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Splining over Four Time Intervals

[1]

To draw a comparison with Table 2.10, one has to go through the splining process over the time intervals (0, 0.3333), (0.3333, 0.4167), (0.41677, 0.7500) and (0.7500, 1.2500). As before, one has to first assume that one is interested in fitting a function in the time interval (), in the time interval (), in the time interval (), and in the time interval () where, , ,, and . Again, one has to simultaneously guess at the zero rates for times, , , and and then apply the cubically splined zero curve to ensure the reproduction of the market rates for the overnight, 1-month, 2-month, 3-month, 6-month, and 12-month instruments.

Letting the appropriate zero rates at times , , , be , , , respectively, one gets

(2.14a)

(2.14b) (2.14c)

(2.14d)

(2.14e)

(2.14 f)

(2.14g)

(2.14h)

where , , , and are guessed zero rates used to reproduce the given market rates by using interpolated cubic spline values.

Since equations (2.14a) to (2.14h) have 16 unknowns and only 8 equations, one would need another 8 independent equations with these unknowns so as to arrive at unique solutions for the cubic polynomial coefficients. Using reasoning similar to that used to fit cubic splines over two time intervals, one can arrive at the following equations

(2.141)

(2.14b)

(2.14k)

(2.141)

interested to note that in splining over the actual 5 intervals, one would arrive at a zero curve that is not that different from that obtained when splining over 4 time intervals. The reader is left to see this as an exercise.

(2.14m)

(2.14n)

(2.14ο)

(2.14p)

To solve these equations, using matrix notation, one can rewrite them as MX = Y, where

and, M is the 16 X 16 matrix which is given in Figure 2.3.

Since the system of equations can be easily solved for(for i= 0,1,2,3) onceare known, one now needs to determine the values of these rates. Using the fact that zero rates in the intervals (0, 0.3333), (0.3333, 0.4167), (0.4167, 0.75), and (0.75,1.25) can be interpolated using the cubic splines

one can easily arrive at equations (2.14q) to (2.14u).

(2.14q)

(2.14r)

(2.14s)

(2.14t)

(2.14u)

Using the relationship between spot and forward rates, it can be seen that

(2.15a)

(2.15b)

The Form of the 16 × 16 Matrix

FIGURE 2.3 The Form of the 16 × 16 Matrix

(2.15c)

(2.15d)

(2.15e)

where

and the corresponding zero rates at these times are given by, , , and respectively.

Solving equations (2.14a) to (2.14u) subject to equations (2.15a) to (2.15e) yields the cubic polynomial coefficients as given in Table 2.15.

TABLE 2.15 Splining Coefficients When Fitting over Four Time Intervals

TABLE 2.16 Zero Rates Obtained Using All Eurodollar Futures Contract and Splining

Using Table 2.15, one easily obtains the zero rates as presented in Table 2.16.

  • [1] It is more accurate to spline over the five time intervals (0,0.3333), (0.3333,0.4167), (0.4167,0.5000), (0.5000,0.7500), and, (0.7500,1.2500). Since “0.5000” refers to both the settlement time for the 3-month Eurodollar futures contract and the maturity time of the 6-month Eurodollar futures contract, for the purposes of keeping my example simple and easy to follow, I collapsed the time intervals (0.4167,0.5000) and (0.5000,0.7500) to the interval (0.4167,0.7500). As a consequence, one only needs to do the splining over the time intervals (0,0.3333), (0.3333,0.4167), (0.4167,0.7500), and, (0.7500,1.2500). The reader would be
 
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