I have two tasks:

  1. Given country's CDS spread draw implied probability of default.
  2. Given probability of default calculate CDS spread.

If possible, refer to any papers.

  • $\begingroup$ Many answers to this old question already, I'll just point out that the conversion to/from roak-neutral (not physical) probability of default and CDS quote also needs the recovery assumption, and that some numbers will change somewhat depending on the recovery assumption. $\endgroup$ Jan 19 at 14:04

4 Answers 4


Risk-neutral default probability implied from CDS is approximately $P=1-e^\frac{-S * t}{1-R}$, where $S$ is the flat CDS spread and $R$ is the recovery rate.

The CDS Spread can be solved using the inverse:

$$S=\ln(1-P) \frac{R-1}{t}$$

  • $S$ is the spread expressed in percentage terms (not basis points)
  • $t$ are the years to maturity
  • $R$ is the recovery rate in percentage terms

Hulls equation is a gross simplification. This equation is not perfect, but is far more accurate and works for all tenor points. It generally works well except when approaching boundary conditions (distressed credits).

  • 2
    $\begingroup$ Do you have a reference for this? Many thanks. $\endgroup$
    – CFW
    Aug 4, 2017 at 11:44

I believe the answer can be further improved for all those being directed here by google after 3 years.

A common way to model the default probability is by the hazard rate. As @Bob correctly mentions, a traditional requirement is for it to satisfy (see Option Futures and Other Derivatives section 23.4 in which the author discusses also other more exact approximations): $$\lambda(t)=\frac{S(t)}{1-R}.$$ This is associated with the default probability by (see Poisson Process): $$P(t,t+h)=\lambda(t)h+o(h)\,,$$ with $P(t,t+h)$ the probability of a default occurring between $t$ and $t+h$. Therefore: $$P(0,T)=\int_0^T(1-P(0,t))P(t,t+dt)=\int_0^T\lambda(t)(1-P(0,t))dt\,,$$ where the first term of the integral is "default has not occurred so far" and the second is "default occurs on the next time step". This means that $P$ satisfies: $$\frac{dP(0,t)}{dt}=\lambda(t)(1-P(0,t)).$$ If the CDS is assumed to be constant then $\lambda$ is constant and a solution would be: $$P(0,t)=1-\exp\left(\frac{-St\,\,\,}{1-R}\right).$$ Equivalently solution for the CDS is: $$S=\frac{R-1}{t}\log(1-P(0,t)).$$

  • $\begingroup$ Noting that to convert the derivation above to approximation $\lambda(t)=\frac{S(t)}{1-R}$ keep in mind $\exp{(x)}\approx 1+x$ for small $x$. $\endgroup$
    – cpage
    Aug 13, 2020 at 14:34
  • 1
    $\begingroup$ Its possible that the P(0,T) inside the integral is instead P(0,t) $\endgroup$
    – david.t_92
    Jan 19 at 10:58
  • $\begingroup$ Recovery rate does not make sense for me, the higer the recovery rate the higer the probability of default (I think it should be in the oposite way) $\endgroup$
    – david.t_92
    Jan 19 at 11:32
  • $\begingroup$ @david.t_92 If you raise the recovery rate, yet your spread is equal, then that means that you expect the country to default more likely $\endgroup$ Jan 19 at 12:36
  • $\begingroup$ Yes, Credit spread = (1 - Recovery Rate) * Default probability, 2 bonds with same credit spread the one with the highest rr have the highest default probability as with this one the probability to recover something if default is higher Thanks a lot @DiegoFMedina $\endgroup$
    – david.t_92
    Jan 19 at 16:38

The chapter in Hull on Credit Risk gives the same formula as emcor as a first approximation with a justification:

Consider first an approximate calculation. Suppose that a bond yields 200 basis points more than a similar risk-free bond and that the expected recovery rate in the event of a default is 40%. The holder of a corporate bond must be expecting to lose 200 basis points (or 2% per year) from defaults. Given the recovery rate of 40%, this leads to an estimate of the probability of a default per year conditional on no earlier default of $0.02/(1-04)$, or 3.33%. In general $$ \bar{\lambda} = \frac{s}{1-R}$$ where $\bar{\lambda}$ is the average default intensity (hazard rate) per year, $s$ is the spread of the corporate bond yield over the risk-free rate, and $R$ is the expected recovery rate.

This formula can easily be rewritten as $$s = \bar{\lambda} (1 -R)$$ as pointed out by @emcor.

User @lakesh pointed in a deleted question to a blog by Donald van Deventer that analyses this formula and he rejects it.

Both Hull and van Deventer remark that this formula is an imperfect approximation. It makes sense but there are some caveats and a number of improvements can be made and Hull gives one you can readily do yourself. What is best probably depends on the goal of the study.

  • $\begingroup$ I came across your answer after remembering this equation but wanting to think it through a bit more. I understand th broad intuition, but when I start to vary R I start to have some questions. Why is it that as R->0 (which should imply more risk?) does lambda become lower. If I assume 0 recovery, i have 100% loss at default. As R->1 lambda->large. Is this not unintuitive? I may be bungling something obvious. $\endgroup$
    – jason m
    Apr 30, 2018 at 17:34
  • $\begingroup$ @jasonm Better for visibility and our formatto have this as a separate question (maybe referencing this). $\endgroup$
    – Bob Jansen
    Apr 30, 2018 at 19:03

From this research from Deutsche Bank, : $$p_{def}=\frac{CDS_{spread}}{1-Rec}$$ $$\Leftrightarrow CDS_{spread}=p_{def}(1-Rec)$$ where $Rec$ is the recovery rate in case of default.

  • 5
    $\begingroup$ I just wanted to add that emcor's answer is almost correct but not quite. What he wrote as $p_{def}$ is not actually the probability of default, but rather the probability per unit time, or the "default intensity", i.e. if the probability of defaulting by time t is Q(0,T), then we have the diff eq $dQ(0,T) = p_{def}Q(0,T)dt$ with $Q(0,0)=0$ so that $Q(0,T) = exp(-p_{def}T)$ $\endgroup$
    – Paul
    Dec 26, 2014 at 1:32
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    $\begingroup$ Hi Paul, welcome to Quant.SE! I converted your answer to a comment for you. When you gather some reputation you will be able to does this yourself. $\endgroup$
    – Bob Jansen
    Dec 26, 2014 at 15:29

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