Applying Ito’s Lemma to the Black–Derman–Toy Short-Rate Model

Summary

The document shows how to convert the Black–Derman–Toy model, specified as a stochastic differential equation for the logarithm of the short rate, into an equation for the short rate itself. It applies Ito’s lemma to the exponential transformation from log rate to rate. The resulting drift includes the original log-rate drift and an additional half-variance term, while the diffusion coefficient is the short rate multiplied by the model volatility.

The answer treats the time-dependent parameters as deterministic and notes that the transformed rate follows a lognormal form under that assumption. In an equation written as dr = A dt + B dW, both coefficients generally depend on time and the current rate. A second response disputes the sign of one drift term but does not provide a derivation, so that assertion should not outweigh the explicit Ito calculation without checking the model convention. This is a mathematical transformation, not a discussion of calibration or empirical performance.

Key ideas

• Apply Ito’s lemma to the exponential of the log short rate to derive the dynamics of the rate itself.
• The transformed diffusion coefficient is the short rate multiplied by the model volatility.
• The drift gains a half-variance adjustment from Ito’s lemma.
• The drift and diffusion coefficients generally depend on both time and the current short rate.
• The stated lognormal implication assumes deterministic time-varying model parameters.

Tags

# How to express the Black Derman & Toy Model in a $dr=A\,dt+B\, dW$ form?


# How to express the Black Derman & Toy Model in a $dr=A\,dt+B\, dW$ form?












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.

## Answer by Gordon (score 3)

https://quant.stackexchange.com/a/15881

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*}

## Answer by Matthew (score 0)

https://quant.stackexchange.com/a/17714

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.