coupon-bearing bond. It can therefore be valued analytically using the Ho—Lee or Hull-White model.
23.11_ INTEREST RATE TREES
 
An interest rate tree is a discrete-time representation of the stochastic process for the short rate inseh the csean wae a gtack rice tenn ten dicneste tiene ree eoonntoatiqn af the eennece fallawed hymuch the samc way as a stock price tree is a disercte-time representation of the process followed by
10 This emphasizes the fact that the initial zero curve does not have to be differentiable to use the Ho—Lee andHull-White models.Interest Rate Derivatives: Models of the Short Rate 551
a stock price. If the time step on the tree is dt, the rates on the tree are the continuouslycompounded 5t-period rates. The usual assumption when a tree is constructed is that the dt-periodrate, R, follows the same stochastic process as the instantaneous rate, r, in the correspondingcontinuous-time model. The main difference between interest rate trees and stock price trees is inthe way that discounting is done. In a stock price tree, the discount rate is usually assumed to be
the same at each node. In interest rate trec the discount rate from node to nodeing Same at eacn node. if an interest rate trec, tne aiscount rate varies irom node to node,
It often proves to be convenient to use a trinomial rather than a binomial tree for interest rates.The main advantage of a trinomial tree is that it provides an extra degree of freedom, making iteasier for the tree to represent features of the interest rate process such as mean reversion. Asmentioned in Section 18.8, using a trinomial tree is equivalent to using the explicit finite differencemethod.
Itustration of Use of Trinomial Trees
To illustrate how trinomial interest rate trees are used to value derivatives, we consider the simpleexample shown in Figure 23.6. This is a two-step tree with each time step equal to one year inlength, so that 6t = 1 year. We assume that the up, middle, and down probabilities are 0.25, 0.50,and 0.25, respectively, at each node. The assumed St-period rate is shown as the upper number ateach node.
The tree is used to value a derivative that provides a payoff at the end of the second time step of
max{100(R — 0.11), 0]
where R is the St-period rate. The calculated value of this derivative is the lower number at each
 
E 14%312% B Big1LU40% A 10% S 10%0.35 023 08% 5 H gaa 0| 6%0
Figure 23.6 Example of the use of trinomial interest rate trees:upper number at each node is rate; lower number is value of instrument
'\ We explain later how the probabilities and rates on an interest rate tree are determined.552 CHAPTER 23
(a) (b) (e)Figure 23.7 Alternative branching methods in a trinomial tree
node. At the final nodes, the value of the derivative equals the payoff. For example, at node E, thevalue is 100 x (0.14 — 0.11) = 3. At earlier nodes, the value of the derivative is calculated using therollback procedure explained in Chapters 10 and 18. At node B, the one-year interest rate is 12%.This is used for discounting to obtain the value of the derivative at node B from its values atnodes E, F, and G as
(0.25 x 340.5 x 140.25 x Ne?! = 1.11
At node C, the one-year interest rate is 10%. This is used for discounting to obtain the value of thederivative at node C as
(0.25 x 140.5 x 040.25 x Oje!*! = 0.23At the initial node, A, the interest rate is also 10% and the value of the derivative is
(0.25 x 1.11 +0.5 x 0.23 +.0.25 x O)e°'™! = 0,35
Nonstandard Branching
It sometimes proves convenient to modify the standard branching pattern, which is used at allnodes in Figure 23.6. Three alternative branching possibilities are shown in Figure 23.7. The usualbranching is shown in Figure 23.7a. It is “up one/ straight along /down one”. One alternative tothis is “up two/up one/straight along”, as shown in Figure 23.7b. This proves useful forincorporating mean reversion when interest rates are very low. A third branching pattern, shownin Figure 23.7c, is “straight along / one down / two down’’. This is useful for incorporating meanreversion when interest rates are very high. We illustrate the use of different branching patterns inthe following section.
23.12 A GENERAL TREE-BUILDING PROCEDURE
 
 
Hull and White have proposed a robust two-stage procedure for constructing trinomial trees toreptesent a wide range of one-factor models.!” This section first explains how the procedure can be
® See J. Hull and A. White, “Numerical Procedures for Implementing Term Structure Models I: Single-FactorModels,’ Journal of Derivatives, 2, no. 1 (1994), 7-16; J. Hull and A. White, “Using Hull!-White Interest Rate Trees,”Journal af Derivatives, Spring 1996, 26-36.Interest Rate Derivatives: Models of the Short Rate 553
used for the Hull-White model in Section 23.9 and then shows how it can be extended to representother models.
First Stage
The Hull-White model for the instantaneous short rate r isdr = [@(t) — ar]dt + adz
For the purposes of our initial discussion, we suppose that the time step on the iree is cousiant audequal to ér.We assume that the dt rate, R, follows the same process as r.
dR = [0(t) ~ aR] dt + odz
Clearly, this is reasonable in the limit as dr tends to zero. The first stage in building a tree for thismodel is to construct a tree for a variable R* that is initially zero and follows the process
dR* = —aR* dt+odz
This process is symmetrical about R* = 0. The variable R*(¢ + dt) — R*(t) is normally distributed.If terms of order higher than dt are ignored, the expected value of R*(t + dt) ~ R*(t) is —aR*(t) dtand the variance of R*(t + 6t) — R*(t) is a” ét.
We define 5R as the spacing between interest rates on the tree and set
6R = av3 dt
This proves to be a good choice of 6R from the viewpoint of error minimization.
Our objective during the first stage is to build a tree similar to that shown in Figure 23.8 for R*.To do this, we must resolve which of the three branching methods shown in Figure 23.7 will applyat each node. This will determine the overall geometry of the tree. Once this is done, the branchingprobabilities must also be calculated.
Define (i, j) as the node where t = i St and R* = j5R. (The variable i is a positive integer and jisapositive or negative integer.) The branching method used at a node must lead to the probabilities onall three branches being positive. Most of the time, the branching in Figure 23.7a is appropriate.When a > 0, it is necessary to switch from the branching in Figure 23.7a to the branching inFigure 23.7c for a sufficiently large j. Similarly, it is necessary to switch from the branching inFigure 23.7a to the branching in Figure 23.7b when / is sufficiently negative. Define j,,., aS the valueof j where we switch from the Figure 23.7a branching to the Figure 23.7c branching and ji, as thevalue of j where we switch from the Figure 23.7a branching to the Figure 23.7b branching. Hull andWhite show that probabilities are always positive if we set jmax equal to the smallest integer greaterthan 0.184/(a 2) atid jin cqual to —jmax.!? Define p,, fm, and py as the probabilitics of the highest,middle, and lowest branches emanating from the node. The probabilities are chosen to match theexpected change and variance of the change in R* over the next time interval 5t. The probabilitiesmust also sum to unity. This leads to three equations in the three probabilities.
As already mentioned, the mean change in R* in time St is —aR* 6t and the variance of the
8 The probabilities are positive for any value of jmax between 0,184/(a 5t) and 0.816/(a6t) and for any value of joinbetween —0.184/(a6t) and —0.816/(ad5t). Changing the branching at the first possible node proves to becomputationally most efficient.554 CHAPTER 23
 
 
OK
 
Node: A B Cc D E F G H IR 0.000% 1.732% 0.000% —1.732% 3.464% 1.732% 0.000% —-1.732%  —3.464%
Pu 0.1667 0.1217 0.1667 0.2217 0.8867 0.1217 0.1667 0.2217 0.0867Pn 0.6666 0.6566 0.6666 0.6566 0.0266 0.6566 0.6666 0.6566 0.0266Pa 0.1667 0.2217 (0.1667 0.1217 0.0867 0.2217 0.1667 0.1217 0.8867
Figure 23.8 Tree for R* in Hull-White model (first stage)
change is o7 6t. At node (i, j), R* = j5R. If the branching has the form shown in Figure 23.7a,the p,, Pm» and py at node (i, j) must satisfy the following three equations:p, dR — pg dR — —aj 6R 5tPy OR? + pg SR? = 0 dt +a" 7? bR St"Put Pm + Pa = 1
Using 5R = 0/3 dt, we find that the solution to these equations is
_,  & 78? —aj itPy = gtPm =i Por
a7 bt? + aj st+ 2
ar
Pa=Similarly, if the branching has the form shown in Figure 23.7b, the probabilities are242 52 .aj &t° + ajétPu = pty
Pm = —b—@ 7? bt? — 2a jt
a’? bt" + 3aj ot
a=it 2
ac]interest Rate Derivatives: Models of the Short Rate 555
Finally, if the branching has the form shown in Figure 23.7c, the probabilities are
2:2 9,2 +a’ j° dt — 3aj otPu pee
Pm = ha? 7? 60 + 2a jota’j? 6 — ajét
Pa = 3+ 2
To illustrate the first stage of the tree construction, suppose that o = 0.01, a = 0.1, and 6t = I year.In this case, 8R = 0.013 = 0.0173, Jmax is set equal to the smallest integer greater than 0.184/0.1,and jmin = —Jmax- This means that jmax = 2 and jmin = —2 and the tree is as shown in Figure 23.8.The probabilities on the branches emanating from each node are shown below the tree and arecalculated using the equations above for p,, Pm, and pg.
Note that the probabilities at each node in Figure 23.8 depend only on j. For example, theprobabilitics at node B are the same as the probabilities at node F. Furthermore, the tree issymmetrical. The probabilities at node D are the mirror image of the probabilities at node B.
Second Stage
The second stage in the tree construction is to convert the tree for R* into a tree for R. This isaccomplished by displacing the nodes on the R*-tree so that the initial term structure of interestrates is exactly matched. Definea(t) = R(t) — R*(t)BecausedR = [0(t) — aR] dt + adz
and
dR* = —aR* dt +adzit follows that
da = [A(t) — aa(t)| dt
If we ignore the distinction between r and R, equation (23.18) shows that the solution to this is
2a(t) = F(0, t) + sol ety (23.27)
As a tends to zero, this becomes a(t) = F(0, t) + 0717/2.
Equation (23.27) can be used to create a tree for R from the corresponding tree for R*. Theapproach is to set the interest rates on the R-tree at time i6t to be equal to the correspondinginterest rates on the R*-tree plus the value of at time / 4t and to keep the probabilities the same
The tree for R produced using equation (23.27), although satisfactory for most purposes, is notexactly consistent with the initial term structure. An alternative procedure is to calculate the a’siteratively so that the initial term structure is matched exactly. We now explain this approach. Itprovides a tree-building procedure that can be extended to models where there are no analyticresults. Furthermore it is applicable to situations where the initial zero curve is not differentiableeverywhere.
Define a; as a(i St), the value of R at time i St on the R-tree minus the corresponding value of R*at time 7 6¢ on the r*-tree. Define Q, ; as the present value of a security that pays off $1 if node (i, j)556
Table 23.1 Zero rates for example in Figure 23.8
 
Maturity Rate (%)0.5 3.4301.0 3.8241.5 4.1832.0 4.5122.5 4.8123.0 5.086
CHAPTER 23
is reached and zero otherwise. The a; and Q; ; can be calculated using forward induction in such away that the initial term structure is matched exactly.
Illustration of Second Stage
Suppose that the continuously compounded zero rates in the example in Figure 23.8 are as shownin Table 23.1. The value of Qo, is 1.0. The value of ap is chosen to give the right price for a zero-coupon bond maturing at time 6f. That is, a is set equal to the initial St-period interest rate.Because dt = | in this example, ay = 0.03824. This defines the position of the initial node on theR-tree in Figure 23.9. The next step is to calculate the values of Q11, Qo, and Q;_,. There is aprobability of 0.1667 that the (1, 1) node is reached and the discount rate for the first time step
Node:
R
P uPmPa
A
3.824%0.16670.66660.1667
B
6.937%0.12170.65660.2217
 
 
 
Cc D E F G H5.205% 3.473% 9.716% 7.984% 6.252% 4.520%0.1667 0.2217 0.8867) 0.1217 0.1667, 0.22170.6666 0.6566 0.0266 0.6566 0.6666 0.65660.1667 = 0.1217 0.0867) 0.2217 (0.1667 0.1217
Figure 23.9 Tree for R in Huli-White model (the second stage)
I
2.788%0.08670.02660.8867Interest Rate Derivatives: Models of the Short Rate 557
is 3.82%. The value of Q,, is therefore 0.1667¢ °°** = 0.1604. Similarly, Q,9 = 0.6417 andQ;-) = 0.1604.
Once Q,,;, Gio, and Q,,-; have been calculated, we are in a position to determine a. This ischosen to give the right price for a zero-coupon bond maturing at time 2 61. Because 6R = 0.01732and 6t = 1, the price of this bond as seen at node B is e170"), Similarly, the price as seen atnode C is e~*! and the price as seen at node D is e@1-""!9)_ The price as seen at the initial nodeA is therefore
Qype #4001732) Oye + Q, e717 0.01732) (23.28)
From the initial term structure, this bond price should be 7 0045122 _ 99137, Substituting for theQ’s in equation (23.28), we obtain
0.160427) 4 9.641767"! 4.0.1604e701 9 — 9.9137or
 
€!(0.1604e7°"'7? +. 0.6417 + 0.160429!) = 0.9137or
 
Cm“ 201732 +g 6417 +0. 1604e°9a= In
ona ) = 0.05205
This means that the central node at time df in the tree for R corresponds to an interest rate of5.205% (see Figure 23.9).
The next step is to calculate Q >, O21, Q29, 2,1, and Q» _>. The calculations can be shortenedby using previously determined Q values. Consider Q>, as an example. This is the value of asecurity that pays off $1 if node F is reached and zero otherwise. Node F can be reached only fromnodes B and C. The interest rates at these nodes are 6.937% and 5.205%, respectively. Theprobabilitics associated with the B-F and C-F branches are 0.6566 and 0.1667. The valuc at node Bof a security that pays $1 at node F is therefore 0.6566e-°°", The value at node C is0.1667e°°°5. The variable Q>, is 0.6566e~°37 times the present value of $1 received atnode B plus 0.1667e~°°5 times the present value of $1 received at node C, that is,
Or, = 0.6566e °°? x 0.1604 + 0.1667¢ °°*5 x 0.6417 = 0.1998
Similarly, Qo, = 0.0182, Q25 = 0.4736, Q2_, = 0.2033, and Q)_) = 0.0189.The next step in producing the R-tree in Figure 23.9 is to calculate a. After that, the Q; ;’s canthen be computed. We can then calculate a3, and so on.
Formulas for a's and Q’s
To express the approach more formally, we suppose that the Q, ;’s have been determined for i < m{(m 20). The next step is to determine «,, so that the tree correctly prices a zero-coupon bondmaturing at (m+ 1)édt. The interest rate at node (m, j) is a» + j6R, so that the price of a zero-coupon bond maturing at time (m + 1) dt is given by
Poot = S> Op jexPL—(0m + 75R) di] (23.29)
Jahn558 CHAPTER 23
where 2,, is the number of nodes on each side of the central node at time m 6t. The solution to thisequation is—j5R5In ig Om je — I PragOm = St
Once @,, has been determined, the Q; ; for i = m+ 1 can be calculated using
Omni = Y> Omegks J) expl—(m + KAR) St]k
where g(k, j) is the probability of moving from node (m, k) to node (m + 1, j) and the summation istaken over all values of k for which this is nonzero.
Cyt nine £5Extension to
The procedure that has just been outlined can be extended to more general models of the formdf(r) = (6) —af(n}dt+odz (23.30)
This family of models has the property that they can fit any term structure.'*As before, we assume that the dt-period rate, R, follows the same process as r:
af(R) = [0(t) — af(R)] dt + 0 dzWe start by setting x = f(R), so thatdx = [O(t) — ax]dt+odz
The first stage is to build a tree for a variable x* that follows the same process as x except that0(t) = 0 and the initial value of x is zero. The procedure here is identical to the procedure already_ outlined for building a tree such as that in Figure 23.8.
As in Figure 23.9, we then displace the nodes at time 7 dt by an amount q; to provide an exact fitto the initial term structure. The equations for determining a, and Q, , inductively are slightlydifferent from those for the f(R) = R case. Qo = 1. Suppose that the Q, ;’s have been determinedfor i <m (m 3 0). The next step is to determine @,, so that the tree correctly prices an (m + 1) dtzero-coupon bond. Define g as the inverse function of f so that the St-period interest rate at the jthnode at time m St is
G(Xn, + J 8x)
The price of a zero-coupon bond maturing at time (m + 1) 6t is given by
Prat = Yo Om jexpl—Gloim + 75x) 54 (23.31)
J=—lMn
'4 Not all no-arbitrage models have this property. For example, the extended-CIR model, considered by Cox,Ingersoll, and Ross (1985) and Hull and White (1990), which has the form
dr = [0(t) — arldt + ofrdz
cannot fit yield curves where the forward rate declines sharply. This is because the process is not well defined whenQt) is negative.interest Rate Derivatives: Models of the Short Rate 559
 
 
 
Node: A B Cc D E F G H I
x 3.373 2.875 3.181 3.487 2.430 2.736 -—3.042 -3.349 —-3.655R 3.430% 5.642% 4.154% 3.058% 8.803% 6.481% 4.772% 3.513% 2.587%Pu 0.1667 0.1177, 0.1667) 0.2277, «0.8609 0.1177 0.1667 0.2277 (0.0809Pm 0.6666 0.6546 0.6666 = 0.6546 = 0.0582 0.6546 = (0.6666 = 0.6546 0.0582Pa 0.1667 0.2277 0.1667 0.1177 0.0809 0.2277 0.1667 0.1177 0.8609
Figure 23.10 Tree for lognormal model
This equation can be solved using a numerical procedure such as Newton—Raphson. The value aof a when m = 0 is f(R(0)).Once @,, has been determined, the Q; ; for i = m+ 1 can be calculated using
Omi.) = Y_ OmaMk, i) expl—G(en + kx) 81]k
where q(k, j) is the probability of moving from node (m, k) to node (m + 1, j) and the summation istaken over all values of k where this is nonzero.Figure 23.10 shows the results of applying the procedure to the model
d1n(r) = [0(0) — aln(n)] dt + odz
when a = 0.22, o = 0.25, 5t = 0.5, and the zero rates are as in Table 23.1.
Choosing f(r)
The main alternatives when choosing a model of the short rate are f(r) =r and f(r) =In(r). Inmost circumstances these (wo models appear to perform about the same in fitting market data onactively traded instruments such as caps and European swap options. The main advantage of thef(r) = r model is its analytic tractability. Its main disadvantage is that negative interest rates arepossible. In most circumstances, the probability of negative interest rates occurring under the560 CHAPTER 23
model is very small, but some analysts are reluctant to use a model where there is any chance at allof negative interest rales. The f(r) = lar model has no analytic tractability, but has the advantagethat interest rates are always positive. Another advantage is that traders naturally think in termsof o’s arising from a lognormal model rather than o’s arising from a normal model.
There is a problem in choosing a satisfactory model for low-interest-rate countries (At the timeof writing, Japan is one such country.) The normal model is unsatisfactory because, when theinitial short rate is low, the probability of negative interest rates in the future is no longer negligible.The lognormal model is unsatisfactory because the volatility of rates (i.e., the o parameter in thelognormal model) is usually much greater when rates are low than when they are high. (Forexample, a volatility of 100% might be appropriate when the short rate is less than 1%, while 20%might be appropriate when it is 4% or more.) A model that appears to work well is one where f(r)is chosen so that rates are lognormal for r less than 1% and normal for r greater than 1%.°
Using Analytic Results in Conjunction with Trees
When a tree is constructed for the f(r) = r version of the Hull-White model, the analytic results inSections 23.9 and 23.10 can be used to provide the complete term structure and European optionprices at cach node. It is important to recognize that the interest rate on the tree is the 5t-periodrate, R. It is not the instantaneous short rate, r. We should therefore calculate bond prices usingequation (23.22), not equation (23.19).
Example 23.3 As an example of the use of analytic results, we use the zero rates in Table 23.2.Rates for maturitics between those indicated are generated using linear interpolation.
We will price a three-year (= 3 x 365 day) European put option on a zero-coupon bond that willexpire in nine years (= 9 x 365 days). Interest rates are assumed to follow the Hull-White (f() =n)model. The strike price is 63, a = 0.1, and o = 0.01. We constructed a three-year tree and calculated
Table 23.2 Zero curve with all rates continuously compounded
 
Maturity Days Rate (%)3 days 3 5.017721 month 31 4.982842 months 62 4.972343 months 94 4.961576 months 185 4.990581 year 367 5.093892 years 731 5.797333 years 1,096 6.305954 years 1,461 6.734645 years 1,826 6.948166 years 2,194 7.088077 years 2,558 7.275278 years 2,922 7.308529 years 3,287 7.39790
10 years 3,653 7.49015
'S See J. Hull and A. White,‘Taking Rates to the Limit,” R/SK, December 1997, pp. 168-69.Interest Rate Derivatives: Models of the Short Rate 561
Table 23.3 Value of a three-year put option on a nine-yearzero-coupon bond with a strike price of 63: a= 0.1 and a = 0.01;zero curve as in Table 23.2
Steps Tree Analytic
 
10 1.8658 1.809330 1.8234 1,809350 1.8093 1.8093100 1.8144 1.8093200 1.8097 1.8093
500 1.8093 1.8093
zero-coupon bond prices at the final nodes analytically as described in Section 23.9. As shown inTable 23.3, the results from the tree are consistent with the analytic price of the option.
This example provides a good test of the implementation of the model because the gradient of thezero curve changes sharply immediately after the expiration of the option. Small errors in theconstruction and use of the tree are liable to have a big effect on the option values obtained. (Theexample is used in Sample Application G of the DerivaGem Application Builder software.)
Tree for American Bond Options
The DerivaGem software accompanying this book implements the normal and the lognormalmodel, as well as Black’s model, for valuing European bond options, caps/floors, and Europeanswap options. In addition, American-style bond options can be handled. Figure 23.11 shows thetree produced by the software when it is used to value a 1.5-year American call option on a 10-yearbond using four time steps and the lognormal model. The parameters used in the lognormal modelare a = 5% and o = 20%. The underlying bond lasts 10 years, has a principal of 100, and pays acoupon of 5% per annum semiannually. The yield curve is flat at 5% per annum. The strike priceis 105. As explained in Section 22.2, the