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If a zero coupon bond price at time $t$, with maturity $T$ ($t<T$), is denoted by

$B(t;T) = B(T;T) e^{(-\int_{t}^{T} r(s) ds)}$

where $r(t)$ is a known interest rate.

How does this transform to $r(T) = - \frac{1}{B(t;T)} \frac{\partial B(t;T)}{\partial T}$

I know that $B(T;T)=1$ and we can rearrange, but I don't understand how to obtain partial differentials from integrals?

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If interest rates are deterministic (i.e. time-dependent but non-ranom), then \begin{align} B(t,T) &= \exp\left( - \int_t^T r(s)\mathrm{d}s\right) \\ \Leftrightarrow \int_t^T r(u)\mathrm{d}u &= -\ln B(t,T). \end{align} Differentiating both sides with respect to $T$ according to the Leibniz rule yields \begin{align*} r(T) &= -\frac{\partial \ln B(t,T)}{\partial T}\\ &= -\frac{1}{B(t,T)} \frac{\partial B(t,T)}{\partial T}. \end{align*} The latter line uses the chain rule. Recall that $\frac{\partial B(t,T)}{\partial T}<0$ which gives you positive interest rates. After all, bond prices typically decrease with maturity.

If interest rates are stochastic, the first equation requires a conditional expectation. However, note that by definition, the instantaneous forward rate $f(t,T)$ satisfies \begin{align*} f(t,T) &= -\frac{\partial \ln B(t,T)}{\partial T} \\ &=-\frac{1}{B(t,T)}\frac{\partial B(t,T)}{\partial T}. \end{align*} The latter equality follows again from the chain rule. The first line allows you to write again \begin{align*} B(t,T) = \exp\left(-\int_t^T f(t,s)\mathrm{d}s\right). \end{align*} This holds even if interest rates are stochastic. Furthermore, note that $\lim\limits_{T\to t} f(t,T) = r(t)$, which is the instantaneous short rate.

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