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?


1 Answer 1


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

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


$$ Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0, $$ $$ \tau = \inf \; \{t\geq 0 | Y_t = 0 \}, $$ 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:

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


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