Given two methods to calculate the 1 year conditional probability of default of a zero coupon bond, I've come up with slightly different but close results.

From my approaches below, is it reasonable for the results to be off? Is the amount they are off considered large? Am I missing something?


1 year zero coupon bond with a face value of 1 million trading at 80% of face value. Assuming 0 recovery and a risk free rate of 5%.

1 year conditional (on no prior defaults) probability of default:

Method 1

Obtain the probability of default from a hazard rate (instantaneous conditional probability of default)

$Bond Return = (\frac{Face}{Price})^{1/maturity} -1 = 25\%$

$Spread = 25\% - 5\% = 5\%$

$\lambda = \frac{spread}{1-Recovery} = 20\%$

$\pi_{1 year} = 1 - e^{-\lambda} = 18.13%$

Method 2

Equate the future value of a risky bond with yield (y) and default probability ($\pi$) to a risk free asset with yield ($R_f$)

$1 + R_f = (1-\pi)*(1+R_f+z)+\pi*Recovery$

Where z is the spread.

Given the above (sourced from a GARP FRM practice exam), the result is 16%.


Let $\tau$ be the default time, $\lambda$ be the constant hazard rate, and $T=1.0$ be the bond maturity. The value of the defaultable zero-coupon bond is given by \begin{align*} D(0, T) &= e^{-rT}P(\tau > T). \end{align*} Then the default probability is given by \begin{align*} P(\tau \le T) &= 1- P(\tau > T)\\ &=1-D(0, T) \times e^{rT}\\ &=1-0.8 \times e^{0.05}\\ &=0.158983. \end{align*} The Method 2 result appears much closer.

$$$$ The mismacth in your Method 1 is caused by the inconsistency of the bond return and the hazard rate, that is, one is simple compound, while the other is continuous. If you define your bond reurn by \begin{align*} BondReturn &= -\frac{1}{maturity} \ln \frac{price}{face}\\ &= -\ln (0.8) = 0.22314, \end{align*} then you have hazard rate \begin{align*} \lambda = spread= 0.22314 -0.05 = 0.17314. \end{align*} and the default probability \begin{align*} 1-e^{-\lambda} = 0.158983. \end{align*} In Method 2, both the spread and bond return are assumed to be simple, which may not be accurate, but, at least, consistent.

  • $\begingroup$ Thanks! Arg I need more characters to post this comment. $\endgroup$ – AfterWorkGuinness Nov 18 '15 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.