# Extension of HJM to multiple factors

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-

$$df(t,T)=\sigma(t,T)\int_0^T\sigma(t,\tau)d\tau+\sigma(t,T)dW^Q$$

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

$$\frac{dZ(t,T)}{Z(t,T)}=r_tdt+\sigma_Z(t,T)dW^Q$$

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

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.

$$df(t,T)=m(t,T)dt+\sigma_1dW_1^Q+\sigma_2dW_2^Q$$

The evolution of the zero price process would be-

$$\frac{dZ(t,T)}{Z(t,T)}=r_tdt+\rho\sigma_Z(t,T)dW_1^Q+\sqrt{1-\rho^2}\sigma_Z(t,T)dW_2^Q$$

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-

$$\sqrt{\sigma_1^2(t,T)+\sigma_2^2(t,T)}\int_t^T\sqrt{\sigma_1^2(t,s)+\sigma_2^2(t,s)}ds$$

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

$$\sum_k\sigma_k(t,T)\int_t^T\sigma_k(t,s)ds$$

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?

• 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 $\frac{dZ(t,T)}{Z(t,T)}=r_tdt+\rho\sigma_Z(t,T)dW_1^Q+\sqrt{1-\rho^2}\sigma_Z(t,T)dW_2^Q$. – Gordon Jul 23 '18 at 19:20
• 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. – Amrit Prasad Jul 24 '18 at 3:03
• I meant that with different $\sigma_1$ and $\sigma_2$, try to derive the forward rate dynamics and see what you can obtain. – Gordon Jul 24 '18 at 13:32