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Splining over All Time Intervals

To draw a comparison with Table 2.12, one has to go through the process over the time intervals (0, 0.3333), (0.3333, 0.4167), (0.4167, 0.7500), (0.7500, 1.2500), (1.2500, 2.0000), and (2.0000, 5.0000). As before,I will assume that one is interested in fitting a function (for i= 0,1,2,3,4,5) in the time intervals (), (), (), (), (), and () respectively, where,,, 1.25,, and. As before, one has to simultaneously guess at the zero rates for,,,,, andand then apply the cubically splined zero curve to ensure that one is able to reproduce the market rates for the overnight, 1-month, 2-month, 3-month, 6-month, 12-month, 24-month, and 60-month instruments. By letting the appropriate zero rates at times,,,,,,be given by ,,,,,respectively, one gets

(2.16a)

(2.16b)

(2.16c)

(2.16d)

(2.16e)

(2.16f)

(2.16g)

(2.16h)

(2.161)

(2.16j) (2.16k)

(2.161) (2.16m) (2.16n)

(2.16o)

(2.16p)

(2.16q)

(2.16r)

(2.16s)

(2.16t)

(2.16u)

(2.16v)

(2.16w)

(2.16x)

Where ,, , , , andare guessed zero rates.

As before, to solve these equations, one has to use the matrix notation to rewrite equations (2.16a) to (2.16x) as MX = Y, where

and, M is given by the (24 x 24) matrix as shown in Figure 2.4. Since the system of equations can be solved for(for i = 0,1,2,3, 4, 5) once the guessed zero rates are known, it remains for one to determine these guessed zero rates. Using the fact that zero rates in the

The Form of the 24 × 24 Matrix

FIGURE 2.4 The Form of the 24 × 24 Matrix

intervals (0, 0.3333), (0.3333, 0.4167), (0.4167, 0.75), (0.75, 1.25), (1.25, 2), and (2, 5) can be interpolated using the cubic splines

one can arrive at the following

(for k = 0.0833, 0.1667, 0.25, 0.5, 1, 1.5, 2.5, 3, 3.5, 4, 4.5)

Using the spot and forward rate relationship in equation (2.7), it can be seen that

(2.17a)

(2.17b)

(2.17c)

(2.17d)

(2.17e)

Using the swap pricing formula given in equation (2.6), one can also get

(2.17f)

(2.17g)

where the discount factor associated with time ί(• is given by the expression

TABLE 2.17 Splining Coefficients for the Entire Fit

Solving equations (2.16a) to (2.16x) – subject to equations (2.17a) to (2.17g) – yields the numbers in Table 2.17.

With the values in Table 2.17, one can then obtain the zero rates shown in Table 2.18.

The information in Table 2.12 and Table 2.18 can be alternatively represented as in Figure 2.5.

As can be seen from Figure 2.5, the zero curve obtained using a cubic-splining approach has a much smoother fit (by the nature of the construction) – in the process helping one avoid kinks in a zero rate term structure. As mentioned, when calculating the present value of instruments

TABLE 2.18 Splined Zero Rate Term Structure

Comparison of Zero Curves Obtained Using Linear Interpolation and Cubic Splining Methods

FIGURE 2.5 Comparison of Zero Curves Obtained Using Linear Interpolation and Cubic Splining Methods

with cash flows lying in the two- to five-year bucket, one can expect a big discrepancy arising from the use of the two methods. Valuing a four-year at-market swap using the linearly interpolated zero curve yields an annual swap rate of 2.67 percent. This same structure, when valued using a zero curve obtained using a cubic splining, yields a value of 3.32 percent, creating a difference of 64 annual basis points. Put another way, this causes a difference of about $24,340 when the notional size of the swap is $1,000,000 (a 2.4 percent difference). Clearly, the higher the notional amount of the swap, the higher the present value of this discrepancy.

APPENDIX: FINDING SWAP RATES USING A FLOATING COUPON 00ND APPROACH

To do this, I will assume that one is interested in valuing an n-period fixed- floating swap in which the floating rates are set at times tl, t2, ..., tn and settled at times t2, t3, ..., tn+1. This interest swap can be represented as shown in Figure 2.6a where N is the notional principal of the swap (based on which interest cashflows are calculated), R is the swap rate that one is trying to determine, and L(- (for i = 1, 2, ..., n) is the ith Libor (floating rate) that is yet to be set in the future and hence denoted by “?” as the dotted cashflows.

Cash Flows in a Fixed-Floating Swap

FIGURE 2.6A Cash Flows in a Fixed-Floating Swap

Cash Flows in a Fixed-Floating Swap with Principals

FIGURE 2.6B Cash Flows in a Fixed-Floating Swap with Principals

Adding the notional principals to both the commencement and end of swap, one can then arrive at Figure 2.6b.

The cash flows in Figure 2.6b can now be decomposed into Figures 2.6c and 2.6d.

The value of the floating rate bond in Figure 2.6d is 0 (simply because any bond whose floating rate coupons are ALL based on the then-prevailing market rates simply has no value to the holder of the bond). The consequence of this is the simplification of Figure 2.6a to Figure 2.6c. Given this backdrop, it just remains to observe that for Figure 2.6c to hold true all the present values of the cash flows must net out to 0. More precisely one would need that

(2.18)

Cash Flows of a Fixed Rate Bond

FIGURE 2.6C Cash Flows of a Fixed Rate Bond

Cash Flows of a Floating Rate Bond

FIGURE 2.6D Cash Flows of a Floating Rate Bond

where N represents the notional principal associated with the swap, SR represents the swap rate that one is trying to determine, and both and (for i = 1, 2, ... ,n + 1) are as defined in equation (2.6).

It easy to see that equation (2.18) can be simplified to obtain equation (2.6).

 
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