# Bond Prices in terms of short and forward rates

Of course, a pure discount bond price $P(t,T)$ may be stated in terms of its yield $R(t,T)$ as $$P(t,T) = e^{-R(t,T)(T-t)}.$$ Let's assume both the (instantaneous) short rate $r(t)$ and (instantaneous) forward rate $f(t,T)$ are deterministic functions. The relationships to discount bond prices are \begin{align} r(t) & = -\frac{\partial}{\partial T} \log P(t,t), \\ f(t,T) & = -\frac{\partial}{\partial T} \log P(t,T). \end{align} From this, it is clear that $$P(t,T) = \exp\left(-\int_t^T f(t,u) \, du\right) \qquad (1).$$

On the other hand, the bond price is often stated in terms of risk-neutral expectations using the short rate, such as $$P(t,T) = E_Q\left(\exp\left(-\int_t^T r(u) \, du\right) \mid \mathcal{F}_t\right),$$ and since I am assuming $r(t)$ is deterministic, it should be that $$P(t,T) = \exp\left(-\int_t^T r(u) \, du\right) \qquad (2).$$ Comparing Eqns (1) and (2), it seems like $$\int_t^T r(u) \, du = \int_t^T f(t,u) \, du.$$

Does this even make sense? Furthermore, since $P(t,T) = e^{-R(t,T)(T,t)}$, we would get $$R(t,T) = \frac{1}{T-t}\int_t^T f(t,u) \, du = \frac{1}{T-t}\int_t^T r(u) \, du.$$ That is, the yield is both the average of the instantaneous forward rate (this is true), and the average of the instantaneous spot rate. Is this latter statement true?

Since the interest rate is deterministic, for $t< u \le T$, \begin{align*} f(t, u) &= -\frac{\partial}{\partial u} \ln P(t, u)\\ &=-\frac{\partial}{\partial u} \ln \left(E\left(\exp\left(-\int_t^u r(s)\, ds \right) \mid \mathcal{F}_t\right) \right)\\ &=-\frac{\partial}{\partial u} \ln \left(\exp\left(-\int_t^u r(s)\, ds\right) \right)\\ &=\frac{\partial}{\partial u}\int_t^u r(s)\, ds \\ &= r(u). \end{align*} Consequently, \begin{align*} \frac{1}{T-t}\int_t^T f(t, u)\, du = \frac{1}{T-t}\int_t^T r(u)\, du. \end{align*}