The Merton model says we have a geometric brownian motion $V(t)$ with drift $\mu$ and volatility $\sigma$. Thus $$V(t)=V(0) \exp\left(\sigma W(t)+(\mu-\frac{1}{2}\sigma^2)t\right)$$ where $W(t)$ is a brownian motion. Now at time $T$ if $V(T) \leq B$ for some given constant $B$ we say a default has occurred. I am trying to find a family of functions $\mu =f(\sigma\;|\;B,T,V_0)$ for which the probability of default is constant.

I'm a stuck on this, can someone help? We know that $\ln V(T)$ is normally distributed since $V(t)$ is a geometric brownian motion, but when I apply this I just get an integral equation I get stuck on. I also tried using the reflection principle for brownian motion, but I can't seem to get it to help.


1 Answer 1


Since (for $t\gt 0$) $W(t)/\sqrt{t}$ has a standard Normal distribution, independent of $t$, then

$$\Pr(V(t)\le B)= \Pr\left(\frac{W(t)}{\sqrt{t}} \le \frac{\log\left(B/V(0)\right) - t\mu(t) + t \sigma^2/2}{\sigma \sqrt{t}}\right)$$

is a constant in $t$ if and only if the right hand side is, whence

$$\mu(t) = \frac{1}{t}\log \frac{B}{V(0)} - \frac{1}{\sqrt{t}}\lambda \sigma + \sigma^2/2$$

for some real number $\lambda$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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