. In general, for par bond coupon rate $c$ and CDS spread $s$ $$c-s=\frac{\int_0^T P(t)\mathrm d(-B(t))}{\sum_{i=1}^n \delta_iB_iP_i} \ge 0,$$ where $P(t)$ is the survival probability and $B(t)$ the discount factor at time $t$, $t_i$ is the $i$'th coupon date and $P_i=P(t_i)$ and $B_i=B(t_i)$. It is true $B(t)\searrow 0 \Longrightarrow c-s\searrow 0$, for any given $P$. However, once can device decreasing $P$ and $B$ such that $c-s$ is unbounded from above in the set of admissible $P$ and $B$ (decreasing positive function on $[0,\infty)$ taking value $1$ at $t=0$) for any given $\{\delta_i>0\}_{i=1}^n$. Consider very small $P_i$.

The riskless par bond coupon rate of the same coupon schedule is $$c_0=\frac{\int_0^T \mathrm d(-B(t))}{\sum_{i=1}^n \delta_iB_i}.$$ So $$c-s-c_0=\frac{\int_0^T P(t)\mathrm d(-B(t))}{\sum_{i=1}^n \delta_iB_iP_i}-\frac{\int_0^T \mathrm d(-B(t))}{\sum_{i=1}^n \delta_iB_i}$$ We already know from the previous paragraph that the above expression is unbounded from above. To explore the range of the above expression, consider the following case. $$ P(t) = \begin{cases} 1, & t=0 \\ P_1, & t\in (0,t_1] \\ 0, & t\in (0,\infty) \end{cases}, \quad B(t) = \begin{cases} 1, & t=0 \\ B_1, & t\in (0,t_1] \\ 0, & t\in (0,\infty) \end{cases}. $$ then $$c-s-c_0 = \frac{1}{\delta_1B_1}-1-\frac{1}{\delta_1B_1}=-1.$$ So $c-s-c_0$ ranges at least from $-1$ to positive infinity.

Therefore, all we can say is that in the very low interest rate regime like recently, $c$ is not much higher than $s$. Then what additional conditions are imposed or specific models are assumed to justify the folklore statement that $c-s\approx c_0$? Can someone provide a mathematical derivation or supply some references to that effect? Many authors have cited Darrell Duffie's paper as the source of the claim. However, I do not see a mathematical derivation or justification there --- the aforementioned link is a draft version not the published one and perhaps therein lies the rub.

One of the papers making such claim is this. There Assumption 4. is the most pertinent, acknowledging the assumption of constant interest rate, presumably implying $B(t)=e^{-rt}$ with $r$ a positive constant, just as I have stated in the first paragraph. How good an approximation is this? This and this are two other papers amongst many making the claim. None of these authours are "dubious" folks.

Can anyone elucidate?

Answer 1

In many years working in the credit markets, I never encountered anyone making an approximation of CDS spread being equal to risky par bond yield.

If we approximate CDS coupon payments as a continuous stream $s$, default intensity as a constant $h$, and we assume that discount factors come from a constant risk-free rate $r$, then the CDS pricing formula becomes $$ \begin{align} S&=\int_0^T (1-\delta)e^{-(r+h)t} h dt - \int_0^T s e^{-(r+h)t} dt \\ &= \frac{ \left( 1-e^{-(r+h)T} \right)}{r+h} \left(h(1-\delta)-s \right) \end{align} $$ and the value of a risky bond paying coupons at continuous rate $c$ is $$ \begin{align} V&=e^{-(r+h)T} + \int_0^T c e^{-(r+h)t} dt + \int_0^T \delta e^{-(r+h)t} h dt \\ &=e^{-(r+h)T}+\frac{c}{r+h}\left(1-e^{-(r+h)T}\right) +\frac{h\delta}{r+h}\left(1-e^{-(r+h)T}\right). \end{align} $$

If we take the old-school approximation that $\delta=0$, $s=h$, and also

$$c=r+h$$

then these formulae simplify to $$ S = \left(h -s \right) \frac{ \left( 1-e^{-(r+h)T} \right)}{r+h} = 0 $$ and $$ V=1. $$

Thus, the old-school approximations people apply are:

• The fair CDS coupon $s$ is the credit spread $h$ (not the risky par bond yield $r+h$), and
• The bond trades at par when the yield $y$ is the risk-free rate plus the credit spread, and this happens when the coupons are $c=y=r+h$

Note that professionals in CDS markets rarely use these $\delta=0$ approximations, but even in 2014 you do still run into some risky bond market people who use them.

The constant interest rate assumption is commonly used by all sorts of participants, since even when interest rates are not constant it gives you the right answer for DV01 and CD01 sensitivities, to first order. Software libraries (like this) handle the computations when full precision is desired.

Answer 2

I have the answer, thanks to none other than Prof. Darrel Duffie, who points out the claim is for floating rate coupon rather than fixed one.

Here is the formulation. The coupon of a floating rate bond is the LIBOR rate paid plus a spread. Let $B_i^j$ be the discount factor between time $t_i$ and $t_j$. The LIBOR rate between coupon date $i-1$ and $i$ is $l_i = \frac{1}{\delta_i}\big(\frac{1}{B_{i-1}^i}-1\big)$. Assume the short interest rate process $r$ is independent of the reference entity default time $\tau$. Using all the same notations as in the question, the par risky floating rate bond satisfies \begin{align} 1 &= \mathbb E\Big[ \sum_i\delta_i (l_i+c)e^{-\int_0^{t_i}r }\mathbf 1_{\tau>t_i} \Big]+B(T)P(T)+R\int_0^T B(t)\mathrm d(-P(t)) \\ &= \sum_i (B_{i-1}-B_i+c\delta_iB_i)P_i+B(T)P(T)+R\int_0^T B(t)\mathrm d(-P(t)) \\ &= c\sum_i \delta_iB_iP_i+\sum_i (B_{i-1}-B_i)P_i+1-\int_0^T P(t)\mathrm d(-B(t))-L\int_0^T B(t)\mathrm d(-P(t)), \end{align} where the last equality stems from integration by parts, and $R$ is the recovery rate and $L$ the loss rate of the reference entity, and of course $R+L=1$. The CDS rate is still $$s = \frac{L\int_0^T B(t)\mathrm d(-P(t))}{\sum_i \delta_iB_iP_i}.$$ Therefore $$c-s = \frac{\int_0^T P(t)\mathrm d(-B(t))-\sum_i (B_{i-1}-B_i)P_i}{\sum_i \delta_iB_iP_i}.$$ The numerator is just the corresponding Riemann-Stieltjes integral in the question minus its approximating sum. Since $P$ increases with respect to $B$, $$0\le \int_0^T P(t)\mathrm d(-B(t))-\sum_i (B_{i-1}-B_i)P_i \le \sum_i (B_{i-1}-B_i)(P_{i-1}-P_i).$$ Therefore $\max\limits_i(B_{i-1}-B_i)\searrow 0 \bigwedge \max\limits_i(P_{i-1}-P_i)\searrow 0 \Longrightarrow c-s\searrow 0$.

In generally, for $a_i\ge 0$ and $b_i\ge 0,\ \forall i$, we have \begin{align} \Big( \sum_i a_ib_i\Big)^2 &= \Big(\sum_i (a_ib_i)^\frac{1}{2}a_i^\frac{1}{2}b_i^\frac{1}{2}\Big)^2 \\ &\le \sup\limits_i (a_ib_i)\Big(\sum_i a_i^\frac{1}{2}b_i^\frac{1}{2}\Big)^2 \\ &\le \sup\limits_i (a_ib_i)\sum_i a_i\sum_ib_i. \end{align} Therefore, we have stronger result from a weaker premise $\max\limits_i(B_{i-1}-B_i)(P_{i-1}-P_i)\searrow 0 \Longrightarrow c-s\searrow 0$.

The difference $c-s$ could be large when either default density is large in some period or the interest rate changes dramatically.