Backward evaluation through a tree or lattice is a powerful method for evaluating financial instruments. There are times when a dual method—a forward recursion— is even better. This forward method is particularly useful for determining the term structure based on a short rate lattice

In Section 14,4 we saw that a short rate lattice completely determines the term structure, This term structure can be computed by finding the prices of zero-coupon bonds for each maturity using the backward evaluation method, However, separate recursions and separate price lattices are required for each of these maturities Hence if there are n periods, n separate recursions must be made in order to compute the entire term structure For large values of n the number of single-node evaluations is approximately >z3/6, as compared to ir/2 for one pass through the entire tree.4 The forward process described next requires only a single recursion

"lA recursion al period j — 1 requires j single evaluations Hence to evaluate a bond of maturity k requires 1 4- 2 + + k — (k + 1 )k/2 separate evaluations Since this must be done for all n maturities the total is ^ + = {"(« + f)2/6|n + 1/0» + 1)1 For one pass through the entire tree the number of'

The forward recursion is based on calculating elementary prices. The elementary price Po(k, .v) is the price at lime zero of a security that pays one unit at time k and state s, and pays nothing at any other time or state. The prices Po(k, s) are termed elementary prices because they are the prices of elementary securities that have payoff at only one node We could find PQ(k, v) for any fixed k and s by assigning a 1 at the node (A\ s) in the lattice and then working backward to time zero. Alternatively, we can work forward.

Suppose that elementary prices have been found for all nodes in the lattice for times from 0 through k. Consider a node of the form (£ +1, .s), where $ ^ 0, v ^ k+1; that is, s is not the bottom or the top node of the lattice at time k + 1. This situation is illustrated in Figure 14.9 Such a node has two predecessor nodes (nodes leading to it), namely, (k, s — 1) and (k,s). Suppose that a security pays one unit at node (k 4 1, 0 and nothing elsewhere. If we were to work backward in the lattice, this security would have values 5dk.x-.\ and 5<:4,|V at the respective predecessor nodes, where r4.y-i anc* (k s are the one-period discount factors (determined from the short rates at those nodes).

At time zero the values at these two predecessor nodes are worth, by definition of the elementary prices, 5dk.x-\Pa(k, s — 1) and ,5dk.sP[)(k, v), respectively. The total value at time zero is the sum of these two, and this is the elementary price at (¿4 1, ,v), Thus Pi)(k 4 1, -s) = 5dk.x~\Po(k, s ~~ 1) 4 .5dkAPa(k, This is a forward recursion because the value at time k 4- I is expressed in terms of values at time k If v = 0 or k 4-1, there is only one predecessor node, and the result is modified accordingly. Overall we obtain the three forms of the for ward equation, depending on whether the node is in the middle, at the bottom, or at the top of the lattice,

Po(k 4- 1, v) = i|[dk Pu(k, s - 1 )+dkiSP0{k,s)l 0 < s- < k + I (14 2a)

Although we derived this equation through intuitive reasoning, it is possible to derive it algebraically from the backward equation. This forward equation is just a different way of organizing the fundamental risk-neutral pricing equations.

The price of any interest rate security can be found easily once the elementary prices are known We simply multiply the payoff at any node (k,s) by the price P{)(k, s) and sum the results over all nodes that have payoffs. For example, the price at time zero of a zero-coupon bond with value 1 that matures at time n is

P{)(k 4 1,0) = \dk0P0(k,0), P0(k+\,k+\) = ykkP0(k,k), s = 0

FIGURE 14.9 Construction of forward equation. The elemen-(A- + 1, s) tary price for node (k -f- 1, s) can be expressed as a combination of the elementary prices for the two predecessor nodes

2599 | |||||

Short rate |
.1999 |
1799 | |||

15,38 |
1384 |
1246 | |||

I i 8 .3 |
1065 |
0958 |
.0862 | ||

0910 |
0819 |
0737 |
.0663 |
.0597 | |

0700 06.30 |
0567 |
0510 |
0459 |
0413 | |

0069 | |||||

Elementary prices |
0173 |
.0468 | |||

0415 |
0943 |
1302 | |||

.0958 |
1754 |
2028 |
1894 | ||

2142 |
.2963 |
2757 |
2155 |
1527 | |

4673 |
4.340 |
3046 |
191.3 |
1134 |
0648 |

l 0000 467.3 |
.2198 |
1040 |
0495 |
0237 |
0114 |

Bond prices 9346 |
8679 |
8006 |
7334 |
6670 |
602! |

Spot rates 0700 0734 0769 .0806 0844 0882

FIGURE 14,10 Use of elementary prices to find term structure.. The elementary prices are determined by a single forward sweep through the lattice The sum of any column then gives the price of a zero-coupon bond of that maturity. Note that a short rate applies over the coming year while a spot rate applies to the previous years Hence the initial short rate and the initial spot rate, although equal, are one column apart

The forward equation can be used to find the entire term structure corresponding to a short rate tree by a single forward recursion—because all zero-coupon bond prices can be determined.

Example 14.6 {The simple lattice) Let us apply the forward equation to Example 14.1, The elementary price lattice can be calculated directly from the short rate lattice It is shown in Figure 14.10 together with the resulting zero-coupon bond prices and the derived term structure

As an example of the calculation, both terms in the second column are derived from the single predecessor node; and these terms are equal to one-half times the discount rate at the first period times the elementary price at 0, which is 1 Hence these values are .5/1 07 = 4673. The figures directly below the lattice are the sums of the elements above them These values correspond to prices of zero-coupon bonds The final figures below the lattice make up the term structure, expressed as spot rates computed directly from the bond prices above. The values agree with those computed in Example 14.1 by the more laborious process

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