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I am looking at the Black Derman Toy local short rate model as $$d\log r(t)=\alpha(t)(\theta (t)-\log r(t))dt+\sigma dW(t)$$ under RN measure. I would like to derive the bond price PDE. For that I consider $f(t,r(t))$ to be a price of a bond and by Feynman Kac I want to write $d[D(t)f(t)]$ and since this has to be a martingale($D(t)$ is a discount factor), I can set $dt$ term to zero. This is where I am stuck. I don't know what $dr(t)$ is!

I start by $$d(D(t)f(t))=fdD+Ddf+dfdD$$ where $dD=-rDdt$, and $df=f_tdt+f_rdr+0.5f_{rr}drdr$, and I need $dr$, can I get this from the $d\log r(t)$ in some way?

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Note that $r(t)=e^{\ln r(t)}$. Then \begin{align*} dr(t) &= e^{\ln r(t)} d\ln r(t) + \frac{1}{2}e^{\ln r(t)}\langle d\ln r(t), d\ln r(t)\rangle\\ &=r(t) \big[\alpha(t)(\theta (t)-\log r(t))dt+\sigma dW(t) \big] + \frac{1}{2}\sigma^2 r(t) dt\\ &=r(t)\Big[\Big(\frac{1}{2}\sigma^2 + \alpha(t)(\theta (t)-\log r(t))\Big)dt + \sigma dW(t)\Big]. \end{align*}

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  • $\begingroup$ thanks. But in this case, can $r(t)$ be negative? What is the distribution of $r(t)$? And if I were to solve this PDE numerically, where would I place the boundaries? $\endgroup$ – Medan Aug 2 '16 at 18:42
  • $\begingroup$ @Medan: For Derman model, $r(t)$ is log-normal, which can not be negative, otherwise, you won't be able to take logarithm. $\endgroup$ – Gordon Aug 2 '16 at 18:55

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