The HJM model calibrates the entire forward curve using the existing yield curve data and this results in the following expression for its instantaneous forward rate-


The above can be derived using the following expression for the zero coupon bond-


$$f(t,T,T+\tau)=\frac{\ln Z(t,T)-\ln Z(t,T+\tau)}{\tau}\implies\\ df(t,T,T+\tau)=\frac{\sigma_Z^2(t,T+\tau)-\sigma_Z^2(t,T)}{2\tau}dt+\frac{\sigma_Z(t,T)-\sigma_Z(t,T+\tau)}{\tau}dW^Q$$

Taking limit $\tau\rightarrow0$, we get-

$$df(t,T)=\sigma_Z(t,T)\frac{\partial \sigma_Z(t,T)}{\partial T}dt-\frac{\partial \sigma_Z(t,T)}{\partial T}dW^Q$$

Setting $\frac{\partial \sigma_Z(t,T)}{\partial T}=\sigma(t,T)$ gives us the first expression. We can extend the HJM similarly for two independent factors.


The evolution of the zero price process would be-


After manipulating the above similar to the 1 factor case, we get-

$$df(t,T)=\sigma_Z(t,T)\frac{\partial \sigma_Z(t,T)}{\partial T}dt-\rho\frac{\partial \sigma_Z(t,T)}{\partial T}dW_1^Q-\sqrt{1-\rho^2}\frac{\partial \sigma_Z(t,T)}{\partial T}dW_2^Q$$

Setting the loadings on the Brownian motion increments equal to $\sigma_i(t,T)$ gives us-

$$\frac{\partial \sigma_Z(t,T)}{\partial T}=\sqrt{\sigma_1^2+\sigma_2^2}$$

Thus the drift of the instantaneous forward rate should be given by-


However that's not the case. The canonical expressions for the drift are-


The risk-neutral drift is a function of the the partial standard deviations rather than the total standard deviation as was implied by my derivation. Can someone please explain why this is the case?

  • $\begingroup$ Have a try by assuming that $\frac{dZ(t,T)}{Z(t,T)}=r_tdt+\rho\sigma_1(t,T)dW_1^Q+\sqrt{1-\rho^2}\sigma_2(t,T)dW_2^Q$, rather than $\frac{dZ(t,T)}{Z(t,T)}=r_tdt+\rho\sigma_Z(t,T)dW_1^Q+\sqrt{1-\rho^2}\sigma_Z(t,T)dW_2^Q$. $\endgroup$ – Gordon Jul 23 '18 at 19:20
  • $\begingroup$ But that assumes that the total volatility of the zero coupon bond is $\sqrt{\sigma_1^2+\sigma_2^2}$, which also happens to be the total volatility of the forward rate. I don't think that assumption is valid in general. $\endgroup$ – Amrit Prasad Jul 24 '18 at 3:03
  • $\begingroup$ I meant that with different $\sigma_1$ and $\sigma_2$, try to derive the forward rate dynamics and see what you can obtain. $\endgroup$ – Gordon Jul 24 '18 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.