Skip to content
All library documents

Why Black–Cox Default Intensity Is Zero Before Barrier Hitting

Article Quant Q&A · Author: Pollo Gi
Read the original

Summary

The note examines the default intensity in the Black–Cox structural credit model, where default occurs when a geometric Brownian firm value first reaches a fixed absorbing barrier. It rewrites the distance to default as a Brownian motion with drift and gives a conditional first-passage probability for default by a future time, conditional on the current information.

Differentiating that probability with respect to the future time yields a conditional default-time density. The expression simplifies using an identity for the normal density, and its limit as the future time approaches the present is zero when the firm remains above the barrier. This supports the conclusion that the instantaneous intensity is zero before barrier contact; at contact, default is immediate under the absorbing-barrier definition, corresponding to the stated zero-or-infinite behavior. The note sketches the calculation rather than spelling out all regularity conditions or the boundary case in detail.

Key ideas

  • The Black–Cox model defines default as the first time firm value reaches a constant barrier.
  • The distance to default can be represented as a Brownian motion with drift and volatility.
  • A conditional first-passage probability can be differentiated to obtain a default-time density.
  • The density tends to zero as the future time approaches the present when the barrier has not been reached.

Tags

Fixed incomeStatisticsRisk management
Full text
# Default intensity in Black-Cox model


# Default intensity in Black-Cox model












Consider the model by Black and Cox (Journal of Finance, 1976).

The default intensity function is defined in the usual way: $$h(t) \equiv - \frac{\partial \log P[\tau > t| \mathcal{F}_t]}{\partial t}$$ where $\tau$ is the first hitting time of a constant absorbing barrier $V_b$ of a Geometric Brownian motion $V_t$, and $\mathcal{F}_t$ is the filtration up to time $t$. In the Black-Cox model, $h(t) \in \{0, \infty\}$. How can one prove this?

## Answer by ir7 (score 5, accepted)

https://quant.stackexchange.com/a/63245

As shown in Credit Risk Modeling Notes (Bielecki, Jeanblanc, Rutkowski), Corollary 1.3.1, for $t < s$, we have:

$$ P(\tau \leq s | {\cal F}_t) = N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right ) + {\rm e}^{-2\nu \sigma^{-2}Y_t} N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right ),$$

where

$$ Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0, $$ $$ \tau = \inf \; \{t\geq 0 | Y_t = 0 \}, $$ and $N$ is the standard normal cdf.

(In your notations, $Y_t$ is the distance to default, $Y_t =\ln (V_t/V_b)$.)

We then calculate the conditional density probability as follows: $$ \frac{\partial P(\tau \leq s | {\cal F}_t)}{\partial s} $$ $$ = n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right) \left( 2^{-1}Y_t \sigma^{-1}(s-t)^{-3/2}- 2^{-1}\nu(s-t)^{-1/2}\right) $$ $$ + {\rm e}^{-2\nu \sigma^{-2}Y_t} n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right) \left( 2^{-1}Y_t \sigma^{-1}(s-t)^{-3/2}+ 2^{-1}\nu(s-t)^{-1/2}\right) $$

$$ = n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right) Y_t \sigma^{-1}(s-t)^{-3/2}, $$

noting that $$ {\rm e}^{-2\nu \sigma^{-2}Y_t} n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right) = n\left( -Y_t \sigma^{-1}(s-t)^{-1/2}-\nu(s-t)^{1/2}\right),$$

where $n$ is the standard normal pdf.

Using L'Hospital we get:

$$ \lim_{s\rightarrow t^+} \frac{\partial P(\tau \leq s | {\cal F}_t)}{\partial s} =0.$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.