Default intensity survival and recovery, financial engineering for quants lesson

A bond promises to pay 100 in four years, but its issuer may default before then. Which probability of survival enters the price of that promise, and what does the holder receive if the issuer fails first? This lesson defines the default time and its survival function, derives survival from a hazard rate, fixes what a recovery convention must specify, prices a zero-coupon bond whose recovery is paid at default, and states the Cox construction for a random intensity.

Key terms

Default time

Definition Fix a probability space $(\Omega, \mathcal{F}, \mathbb{Q})$ in which $\mathbb{Q}$ is a pricing measure, equivalent to the physical measure $\mathbb{P}$ and not necessarily unique. The default time of an issuer is a random variable $\tau$ with values in $(0, \infty]$ , the date at which the issuer first fails to meet an obligation, and the value $\tau = \infty$ means that it never defaults. On this page $\tau$ always denotes this random date, not a time to maturity. The event $\{\tau > t\}$ is survival to $t$ , and its complement $\{\tau \le t\}$ is default by $t$ . Since $\tau > 0$ , the issuer is alive at time $0$ . The same random time also has a law under $\mathbb{P}$ . Two equivalent measures agree on which default events have probability zero, but they can give the same event $\{\tau \le t\}$ different probabilities, and prices use the $\mathbb{Q}$ -law.

In plain words The default time is the random date on which a promise is broken. Every cash flow of a credit-risky claim is attached either to the event that this date comes after the payment date or to the event that it comes before.

Example A zero-coupon bond with face 100 and maturity 4 years pays 100 at year 4 on the event $\{\tau > 4\}$ . On the event $\{\tau \le 4\}$ it pays nothing at year 4, and the payment it makes instead, and its date, are set by its recovery convention. If $\tau$ is exponentially distributed under $\mathbb{Q}$ with rate $0.04$ per year, as in the constructed example of this lesson, then $\mathbb{Q}(\tau \le 4) = 1 - e^{-0.16} \approx 0.147856$ and $\mathbb{Q}(\tau = \infty) = 0$ .

Survival probability

Definition The survival function of $\tau$ under $\mathbb{Q}$ is

S(t) = \mathbb{Q}(\tau > t), \quad t \ge 0 .

On this page $S$ is a survival function, not a stock price. It satisfies $S(0) = 1$ , it is nonincreasing and right-continuous with values in $[0, 1]$ , and $S(t) \to \mathbb{Q}(\tau = \infty)$ as $t \to \infty$ . The default probability by $t$ is $1 - S(t)$ , the probability of default in the interval $(s, t]$ is $S(s) - S(t)$ , and for $s \le t$ with $S(s) > 0$ the conditional survival probability is

\mathbb{Q}(\tau > t \mid \tau > s) = \frac{S(t)}{S(s)} .

The survival function under $\mathbb{P}$ is in general a different function, and the two cannot replace each other.

In plain words The survival function gives, for every horizon, the probability that the issuer is still alive at that horizon. It can only stay level or fall as the horizon lengthens, because surviving to a later date requires surviving to every earlier one.

Example With the constant rate $0.04$ per year of the previous card, $S(t) = e^{-0.04 t}$ and $S(4) = e^{-0.16} \approx 0.852144$ . The probability of default between years 1 and 4 is $e^{-0.04} - e^{-0.16} \approx 0.108646$ , and an issuer alive at year 1 survives to year 4 with probability $e^{-0.12} \approx 0.886920$ . That conditional probability equals $S(3)$ , which is a feature of a constant rate and fails in general for a rate that changes over time.

Hazard rate

Definition Suppose $S$ is strictly positive and continuously differentiable on $[0, T]$ . The hazard rate of $\tau$ is

\begin{aligned} \lambda(t) &= -\frac{S'(t)}{S(t)} \\[0.4em] &= -\frac{d}{dt} \ln S(t), \end{aligned}

measured per year. At each $t \in [0, T)$ it is the conditional default rate over a short interval,

\begin{aligned} &\mathbb{Q}(t < \tau \le t + h \mid \tau > t) \\[0.4em] &\quad = \lambda(t)\, h + o(h) \end{aligned}

as $h \downarrow 0$ , because the left side equals $\bigl(S(t) - S(t + h)\bigr)/S(t)$ . In the other direction, surviving to $t + h$ means surviving to $t$ and then surviving the next interval, so $S(t + h) = S(t)\bigl(1 - \lambda(t) h\bigr) + o(h)$ , which gives the survival equation $S'(t) = -\lambda(t) S(t)$ . Dividing by $S$ and integrating $(\ln S)' = -\lambda$ from $0$ with $S(0) = 1$ gives the unique solution

S(t) = \exp\Bigl(-\int_0^t \lambda(u)\,du\Bigr).

The integral is the accumulated hazard. The default time has the density $f_\tau(t) = \lambda(t) S(t)$ on $[0, T]$ , and $\int_s^t f_\tau(u)\,du = S(s) - S(t)$ . A nonnegative hazard rate corresponds exactly to a nonincreasing $S$ . The formula also defines $S$ from any nonnegative piecewise continuous rate, such as a piecewise constant one, and then $S'(t) = -\lambda(t) S(t)$ holds at every point where $\lambda$ is continuous. When a model is specified by choosing this rate, the rate is called the default intensity. For a deterministic rate the default intensity and the hazard rate are the same function, and a random intensity is the subject of the Cox construction below.

In plain words The hazard rate is the default rate of an issuer that is still alive, expressed per year. It is not a probability. The one-year default probability for a constant rate $\lambda > 0$ is $1 - e^{-\lambda}$ , which is smaller than $\lambda$ , and a rate above one per year is possible. Survival is the negative exponential of the area under the rate curve up to the horizon.

Example At the constructed intensity $\lambda = 0.04$ per year, the one-year default probability is $1 - e^{-0.04} \approx 0.039211$ , slightly below $0.04$ . Doubling the intensity to $0.08$ gives $S(4) = e^{-0.32} \approx 0.726149$ . For the piecewise constant intensity equal to $0.02$ on $[0, 2)$ and $0.05$ on $[2, 5]$ , the accumulated hazard at year 5 is $0.04 + 0.15 = 0.19$ , so $S(5) = e^{-0.19} \approx 0.826959$ . A rate of $2$ per year gives the one-year default probability $1 - e^{-2} \approx 0.864665$ , a clear sign that the rate itself is not a probability.

Recovery convention

Definition A recovery convention states the amount paid to a claim holder at default and the date of that payment. For a zero-coupon bond with face $F$ , maturity $T$ and recovery rate $R \in [0, 1]$ , three common conventions are

recovery of face at default, the amount $RF$ paid at $\tau$ when $\tau \le T$ ,
recovery of face at maturity, the amount $RF$ paid at $T$ when $\tau \le T$ , which has the same value as receiving at $\tau$ the fraction $R$ of a default-free bond with the same face and maturity, the recovery of treasury convention,
recovery of market value, the fraction $R$ of the bond’s own market value just before default, paid at $\tau$ .

The same number $R$ gives three different random cash flows, because the base of the percentage or the payment date differs, and therefore three different prices. This lesson uses recovery of face at default unless it says otherwise. The recovery rate is an input of the pricing problem, and its calibration from market data is outside this lesson.

In plain words A recovery percentage alone does not define a payoff. The contract or the model must also say what the percentage applies to and when the money arrives, and each of these choices changes the value even when the percentage stays the same.

Example Take the constructed inputs $F = 100$ , $T = 4$ years, $r = 0.03$ , $\lambda = 0.04$ per year and $R = 0.35$ , with a constant continuously compounded rate and a constant intensity under $\mathbb{Q}$ . Recovery of face at default gives the price $80.4627$ of the next card. Recovery of face at maturity gives

\begin{aligned} &100\, e^{-0.12}\bigl(S(4) \\[0.4em] &\quad + 0.35\,(1 - S(4))\bigr) \\[0.4em] &\approx 80.1682 . \end{aligned}

For recovery of market value, balancing one short interval of length $h$ gives the condition

\begin{aligned} V(t) &= e^{-rh}\bigl[\lambda h R\, V(t) \\[0.4em] &\quad + (1 - \lambda h)\, V(t + h)\bigr] \\[0.4em] &\quad + o(h) \end{aligned}

on the pre-default value $V(t)$ , which gives $V'(t) = \bigl(r + \lambda(1 - R)\bigr) V(t)$ with $V(T) = F$ . The price is $100\, e^{-(0.03 + 0.04 \cdot 0.65) \cdot 4} = 100\, e^{-0.224} \approx 79.9315$ . One recovery rate gives three prices.

Defaultable bond value

Definition Assume that under $\mathbb{Q}$ the short rate $r(t)$ is deterministic, so that $D(0,t) = \exp\bigl(-\int_0^t r(u)\,du\bigr)$ is a deterministic discount factor, that the hazard rate $\lambda$ is nonnegative and piecewise continuous with survival function $S$ , and that recovery of face with a recovery rate $R \in [0, 1]$ is paid at default. The zero-coupon bond with face $F$ and maturity $T$ pays $F \mathbf{1}_{\{\tau > T\}}$ at $T$ and $RF \mathbf{1}_{\{\tau \le T\}}$ at $\tau$ . These payments sit on disjoint events, and the default time has density $\lambda S$ , so the $\mathbb{Q}$ -expectation of the discounted payments is

\begin{aligned} V_0 &= F\, D(0,T)\, S(T) \\[0.4em] &\quad + RF \int_0^T D(0,t) \\[0.4em] &\qquad \times \lambda(t) S(t)\,dt . \end{aligned}

For a constant rate $r$ and a constant intensity $\lambda$ , write $k = r + \lambda$ . If $k \ne 0$ ,

\begin{aligned} V_0 &= F e^{-kT} \\[0.4em] &\quad + RF \lambda\, \frac{1 - e^{-kT}}{k} , \end{aligned}

and if $k = 0$ , which needs a negative rate, the fraction is replaced by its limit $T$ . With $R = 0$ the price is $F e^{-(r + \lambda)T}$ , so the intensity enters like a spread added to the rate. If $r(t) \ge 0$ for all $t$ , every payment is at most $F$ and every discount factor is at most $1$ , so $V_0 \le F$ . The default-free price $F D(0,T)$ is not in general an upper bound under this convention when rates are positive, as the last question below shows. These formulas are exact in this deterministic model. With random rates and intensities the price is an expectation over their joint paths.

In plain words Price each possible payment on its own and add. The full face is weighted by the probability of surviving to maturity, and the recovery is weighted by the probability of default near each date and discounted from that date. The bond price is not the face reduced by an unexplained percentage.

Example For the constructed inputs, $S(4) \approx 0.852144$ , the survival payment is worth $100\, e^{-0.28} \approx 75.5784$ and the recovery payment is worth

\begin{aligned} &35 \cdot 0.04 \cdot \frac{1 - e^{-0.28}}{0.07} \\[0.4em] &\quad \approx 4.8843 , \end{aligned}

so $V_0 \approx 80.4627$ currency units, and a midpoint rule for the integral gives the same recovery value. Doubling the intensity to $0.08$ moves weight from the full face to the smaller recovery and lowers the price to $64.4036 + 9.0609 \approx 73.4645$ . The price $80.4627$ corresponds to the continuously compounded yield $-\tfrac14 \ln 0.804627 \approx 0.054344$ , a spread of about 243 bp over $r$ , below $(1 - R)\lambda = 0.026$ . A zero-recovery bond with intensity $0.024344$ has the same price, so one bond price cannot separate the intensity from the recovery rate.

Cox construction

Definition Let $(\mathcal{G}_t)$ be a background filtration carrying the market information other than default itself, let $\lambda_t \ge 0$ be a $(\mathcal{G}_t)$ -adapted process with $\Lambda_t = \int_0^t \lambda_s\,ds < \infty$ for every $t$ , and let $\xi$ be a standard exponential random variable, with $\mathbb{Q}(\xi > x) = e^{-x}$ for $x \ge 0$ , independent of the $\sigma$ -algebra $\mathcal{G}_\infty$ generated by all the $\mathcal{G}_t$ . The Cox construction defines

\tau = \inf\{t \ge 0 : \Lambda_t \ge \xi\} .

The path of $\Lambda$ is continuous and nondecreasing, so $\{\tau > t\} = \{\Lambda_t < \xi\}$ , and independence gives the conditional survival probability

\mathbb{Q}(\tau > t \mid \mathcal{G}_\infty) = e^{-\Lambda_t} .

The process $\lambda$ is then the default intensity of $\tau$ . Taking expectations gives $S(t) = \mathbb{E}^{\mathbb{Q}}\bigl[e^{-\Lambda_t}\bigr]$ , an average of exponentials and not the exponential of an average. Because $x \mapsto e^{-x}$ is convex, Jensen’s inequality gives

\mathbb{E}^{\mathbb{Q}}\bigl[e^{-\Lambda_t}\bigr] \ge e^{-\mathbb{E}^{\mathbb{Q}}[\Lambda_t]}

when $\mathbb{E}^{\mathbb{Q}}[\Lambda_t] < \infty$ , with equality only when $\Lambda_t$ is almost surely constant. If the short rate $r_t$ is also $(\mathcal{G}_t)$ -adapted and nonnegative, conditioning on $\mathcal{G}_\infty$ gives the price of a zero-recovery bond,

F\, \mathbb{E}^{\mathbb{Q}}\Bigl[e^{-\int_0^T (r_s + \lambda_s)\,ds}\Bigr] .

The general theory, which characterizes an intensity through the compensator of the default indicator, is outside this lesson.

In plain words Draw a random threshold first, independently of the market. Then accumulate the intensity along the market path, and default happens when the accumulated amount reaches the threshold. Given the path, survival is the familiar negative exponential. Without the path one must average the survival probabilities over paths, and averaging the intensity first never understates the default probability and overstates it whenever the accumulated intensity is random.

Example Let a fair coin, independent of $\xi$ , choose at time $0$ an intensity of $0.02$ or $0.06$ per year for the whole life of the issuer. The mean of $\Lambda_4$ is $0.16$ , yet

\begin{aligned} S(4) &= \tfrac12\bigl(e^{-0.08} + e^{-0.24}\bigr) \\[0.4em] &\approx 0.854872 , \end{aligned}

above $e^{-0.16} \approx 0.852144$ . The hazard rate $-S'(t)/S(t)$ of this survival function is $0.04$ at $t = 0$ and about $0.038403$ at $t = 4$ , because an issuer that has survived is more and more likely to have the low intensity. So the hazard rate of a Cox default time is in general not the expected intensity. With $r = 0.03$ , the zero-recovery bond with face 100 and maturity 4 is worth $50\bigl(e^{-0.20} + e^{-0.36}\bigr) \approx 75.8204$ , not the $75.5784$ of the averaged intensity.

Check yourself

Under $\mathbb{Q}$ an issuer has the constant default intensity $0.03$ per year. Find the probability of survival to year 5, the probability of default within five years, and the probability that an issuer alive at year 4 defaults during year 5. Why is the last number not $0.03$ ?

Show the answer

Survival to year 5 has probability $S(5) = e^{-0.15} \approx 0.860708$ , so default within five years has probability about $0.139292$ . Given survival to year 4, the default probability during year 5 is

\begin{aligned} &1 - \frac{S(5)}{S(4)} = 1 - e^{-0.03} \\[0.4em] &\quad \approx 0.029554 . \end{aligned}

The intensity is a rate that applies at each moment to issuers that are still alive. Over a full year the issuers that have already defaulted can no longer default, so the probability is $1 - e^{-0.03}$ , slightly less than the rate. The unconditional probability of default during year 5, $S(4) - S(5) \approx 0.026212$ , is smaller again, because it also requires survival to year 4.

The intensity is $0.02$ per year on $[0, 2)$ and $0.06$ per year on $[2, 4]$ , the rate is a constant $r = 0.03$ , and a bond with face 100 and maturity 4 years recovers $40$ percent of face at default. Find $S(4)$ and the bond price, and compare it with the price for the constant intensity $0.04$ .

Show the answer

The accumulated hazard at year 4 is $0.04 + 0.12 = 0.16$ , so $S(4) = e^{-0.16} \approx 0.852144$ and the survival payment is worth $100\, e^{-0.12} e^{-0.16} = 100\, e^{-0.28} \approx 75.5784$ . On $[2, 4]$ the survival function is $e^{-0.04 - 0.06(t - 2)}$ , so the recovery value is $40$ times

\begin{aligned} &\int_0^2 0.02\, e^{-0.05 t}\,dt \\[0.4em] &\quad + \int_2^4 0.06\, e^{0.08 - 0.09 t}\,dt \\[0.4em] &= \frac{0.02\,(1 - e^{-0.1})}{0.05} \\[0.4em] &\quad + \frac{0.06\, e^{-0.10}(1 - e^{-0.18})}{0.09} \\[0.4em] &\approx 0.038065 + 0.099369 , \end{aligned}

which is about $5.4974$ . The price is about $81.0757$ . With the constant intensity $0.04$ the survival term is the same, because it depends only on the accumulated hazard at maturity, but the recovery term is $40 \cdot 0.04\,(1 - e^{-0.28})/0.07 \approx 5.5821$ and the price is about $81.1605$ . The rising intensity puts the defaults later, so the same recoveries are discounted more.

A bond with face 100 and maturity 5 years recovers its full face at default, so $R = 1$ , with $r = 0.05$ and the constant intensity $0.10$ per year. Find its price and compare it with the default-free price $100\, e^{-0.25}$ . Is the default-free price an upper bound, and what changes if the recovery is paid at maturity instead?

Show the answer

With $k = 0.15$ the price is

\begin{aligned} &100\, e^{-0.75} \\[0.4em] &\quad + 10 \cdot \frac{1 - e^{-0.75}}{0.15} \\[0.4em] &\approx 47.2367 + 35.1756 \\[0.4em] &= 82.4122 , \end{aligned}

which is $4.5321$ above the default-free price $100\, e^{-0.25} \approx 77.8801$ . With $R = 1$ the holder receives the full face in every state, at the date $\min(\tau, T)$ , so the price is $100\, \mathbb{E}^{\mathbb{Q}}\bigl[e^{-r \min(\tau, T)}\bigr]$ , and an earlier repayment is worth more when $r > 0$ . Under recovery of face at default the default-free price is therefore not an upper bound. With these inputs the two prices coincide at $R \approx 0.8712$ . If the recovery is paid at maturity, the bond with $R = 1$ pays 100 at year 5 in every state and is worth exactly $100\, e^{-0.25}$ , and for $R \le 1$ that price is an upper bound. The bound $V_0 \le 100$ holds under both conventions because $r \ge 0$ .

Default intensity survival and recovery

Key terms

Default time

Survival probability

Hazard rate

Recovery convention

Defaultable bond value

Cox construction

Check yourself

How this lesson connects

Builds on

Used later in