Black Derman & Toy Model

Question

The BDT model is given by

$$d(\ln\,r)=\left(\theta(t)-\frac {d(\ln\sigma(t)}{dt}\ln r\right)\,dt+\sigma(t) \, dW.$$

How can one rewrite the BDT model as $dr=A\,dt+B\, dW$, using It$\hat o$?

Answer 1

Apply Ito's Lemma to $r = f(X) = \exp(X)$ where X = ln(r).

$$df = \frac{\partial f}{\partial X}dX+ \frac{1}{2}\frac{\partial^2 f}{\partial X^2}dX^2$$

In this case

$$\frac{\partial f}{\partial X}= \exp(X)=r,$$ $$\frac{\partial^2 f}{\partial X^2}=\exp(X)= r.$$

Then substitute

$$dX = dln(r) = \mu(r,t)dt +\sigma(t)dW$$

$$dX^2 = [\sigma(t)]^2dt$$