# Interpolation of Zero rate curve

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?

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.