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Deriving CIR Zero-Coupon Bond Dynamics with Itô’s Lemma

Article Quant Q&A · Author: none

Summary

The document asks how to derive the stochastic dynamics of a zero-coupon bond when the short rate follows the Cox–Ingersoll–Ross model and the bond price has an exponential-affine form. A direct Itô expansion differentiates the time-dependent coefficients and appears to leave terms that must cancel to recover the bond’s short-rate drift and diffusion.

The supplied answer instead uses the risk-neutral pricing representation: discounting the bond price by the accumulated short rate gives a martingale. Its quadratic variation follows from the bond’s log-price variation, allowing the diffusion coefficient to be identified without explicitly differentiating the complicated affine functions. The result has drift equal to the short rate times bond price, while the Brownian loading is negative and proportional to the affine rate coefficient and the square root of the short rate. This shortcut relies on the risk-neutral measure and the stated pricing setup; it does not show the algebraic cancellation from direct differentiation.

Key ideas

  • Under risk-neutral pricing, the bond price discounted by the accumulated short rate is a martingale.
  • The bond’s diffusion can be inferred from the quadratic variation of its log price.
  • The bond price drift under this setup is the short rate multiplied by the bond price.
  • The CIR bond’s Brownian loading is negative and depends on its affine coefficient and the square root of the short rate.

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Full text
# Chain rule for Ito's Lemma


# Chain rule for Ito's Lemma












The CIR short rate model is $$dr_t=k(\theta-r_t)dt+\sigma\sqrt{r_t}dW_t$$ under the risk-neutral measure. The bond price is of the form $$P(t,T)=A(t,T)e^{-B(t,T)r_t}$$ where the continuously compounded spot rate is an affine function of the short rate $r_t$. My question is, how should Ito's Lemma be applied to find $dP(t,T)$?

Here is my attempt: $$\ln P(t,T)=\ln A(t,T)-B(t,T)r_t$$ $$d\ln P(t,T)=d\ln A(t,T)-r_tdB(t,T)-B(t,T)dr_t$$ $$(d\ln P(t,T))^2=B(t,T)^2\sigma^2r_tdt$$ \begin{align} d(e^{\ln P(t,T)})&=P(t,T)\bigg(d\ln P(t,T)+\frac{1}{2}(d\ln P(t,T))^2\bigg)\\ &=P(t,T)\bigg(d\ln A(t,T)-r_tdB(t,T)-B(t,T)dr_t+\frac{1}{2}B(t,T)^2\sigma^2r_tdt\bigg)\\ &=\ldots\\ &=r_tP(t,T)dt-B(t,T)P(t,T)\sigma\sqrt{r_t}dW_t \end{align} Although I have followed the steps for Ito's Lemma, I seem to be missing a detail that will allow some terms to cancel out to produce the final line. Moreover the functions $A(t,T)$ and $B(t,T)$ are quite complex and I don't think differentiating them would be a good idea. $$A(t,T)=\bigg[\frac{2h\exp{\{(k+h)(T-t)/2\}}}{2h+(k+h)(\exp{\{(T-t)h\}-1})}\bigg]^{2k\theta/\sigma^2}$$ $$B(t,T)=\frac{2(\exp{\{(T-t)h\}-1)}}{2h+(k+h)(\exp{\{(T-t)h\}-1})}$$ $$h=\sqrt{k^2+2\sigma^2}$$ Source: Brigo & Mercurio, Interest Rate Models, 3.2.3

## Answer by M. Jeunesse (score 2)

https://quant.stackexchange.com/a/30806

You have to go start from the original expression of $P$:

$$P(t,T) = \mathbb{E}[e^{-\int_t^T r^s ds}|\mathcal{F}_t]$$

So if you define :

$$M_t = e^{-\int_0^t r_s ds}P(t,T)$$

this is a martingale.

So since you are in a brownian filtration,

$$dM_t = \sigma_t dW_t$$

It remains to find $\sigma_t$, which will be done by noticing that :

$$\sigma^2_t dt = d<M_t> = (e^{-\int_0^t r_s ds})^2 d<P(t,T)>$$

using your expression of $d<\ln P(t,T)>$ and since $d<P(t,T)>=P(t,T)^2 d<\ln P(t,T)>$ you get (using that $P(t,T)>0$)

$$dP(t,T) = r_t P(t,T) dt + P(t,T)\sqrt{\frac{d<\ln P(t,T)>}{dt}}dW_t$$

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.