In the previous section, I discussed how linear interpolation is used to construct a zero-rate curve. As can be seen from Figure 2.1, using linear interpolation gives rise to kinks on instrument maturity dates – an unwelcome property for the following reasons:

■ Forward rate curves that are derived from kinky zero-rate curves are very often discontinuous at the knot points. This in turn makes some of the interest rate models unusable.

■ Sophisticated interest rate term structure models cannot be reliably calibrated using market-based instruments.

■ Valuing a coupon-bearing instrument using a linear interpolation vis- a-vis a method that does not allow for kinks (e.g., cubic interpolation) could potentially yield significantly different results.

As a consequence of the above, practitioners resort to the use of cubic polynomials to construct a zero-rate curve. More precisely, using the concept of cubic splining, practitioners use market information to construct a smoother zero-rate curve so as to be able to get more accurate prices for nonliquid, market-based instruments. In this section, I will revisit the linear interpolation example and show how the zero-rate curve can be constructed using cubic splining.

Splining over One Time interval

To apply the cubic splining method, first assume that one wants to fit a cubic polynomial in the time interval (0, 0.3333), using the information of the overnight rate and the 1-month continuously compounded 3-month forward rate.

In Step 2 of the previous section, I had assumed that the zero rate at time 0.0833 years can be linearly interpolated using the overnight rate and the guessed zero rate at time 0.3333 years as shown in equation (2.8b). In this section, I will assume that one is interested in fitting the cubic function in the time interval (t0,t0.3333) where t0 = 0, t0.3333 = 0.3333. As in the linear interpolation, one has to first guess a zero rate for io.3333 and then apply the cubically splined zero curve to ensure the reproduction of the continuously compounded forward rate.

Letting the appropriate zero rates at times t0,t0.3333 be given by r0,0^0,0.3333 respectively (where r0,0 = 0.001 and r0,0.3333 is the guessed rate) and making the substitution into the cubic function, one gets

(2.9a)

(2.9b)

One can now solve equations (2.9a) and (2.9b) with the aid of t0.3333, r0,0, r0,0.3333 (which are either known or can be guessed) and easily compute αΰ. Since equation (2.9b) now reduces to an equation in three unknowns, b0,c0, and d0, to compute the value of these unknowns one would need another two equations.

To get the remaining two equations, one can assume that the zero rate curve is instantaneously linear at times 0 and 0.3333 which would imply that the second derivative at times 0 and 0.3333 are both 0. Since the second derivative of is , using the appropriate substitutions at times 0 and 0.3333 one gets

(2.9c)

(2.9d)

From equations (2.9a), (2.9b), (2.9c), and (2.9d), one can easily arrive at the results

Putting this together, one gets the function

(2.10)

for

Using equation (2.10), one can easily interpolate the value of the zero rate at time 0.0833 years and arrive at the value 0.001 +(0.0833). Since the market price of the 1-month Eurodollar futures contract implied a continuously compounded forward rate of 0.0015, one would need equation (2.8a) to hold. Making the substitution for r0,0.0833 in equation (2.8a), one can arrive at

(2.11)

Solving equation (2.11) for, one gets . Using equation (2.10), one can also show that – which matches the numbers in Table 2.8.

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