Deriving Short Rate Dynamics from the Black–Derman–Toy Model

Summary

The document explains how to obtain the short-rate process needed to derive a bond-pricing partial differential equation under the Black–Derman–Toy model. The model specifies the dynamics of the logarithm of the short rate, while the pricing function is expressed in terms of the rate itself. Itô’s lemma bridges those variables by transforming the log-rate process into a process for the rate.

That transformation gives the rate a drift containing both the model’s mean-reversion term and an additional volatility correction, while its diffusion is proportional to the rate. This is the key input for applying Itô’s lemma to the bond price and setting the discounted price’s drift to zero under the risk-neutral measure. The excerpt provides the transformation but does not carry out the full bond-pricing PDE derivation or discuss parameter calibration and boundary conditions.

Key ideas

• Itô’s lemma converts the specified log short-rate process into dynamics for the short rate itself.
• The rate process has a volatility-dependent correction in its drift.
• The resulting rate dynamics can be used in Itô’s lemma for a bond price function.
• Risk-neutral pricing requires the drift of the discounted bond price to vanish.

Tags

# Black Derman Toy short rate and PDE


# Black Derman Toy short rate and PDE












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?

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/28442

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*}

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.