I have the zero rates at certain time nodes, say 3-month, 5-month, 8-month,...,2-yr,... Now I want to interpolate the curve so that the implied one-month forward rates are piecewise linear. That is, between two nodes, the forward rates are linear. I have no idea where to start. Can someone help?


1 Answer 1


Suppose that you interpolate your zero curve, i.e. your discount factors at time $k$, $v_k$, using a log-quadratic approach:

$$\ln(v_i) = \alpha + \beta D_i + \gamma D_i^2 $$

where $v_i$ is a discount factor between two known discount factors, $v_k$ and $v_{k+1}$, and $D_i$ is the day-count-fraction between the dates associated with $v_k$ and $v_i$.

Note that in order to derive the parameters $\alpha$ $\beta$ and $\gamma$ you need boundary conditions for your interpolation, for example endpoints alining and one initially zero derivative:

i.e. if $D_k=0$ then $\alpha = \ln(v_k)$
i.e. for $D_{k+1}$ then $\; \ln(v_{k+1}) = \ln(v_k) + \beta D_{k+1} + \gamma D_{k+1}^2$
i.e a derivative is assumed at start (e.g. zero derivative): $\implies \beta=0$

Consider the discount factor on the day following $v_i$, $v_{i(+1d)}$:

$$\ln(v_{i(+1d)}) = \alpha + \beta (D_i+\frac{1}{360}) + \gamma (D_i+ \frac{1}{360})^2 $$

Then the overnight continuously compounded rate for the date associated with $v_i$ is:

$$ r_i = 360 \ln (\frac{v_i}{v_{i(+1d)}}) = -\beta -\gamma (D_i + \frac{1}{360})$$

You can observe that this is linear in $D_i$ indicating that continuously compounded overnight rates are linearly interpolated between dates $k$ and $k+1$.

Does this mean that your implied one-month forward rates are piecewise linear. No, but its not going to be far off. If you want to derive a curve interpolated in terms of rates then you will need a configured optimiser I expect.


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