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The Black Derman & Toy (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 Ito??

I searched everywhere but no answer.

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  • $\begingroup$ Aren't we going to compute Ito Lemma for f(x) = exp(x) $\endgroup$ – SmallChess Oct 29 '14 at 3:57
  • $\begingroup$ No, because $\exp(x)' = \exp(x)$. $\endgroup$ – Ric Oct 29 '14 at 16:42
  • $\begingroup$ Can we close this question? It seems to be entirely obvious/basic knowledge. $\endgroup$ – Drew Dec 15 '14 at 22:56
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    $\begingroup$ @Drew I think this question is way more brainy than some others out there ...@Student T seeing Gordon's answer your comment is the solution if we let $A$ depend on $r_t$ and $\sigma_t$ ... $\endgroup$ – Ric Dec 16 '14 at 15:14
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    $\begingroup$ @Drew: I think this question is not too bad. $\endgroup$ – vonjd Dec 16 '14 at 15:16
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If we are going to have the form \begin{align*} dr = A dt + BdW_t, \end{align*} Then both A and B are functions of $t$ and $r_t$, otherwise, $r_t$ is normal. However, note that \begin{align*} r_t = \exp\Bigg(\frac{1}{\sigma(t)}\bigg(\int_0^t \theta(s)\sigma(s) ds +\sigma(0)\ln r_0 + \int_0^t\sigma^2(s) dW_s\bigg)\Bigg). \end{align*} That is, $r_t$ is log-normal, assuming that both $\theta(t)$ and $\sigma(t)$ are determinsitic.

From \begin{align*} d(\ln\,r_t)=\Big(\theta(t)-\frac {d(\ln\sigma(t))}{dt}\ln r_t\Big)\,dt+\sigma(t) \, dW_t, \end{align*} we obtain that \begin{align*} dr_t &= d\big(e^{\ln r_t}\big)\\ &= r_t\Big( d \ln r_t + \frac{1}{2}\langle d\ln r_t, \, d\ln r_t\rangle\Big)\\ &= r_t \bigg[\Big(\theta(t)-\frac {d(\ln\sigma(t))}{dt}\ln r_t + \frac{1}{2} \sigma^2(t)\Big)\,dt+\sigma(t) \, dW_t \bigg]. \end{align*}

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  • $\begingroup$ Could you explain a bit how you derive the solution for $r_t$ .. with the exponential (2nd formula). I don't see that this is the solution that easily .. $\endgroup$ – Ric Dec 16 '14 at 15:17
  • $\begingroup$ Hi Richard, you can assume that $r_t = r_0 \exp(a(t)+\sigma(t)W_t) $ and then work out $a(t)$. $\endgroup$ – Gordon Dec 16 '14 at 15:32
  • $\begingroup$ I missed the term $\big(\sigma(t)-\sigma(0)\big)\ln r_0$ in the exponent of the $r_t$ formula, which I am now added. $\endgroup$ – Gordon Dec 16 '14 at 19:55
  • $\begingroup$ My original $r_t$ formula does have some errors, and I have now corrected. The conclusion still holds. $\endgroup$ – Gordon Dec 16 '14 at 20:42
  • $\begingroup$ Changed $\langle dr_t, \, d r_t\rangle$ to $\langle d\ln r_t, \, d\ln r_t\rangle$. $\endgroup$ – Gordon Nov 4 '15 at 14:38
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So I have a "+" sign for the second term (not negative)

dr =r[(θ(t)+ d(lnσ)/dt * lnr + 1/2*σ^2)dt + σdW]

I left out the subscript t's.....

You can let V = log r then Apply Ito and solve for A and B... where B = r*σ

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    $\begingroup$ if you use latex, it will be more readable. $\endgroup$ – Gordon May 9 '15 at 23:44
  • $\begingroup$ I would appreciate that too $\endgroup$ – Tulio Carnelossi May 25 '15 at 5:12

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