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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?

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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*}

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This says that if rates are deterministic, the spot rate follows the forward rates that are initially observed. Makes sense.

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