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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.

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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 $\lambda$.

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