(Reversal of order of integration in forward rates). The forward rate formula (10.3.5) with $v$ replacing $T$ states that
$$
f(t, v)=f(0, v)+\int_0^t \alpha(u, v) d u+\int_0^t \sigma(u, v) d W(u)
$$
Therefore,
$$
-\int_t^T f(t, v) d v=-\int_t^T\left[f(0, v)+\int_0^t \alpha(u, v) d u+\int_0^t \sigma(u, v) d W(u)\right] d v .
$$
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10 Term-Structure Models
(i) Define
$$
\widehat{\alpha}(u, t, T)=\int_t^T \alpha(u, v) d v, \quad \widehat{\sigma}(u, t, T)=\int_t^T \sigma(u, v) d v .
$$
Show that if we reverse the order of integration in (10.7.20), we obtain the equation
$$
\begin{aligned}
& -\int_t^T f(t, v) d v \\
& =-\int_t^T f(0, v) d v-\int_0^t \widehat{\alpha}(u, t, T) d u-\int_0^t \widehat{\sigma}(u, t, T) d W(u) .
\end{aligned}
$$
(In one case, this is a reversal of the order of two Riemann integrals, a step that uses only the theory of ordinary calculus. In the other case, the order of a Riemann and an Itô integral are being reversed. This step is justified in the appendix of [83]. You may assume without proof that this step is legitimate.)
(ii) Take the differential with respect to $t$ in (10.7.21), remembering to get two terms from each of the integrals $\int_0^t \widehat{\alpha}(u, t, T) d u$ and $\int_0^t \widehat{\sigma}(u, t, T) d W(u)$ because one must differentiate with respect to each of the two $t s$ appearing in these integrals.
(iii) Check that your formula in (ii) agrees with (10.3.10).