Matching Forward-Rate Smoothness at an Interpolation Boundary
Summary
The note considers a zero-coupon curve built with different interpolation methods on either side of a shared pillar: linear interpolation in one region and a quartic forward-rate spline in another. The central issue is the behavior at the junction. Continuous interpolated zero-coupon values do not by themselves guarantee a smooth instantaneous forward rate, so the relevant condition depends on what quantity is expected to match across the boundary.
The response distinguishes continuity from smoothness and explains that spline smoothness is obtained by matching derivatives at the joining pillar. To make the forward rate smooth, the derivative conditions for the two pieces must be specified consistently; endpoint derivative choices may need to be supplied to the spline fit or implemented directly. The answer does not give a construction algorithm or address calibration and arbitrage constraints, so it offers a conceptual condition rather than a complete curve-building method.
Key ideas
- Different interpolation schemes can meet continuously at a shared zero-coupon curve pillar.
- Continuity of the interpolated curve is distinct from smoothness of its instantaneous forward rate.
- Smooth joins require matching appropriate derivatives of the neighboring interpolation pieces.
- Spline endpoint derivative conditions may need explicit specification or custom implementation.
- The note does not provide a full curve-fitting procedure or discuss arbitrage constraints.
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# Double interpolation continuity # Double interpolation continuity I am trying to use double interpolation(linear and forward quartic) but the interpolator is the zero coupon so at each time i need to convert the instantaneous forward rate to zc my problem is at the junction suppose i would like to ensure continuity in the instantaneous forward. Let s formulate the problem and how i think it could be solved and if someone can help i will appreciate. Suppose i have (t1,…,tn) with (zc1,…,zcn) and i want to have linear interpolation at the first pillars (t1,…,ti) and (ti,…,tn) forward quartic in such a way i ensure that fwd(ti+,linear) =fwd(ti-,quartic). I think a solution is to introduce a pillar before ti and use it but i don’t know how to do it properly. How can we solve this problem please. Thanks ## Answer by Rylan (score 1) https://quant.stackexchange.com/a/79076 If the problem is continuity, most types of interpolation (incl. linear and quartic splines) ensure that the interpolation is a continuous function -- switching types of interpolation shouldn't cause a discontinuity as long as the linear interpolation goes between $t_0$ and $t_i$ and the quartic interpolation goes between $t_i$ and $t_n$ -- both linear and quartic should result in continuity at $t_i$ (as well as all other points). If you're concerned about smoothness, the way splines remain visually smooth is by ensuring that the derivatives match at the pillars, and in general there needs to be some decision made about what the derivatives must equal at the end points (in your case, $t_i$ and $t_n$. So depending on how exactly you plan to fit this spline, there might be a place where you can input your choices for the derivatives at those points, or you might have to code it yourself. In either case, if I've understood you correctly, matching the derivatives at $t_i$ between the linear section and the spline section should do what you're looking for.
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