# One state variables implies perfect correlation of returns?

In Vasicek's seminal 1977 paper "An equilibrium characterization of the term structure", he states the bond price $P(t,s)$ is a function of the spot rate $r(t)$, $P(t,s) = P(t,s,r(t))$. He then states

"Thus, the value of the spot rate is the only variable for the whole term structure.

Okay, so far so good. He continues,

"Since there exists only one state variable, the instantaneous returns on bonds of different maturities are perfectly correlated.

I'm not sure why that would follow...could I get this translated to math, with a decent explanation? Finally, he states

This means that the short bond and just one other bond completely span the whole of the term structure."

What exactly does this mean, and why would he need it?

Under the Vasicek's model, the price of a zero-coupon bond is given by \begin{align*} P(t, T) = A(t, T)\exp\big(-B(t, T) r_t\big), \end{align*} where $A$ and $B$ are deterministic functions. In particular, $B$ is a positive increasing function (see any books on interest rate models). Then \begin{align*} \ln P(t, T) = \ln A(t, T) - B(t, T) r_t. \end{align*} For any four maturities $T_1$, $T_2$, $T_3$, and $T_4$, where $T_1 < T_2 <T_3 < T_4$, \begin{align*} \ln \frac{P(t, T_2)}{P(t, T_1)} = \ln \frac{A(t, T_2)}{A(t, T_1)} - [B(t, T_2)-B(t, T_1)] r_t, \end{align*} and \begin{align*} \ln \frac{P(t, T_4)}{P(t, T_3)} = \ln \frac{A(t, T_4)}{A(t, T_3)} - [B(t, T_4)-B(t, T_3)] r_t. \end{align*} It is now easy to check that \begin{align*} Var\bigg(\ln \frac{P(t, T_2)}{P(t, T_1)}\bigg) &= [B(t, T_2)-B(t, T_1)]^2 Var(r_t),\\ Var\bigg(\ln \frac{P(t, T_4)}{P(t, T_3)}\bigg) &= [B(t, T_4)-B(t, T_3)]^2 Var(r_t), \end{align*} and \begin{align*} Cov\bigg(\ln \frac{P(t, T_2)}{P(t, T_1)}, \ln \frac{P(t, T_4)}{P(t, T_3)}\bigg) &= [B(t, T_2)-B(t, T_1)] [B(t, T_4)-B(t, T_3)]Var(r_t). \end{align*} Then \begin{align*} Corr\bigg(\ln \frac{P(t, T_2)}{P(t, T_1)}, \ln \frac{P(t, T_4)}{P(t, T_3)}\bigg) &= 1. \end{align*}
As there are three model parameters, any two bonds can be used to calibrate them (there are two $A$s and two $B$s), and then all bond prices are known.
Updates based on poster's comments: Assuming that \begin{align*} dr_t = \alpha(r_t, t)dt + \beta(r_t, t) dW_t, \end{align*} where $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, assuming that the zero-coupon bond price is defined by \begin{align*} P(t, T) = P(t, T, r_t). \end{align*} Then \begin{align*} dP(t, T) &= \frac{\partial P(t, T)}{\partial t}dt + \frac{\partial P(t, T)}{\partial r_t}dr_t + \frac{1}{2}\frac{\partial^2 P(t, T)}{\partial r_t^2}\beta^2(r_t, t) dt\\ &=\bigg(\frac{\partial P(t, T)}{\partial t}+ \frac{\partial P(t, T)}{\partial r_t} \alpha(r_t, t) + \frac{1}{2}\frac{\partial^2 P(t, T)}{\partial r_t^2}\beta^2(r_t, t)\bigg)dt \\ &\qquad\qquad+ \frac{\partial P(t, T)}{\partial r_t}\beta(r_t, t)dW_t\\ &=P(t, T)\big[\mu(r_t, t, T)dt + \sigma(r_t, t, T)dW_t \big], \end{align*} for certain adapted functions $\mu(r_t, t, T)$ and $\sigma(r_t, t, T)$. Moreover, \begin{align*} Var\Big[dP(t, T)/P(t, T) \mid \mathcal{F}_t\Big] = \sigma^2(r_t, t, T) dt. \end{align*} Furthermore, for $T_1$, $T_2$, where $t < T_1 \leq T_2$, \begin{align*} Cov\Big[dP(t, T_1)/P(t, T_1),\, dP(t, T_2)/P(t, T_2) \mid \mathcal{F}_t\Big] = \sigma(r_t, t, T_1)\sigma(r_t, t, T_2) dt. \end{align*} That is, \begin{align*} Corr\Big[dP(t, T_1)/P(t, T_1),\, dP(t, T_2)/P(t, T_2) \mid \mathcal{F}_t\Big] = 1. \end{align*} In other words, the instantaneous returns on bonds of different maturities are perfectly correlated.
• Thanks for the reply...would you mind elaborating a bit? Why is $P(t,T_2)/P(t,T_1)$ considered a return? It seems like for a given bond, say the $T$-bond, a return on it would be $P(t,T)/P(s,T)$ for $s \leq t$, whereas $P(t,T_2)/P(t,T_1)$ seems to just be the ratio of two different bond prices. Also, where does "only one source of randomness" come in here? And I'm still not quite clear how two bonds span the term structure because of that... – bcf Jun 8 '15 at 13:04