How Yield Curve Shock Methods Affect Interpolation and Hedging
Summary
The document compares two ways to apply a small local shock to a LIBOR 3M yield curve built with cubic splines: rebuilding the curve with the spline method, or replacing interpolation between neighboring pillars with linear interpolation. It explains that either approach can produce an unusual curve shape, while the choice also changes how hedges are distributed. Under local linear interpolation, a swap maturing between two pillars is hedged using those nearby swaps; spline-based recalculation can spread hedge exposure across more of the curve.
The answer presents the choice as a matter of modeling preference rather than identifying a universally standard method. It does not provide a quantitative comparison, prescribe controls for curve smoothness, or assess pricing impacts. Its guidance is limited to the stated interpolation and hedge-composition implications.
Key ideas
- A local yield curve shock can be applied through spline recalculation or local linear interpolation.
- Neither approach guarantees a curve shape free of undesirable effects.
- Interpolation choice affects which instruments contribute to a maturity hedge.
- Local linear interpolation tends to keep hedge exposure near adjacent curve pillars.
- Spline recalculation may distribute hedge exposure across more of the curve.
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# Introducing 1bp shocks to yield curve (and interpolation consequences) # Introducing 1bp shocks to yield curve (and interpolation consequences) Let us assume we have a LIBOR 3M curve and that I would like to introduce a small shock up/down of 1bp at a certain point along the curve. I am trying to find out what the best and most efficient way is of doing this, but so far I haven't found a standard approach to be followed. My yield curve is made of cubic splines, and so shifting a point up by 1bp might cause my curve to suffer from unrealistic/undesirable twists around the point in question, rendering the new curve useless. A different approach I have thought of could be shifting up the point by 1bp and then perform linear interpolation in that local area. Thus, if we shift the point at t(i) and I have t(i-1) < t(i) < t(i+1), then I would do linear interpolation for the points between t(i-1) and t(i+1). Has anyone had to solve this problem before? What would be the best approach without making the new curve look funny? ## Answer by dm63 (score 1) https://quant.stackexchange.com/a/32661 The new curve will look a bit strange no matter how you do it, although 1bp is not such a big move that will render it "useless". Two ways of doing it, as you point out, are (a) use a spline method, which will do whatever it will do, including moving some points that are far away from the point in question, and (b) do some linear interpolation, as you specify. The practical effect of this choice is that your hedges look very different. For example, if you do a swap with maturity t(i-0.5), under the second method the hedge will be a mixture of t(i-1) swaps and t(i) swaps, whereas under the first method the hedge could contain swaps from all over the curve in various amounts. This is a matter of preference. Some poeple like the simplicity of (b), but I like (a) better myself.
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