I'm trying to understand which parameter controls the instantaneous correlation in the 2 F HW model. As in, correlation b/w 2 rates observed at the same time. My thinking is as follows:

$$Rate(1)=P(t,x(t),y(t))$$ $$Rate(2)=Q(t,x(t),y(t))$$

Intuitively, if I know $Rate(1)$, more the correlation between short rates $x(t)$ and $y(t)$, the better I can predict $Rate(2)$, and thus correlation must be controlled by the correlation between the Brownian Motions.

However, correlation must also depend on the mean reversion difference, because we're back to perfect correlation (1F HW) if mean reversions of the 2 short rates are the same.

I am wondering if someone has the closed form expression for this correlation. I'd be very glad to see a reference.

For reference above, $P$ and $Q$ are some functions, $x(t)$ and $y(t)$ are the 2 constituent short rates in the model.

Edit: To clarify, I'm asking for the correlation between 2 rates in the same currenct, but of different tenors (say 3M LIBOR and 6M LIBOR)

  • $\begingroup$ Two-factor HW is a model with two stochastic drivers for a single short rate process. It sounds like you are asking for the correlation between two short rates (eg. one for EUR and one for USD)? This will involve four stochastic drivers (two factors per short rate) which will in general have a correlation matrix with 6 off-diagonal terms - is this what you are asking about, or have I misunderstood? $\endgroup$ – StackG Aug 5 '20 at 4:22
  • $\begingroup$ I'm asking about the correlation b/w say the 3M Libor and the 6 M Libor (different tenors). $\endgroup$ – Arshdeep Singh Duggal Aug 5 '20 at 4:32
  • $\begingroup$ Ok, that makes more sense - so in the ideal case, the 3M rate and the 6M rate are both calculated from $- {\frac {\partial} {\partial T}} \ln{P(0,T)}$ where $P(0,T)$ is the ZCB out to that tenor coming from the integrated short rate - you want the autocorrelation of this? $\endgroup$ – StackG Aug 5 '20 at 4:49
  • $\begingroup$ You want correlation of forward-starting libor? I'm pretty sure you do not. I think your $t$ should be $0$, so you exactly want the correlation of the now-to-3M and now-to-6M rate? $\endgroup$ – StackG Aug 5 '20 at 4:59
  • $\begingroup$ Ok without getting too entangled in the terminology, sure you can tell me the correlation of the t=0 case I'd be happy (although strictly speaking I already know these rates, they're not random variables anymore) - but the mechanics of the exercise should be exactly the same if you treat them as random. $\endgroup$ – Arshdeep Singh Duggal Aug 5 '20 at 5:03

To understand correlation in HW2F (or G2++) model it suffices to compute the correlation for the log of bonds at two different maturities. Your intuition is right, that its correlation is not just driven by the correlation of the two Brownian motion, but also by their mean reversions. The model is still different from HW1F as long as the two mean reversions are different. Let us recall the G2++ model dynamics: $$dr_t=\theta_t^\prime+dx_t+dy_t$$ $$dx_t=-ax_t dt +\sigma dW_t^x$$ $$dy_t=-bx_t dt +\eta dW_t^y$$ $$d\langle W^x,W^y\rangle_t=\rho dt$$ When $\rho=-1$, we have $dr_t=[\theta_t^\prime-(ax_t+by_t)]dt+(\sigma+\eta)dW_t^x$, which will be different than $x_t+y_t$ provided $a\ne b$. The term with the smallest mean reversion will indeed be more volatile than the other one. This is theoretically justified by the expression of the correlation: $$Corr(x_t,y_t )=\frac{\sqrt{ab}(1-e^{-(a+b)t})}{(a+b)\sqrt{(1-e^{-2at})(1-e^{-2bt})}}.$$

Now, we can focus on calculating the correlation between the long-term and short-term rates. Let us consider the short and long term rates expressed in terms of log ZCBs: $P(t,T)=\exp⁡{r(t,T)(T-t)}$, and let's compute the correlation between $P(t,T_1)$ and $P(t,T_2)$, with $T_1<T_2$. $$\ln⁡(P(t,T))=-\int_t^T θ_sds-\frac{1-e^{-a(T-t)}}{a}x_t-\frac{1-e^{-b(T-t)}}{b}y_t+\frac{1}{2} V(t,T)$$ with $V(t,T)$ being the conditional variance of the short-rate, \begin{align} V(t,T) &= \frac{\sigma^2}{a^2}\left[(T-t)-2\frac{1-e^{-a(T-t)}}{a}+\frac{1-e^{-2a(T-t)}}{2a}\right]\\ &+\frac{\eta^2}{b^2}\left[(T-t)-2\frac{1-e^{-b(T-t)}}{b}+\frac{1-e^{-2b(T-t)}}{2b}\right]\\ &+2\rho\frac{\sigma\eta}{ab}\left[(T-t)-\frac{1-e^{-a(T-t)}}{a}-\frac{1-e^{-b(T-t)}}{b}+\frac{1-e^{-(a+b)(T-t)}}{a+b}\right] \end{align} Conditionally on the time $\tau$ filtration, $x_t$ and $y_t$ are both normally distributed stochastic processes, whilst the other two terms are deterministic. So, we can study the covariance of the two bonds with different maturity and obtain: \begin{align} Cov_\tau(\ln ⁡P(t,T_1),\ln ⁡P(t,T_2))&=\mathbb{E}_\tau\left[\left(\frac{1-e^{-a(T_1-t)}}{a}\sigma\int_\tau^t e^{-a(t-u)}dW_u^x+\frac{1-e^{-b(T_1-t)}}{b}\eta\int_\tau^te^{-b(t-u)}dW_u^y\right)\left(\frac{1-e^{-a(T_2-t)}}{a}\sigma\int_\tau^te^{-a(t-u)}dW_u^x+\frac{1-e^{-b(T_2-t)}}{b}\eta\int_\tau^te^{-b(t-u)}dW_u^y\right)\right]\\ &=\frac{(1-e^{-a(T_1-t)})(1-e^{-a(T_2-t)})}{2a^3}\sigma^2(1-e^{-2a(t-s)})\\ &+\frac{(1-e^{-b(T_1-t)})(1-e^{-b(T_2-t)})}{2b^3}\eta^2(1-e^{-2b(t-s)}) \\ &+\frac{(1-e^{-a(T_1-t)})(1-e^{-a(T_2-t)})+(1-e^{-b(T_1-t)})(1-e^{-b(T_2-t)})}{ab(a+b)}\rho\sigma\eta(1-e^{-(a+b)(t-s)}). \end{align} $$\Rightarrow \left(\varrho_{(\ln ⁡P(t,T_1),\ln ⁡P(t,T_2))}\right)_\tau=\frac{Cov_\tau(\ln ⁡P(t,T_1),\ln ⁡P(t,T_2))}{\sqrt{V(t,T_1)V(t,T_2)}}$$

At this point you can make some simplifications and realise that regardless of the value of $\rho$, the numerator and denominator of $\left(\varrho_{(\ln ⁡P(t,T_1),\ln ⁡P(t,T_2))}\right)_\tau$ cancel out if $a=b$. You can now use the Libor-Bond relationship to get the correlation between 3M and 6M libors.


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