The forward rate from time $t$ to $T$ ($f_{t,T}$) can be approximated by:

$$ f_{t,T}= \left[ \frac{(1+r_T)^T}{(1+r_t)^t} \right]^{\frac{1}{{T-t}}}-1 \sim \frac{(1+r_T)^T-(1+r_t)^t}{T-t} $$

Why is that the case?


1 Answer 1


You may show it as follows: \begin{align*} f_{t,T}&= \left[ \frac{(1+r_T)^T}{(1+r_t)^t} \right]^{\frac{1}{T-t}}-1\\ &=e^{\frac{1}{T-t} \left[\ln (1+r_T)^T - \ln (1+r_t)^t \right]} -1\\ &\approx e^{\frac{1}{T-t} \left[(1+r_T)^T-1 - \big((1+r_t)^t -1\big)\right]} -1\\ &=e^{\frac{1}{T-t} \left[(1+r_T)^T - (1+r_t)^t\right]} -1\\ &\approx 1+ \frac{1}{T-t} \left[(1+r_T)^T - (1+r_t)^t\right] -1\\ &=\frac{1}{T-t} \left[(1+r_T)^T - (1+r_t)^t\right]. \end{align*}


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