Modelling bonds and interest rates
6.4 Binary tree term structure models
The observed unpredictability of interest rates leads to mathematical mod-els where interest rates and bond prices are stochastic processes. We begin with binomial branching, which as we know has the significant advantages of simplicity and completeness. However, these features are specific to the model of stock prices and we first discuss the modifications needed for bond markets.
The main type of underlying asset studied is a zero-coupon bond. It has finite life-time and becomes deterministic at maturity. Thus we need mod-els where the randomness of prices ceases at maturity. The fact that some future prices are known in advance means that the Cox–Ross–Rubinstein (CRR) model cannot be applied, since there the range of possible values increases with time. In particular, should we apply the uniform binomial tree (i.e. with the same structure of returns at each step) to model bond prices, after some steps we would have values greater than 1 with positive probability, which is unacceptable. So, if we are going to use a binomial model, it has to be more sophisticated.
Another important feature is dependence of the bond price upon matu-rity. Bonds with different maturities are separate objects, like many stocks, but clearly their dynamics are closely correlated. In a simple binomial tree this correlation is perfect, which creates problems. In particular, without due care the model may allow arbitrage opportunities, which is to be avoided.
General binary tree framework
Consider the data available at time 0. In contrast to the CRR model, where only the stock and risk-free asset prices are known at time 0, we now have the entire spot term structure of bond prices, that is, the spot prices
B(0, 1), B(0, 2), . . . , B(0, N).
In practice, many of these unit bond prices will be computed by bootstrap-ping from coupon bond prices.
At times 0 < n < N there remain all the bonds which have not yet matured. At the final time step N only one bond, maturing at N, remains.
Future bond prices are modelled as random variables, except for the prices of maturing bonds, which are equal to 1.
To specify the bond price B(k, n) as a random variable we have to de-scribe a suitable probability space. Let
Ω = {u, d}N
so ω ∈ Ω is a sequence (ω(1), . . . , ω(N)) with ω(k) = u or ω(k) = d.
For ease of notation we will omit the brackets and commas, writing, for example,ω = ddudu rather than ω = (d, d, u, d, u).
We introduce nodesωk = ω(1) . . . ω(k) representing the history of the movements up to time k and in particular providing the exact location on the tree. Given a node ωk, the next possible node is either ωku or ωkd. We will use nodes to describe random variables in the sense that if we write Xωkwe mean X(ω) for any ω where the vector of initial k coordinates coincides withωk. In other words, the notation Xωk implicitly assumes that X is constant on all suchω’s.
Thus for a bond maturing at n we write Bωk(k, n), where k < n ≤ N;
and of course Bωk(k, k) = 1. The tree has binary branching, with ‘up’ and
‘down’ successor nodes and the bond prices will be written as Bωku(k+ 1, n) and Bωkd(k+ 1, n). We adopt the convention that ω0 = ∅ so that the above notation makes sense for k= 0, 1, . . . , N − 1. A simple example best illustrates the idea: N = 3
Buu(2, 3) B(3,3) = 1 Bu(1, 3)
Bud(2, 3) B(3,3) = 1 B(0,3)
Bdu(2, 3) B(3,3) = 1 Bd(1, 3)
Bdd(2, 3) B(3,3) = 1
Strictly speaking, we should include two branches in the last step, but these are reduced to one since the final results are the same and consequently we drop the notation for the last node. The example shows another difference between term structure models and the CRR model which is that we allow the tree not to be recombining, which means that bond prices may be path dependent.
The probabilities of the branches will be denoted by pωk, where
The change of bond prices from time k to k+ 1 is described by the proba-bilities pωk+1:
As a result the ‘up’ and ‘down’ movements at different times are indepen-dent. Note that the probabilities pωku, pωkdare in fact conditional probabil-ities of landing at the specified node provided we start from the nodeωk.
As we noted above, we should discuss the whole family of bonds with various maturities rather than a particular one. An extension of the tree for B(k, 3) presented before would include the bonds maturing at 1 and 2. The initial point is a vector of current bond prices, the whole term structure.
A binomial tree of bond prices, shown here for n = 3, therefore looks like this:
As time progresses, some bonds mature and the vector gets shorter (though some new bonds will be issued and an alternative would be to add B(1, 4) at time 1 and B(2, 4), B(2, 5) at time 2).
Note that unlike in the CRR model we discuss the prices rather than returns. This is a matter of convenience since doing this makes it easier to see that the basic requirement, namely keeping bond prices below 1, is satisfied, so that an apparent violation of the No Arbitrage Principle does not emerge (sell short a bond for B(k, n) > 1 and wait till maturity to pay just the unit).
The next step is to give a concrete example consistent with the No Arbi-trage Principle. This requires some further discussion.
Risk-neutral probabilities and arbitrage
In the CRR model it is easy to give examples since the specification reduces to finding three returns: a risk-free return R and risky stock returns with
−1 < D < U. For a single asset all that is needed to eliminate arbitrage is to make sure that D< R < U. The starting point is similar here but we have to recognise the fact that we are modelling many assets simultanously, namely bonds with various maturities.
Example 6.13
Let N = 3 and take the time step as h = 1/12 (one month). We construct the first step of a tree of unit bond prices. The time 0 prices are known, so suppose they are
B(0, 1) = 0.9966, B(0, 2) = 0.9932, B(0, 3) = 0.9897.
(Typically the bond prices will by computed from LIBOR rates for the corresponding period.)
We seek the following five numbers: p = pu (which determines pd = 1− p), Bu(1, 2), Bu(1, 3), Bd(1, 2), Bd(1, 3) – the last four as shown in the diagram:
Bu(1, 2) Bu(1, 3) B(0, 1) = 0.9966
B(0, 2) = 0.9932 B(0, 3) = 0.9897
Bd(1, 2) Bd(1, 3)
The first step is to find the risk-free return and this is straightforward. This number is provided by the price of the bond maturing in one month:
R= 1− B(0, 1)
B(0, 1) = 0.0034116.
For each of the two assets: B(k, 2), B(k, 3), treated separately, the no-arbitrage restriction on the returns must be satisfied, i.e. writing
U2 = Bu(1, 2) − B(0, 2)
B(0, 2) , U3= Bu(1, 3) − B(0, 3) B(0, 3) , D2 = Bd(1, 2) − B(0, 2)
B(0, 2) , D3= Bd(1, 3) − B(0, 3) B(0, 3) ,
(the subscript here indicates the maturity) we need D2 < R < U2, and D3< R < U3.
With many parameters at hand we could impose some conditions based on the historical data:
E(B(1, 2)) = m2,
Var(B(1, 2)) = σ2, E(B(1, 3)) = m3,
Var(B(1, 3)) = σ3,
and employing the formula for the variance of a binomial random variable we arrive at
pBu(1, 2) + (1 − p)Bd(1, 2) = m2,
p(1− p)(Bu(1, 2) − Bd(1, 2)) = σ2,
and similar equations for B(1, 3). We have a pair of two independent sys-tems of simultaneous equations for the prices of the bonds. To find a unique solution we simplify by taking p= 0.5. Suppose we have estimated the his-torical trends and variances at the following levels
m2= 0.9966, σ2 = 0.00018, m3= 0.9931, σ3 = 0.00021.
This gives the equations
0.5(Bu(1, 2) + Bd(1, 2)) = 0.9966, 0.5(Bu(1, 2) − Bd(1, 2)) = 0.00018, 0.5(Bu(1, 3) + Bd(1, 3)) = 0.9931, 0.5(Bu(1, 3) − Bd(1, 3)) = 0.00021, yielding the prices
Bu(1, 2) = 0.99678, Bd(1, 2) = 0.99642, Bu(1, 3) = 0.99331, Bd(1, 3) = 0.99289.
The solutions can be easily verified against the basic no-arbitrage condi-tions: D2= 0.00324205 < R < U2= 0.00360451, D3 = 0.0032232 < R <
U3= 0.00364757.
Obtaining a no-arbitrage condition for each bond separately is actually not sufficient if we recall our discussion in Section 2.4 where we considered two stocks in the binomial model. The situation here is similar since we also have two risky securities in a binomial model and the name (stock or bond) of the underlying does not matter. Theorem 2.13 says that the risk-neutral probabilities determined by these two assets must coincide to avoid arbitrage. Therefore, writing
q2 = R− D2
U2− D2
, q3= R− D3
U3− D3
we must have
q2= q3.
It turns out that the numbers obtained in the last example do not satisfy this condition:
q2= 0.46777934, q3= 0.44395229.
Exercise 6.15 Find an arbitrage strategy for the above example.
In this example we have assumed that p = 0.5 since we had too many variables (five variables with four conditions only). We could try to refine the example by changing the probability, but as the following exercises show, this does not work in general: only if the historical data satisfy some additional conditions will we have no arbitrage, independently of p; other-wise altering p does not help.
Exercise 6.16 Within the scheme of Example 6.13 show that if B(0, 2)(1+R) = E(B(1, 2)), B(0, 3)(1+R) = E(B(1, 3)) then q3= q2for any p. Formulate a condition linking the expectations and variances of B(1, 2) and B(1, 3) so that q2 = q3for all p. Show that if this condition does not hold, q2 q3for any p∈ (0, 1).
So a modification of the probability does not lead to elimination of ar-bitrage and to rectify the model we have to change at least one of the four prices at time 1 to make q2equal to q3, but this will cause some of the four conditions to be violated.
Exercise 6.17 Change the value of Bu(1, 2) in Example 6.13 so that q2= q3and analyse the expectations and variances after the change.
In practice the estimation of expected returns (or the expected future price, which boils down to the same) is not reliable so we take a different tack in the next attempt.
Example 6.14
We assume that p = 0.5 and we abandon the conditions concerned with the expected bond prices. On the other hand, we assume that the risk-neutral probabilities are not only equal for various maturities but
specifically they are both equal to 0.5, which is much more stringent, but quite convenient.
We take the variances from Example 6.13, so the equations are of the form
B(0, 2)(1 + R) − Bd(1, 2) Bu(1, 2) − Bd(1, 2) = 0.5, 0.5(Bu(1, 2) − Bd(1, 2)) = 0.00018, B(0, 3)(1 + R) − Bd(1, 3)
Bu(1, 3) − Bd(1, 3) = 0.5, 0.5(Bu(1, 3) − Bd(1, 3)) = 0.00021, and we immediately get
Bu(1, 2) = B(0, 2)(1 + R) + 0.00018 = 0.99676840, Bd(1, 2) = B(0, 2)(1 + R) − 0.00018 = 0.99640840, Bu(1, 3) = B(0, 3)(1 + R) + 0.00021 = 0.99328646, Bd(1, 3) = B(0, 3)(1 + R) − 0.00021 = 0.99286646.
The model is of course free of arbitrage due to the assumed equality of risk-neutral probabilities for different maturities.
Another approach to the task of building an example is to use some available market data for derivative securities.
Example 6.15
Let q2 = q3 = 0.5 which gives two conditions which can be conveniently written in the form
B(0, 2)(1 + R) = 1
2(Bu(1, 2) + Bd(1, 2)), B(0, 3)(1 + R) = 1
2(Bu(1, 3) + Bd(1, 3)),
so the construction boils down to finding Bu(1, 2) and Bu(1, 3), say.
Suppose we know the price of a call option written on 2-bond with strike price K = 0.9965 and exercise time k = 1. Consider 100 000 options and assume that the price of the package is 20. With q at hand this gives the
equation to complete the 2-bond details
Next, to make the story more interresting, suppose that we know the price of the following derivative involving both bonds. Let the payoff H(1) be the annual interest on the sum of 100 000 according to the higher rate of the two rates implied by the 2-bond and the 3-bond (payable at time 1). The annual rates on the 2-bond are (simple compounding is used)
Suppose that the price of this derivative security is 4200 and this, as we shall see, requires taking
which agrees with the price assumed.
Multi-factor models
Recall that the trinomial model with two risky assets is typically com-plete with a unique candidate for the risk-neutral probability (common for both assets). Despite its uniqueness we cannot guarantee that this candidate probability will be genuine and nondegenerate (i.e. all branching probabili-ties lie in the interval (0, 1)), as shown by the examples given in Chapter 2.
We now consider a model of this kind with two bonds, as shown in the diagram:
B(0, 2) B(0, 3)
pu=?
pm=?
pd=1− pu− pm
Bu(1, 2) =?
Bu(1, 3) =?
Bm(1, 2) =?
Bm(1, 3) =?
Bd(1, 2) =?
Bd(1, 3) =?
A substantial advantage now is that the bond prices do not have to be per-fectly correlated and we can insist on some nontrivial correlation between the prices, perhaps close to, but not necessarily equal to one. There are eight parameters to find and following the scheme based on historical data, six conditions can be formulated based on:
• 2 expectations,
• 2 standard deviations,
• 1 correlation.
From the point of solvability we are here in a good position having many variables and few equations. However, this requires some ad hoc decisions since typically there will be infinitely many solutions.
Example 6.16
As before, the simplest assumption is to take the three probabilities to be equal, pu = pm = pd = 1/3. This arbitrary choice is again guided by the fact that for option pricing the original probabilities are irrelevant and only the risk-neutral probabilities are used. Let us try to use the data from Example 6.13. We leave it as an elementary numerical exercise to check
that with some accuracy, the following prices fit the scheme Bu(1, 2) = 0.996820,
Bm(1, 2) = 0.996600, Bd(1, 2) = 0.996379, Bu(1, 3) = 0.993310, Bm(1, 3) = 0.993176, Bd(1, 3) = 0.992814,
the correlation coefficient is ρ = 0.964 and the risk-neutral probabilities are qu= 0.3533, qm = 0.2409, qd= 0.4059.
We will not pursue other methods of fitting the tree to the data which were outlined above, such as using the current prices of derivative secu-rities, which is arguably the most promising, since it is related to current market data. However, the work involved in formulating the conditions and solving the equations is similar as before.
Suppose we admit an additional bond B(1, 4) maturing at time 4. In real markets many bonds with different maturities are traded, so this is an in-evitable direction for the development of the model. We are working in the trinomial model, so we have 11 parameters:
• 9 bond prices,
• 2 probabilities, and 12 conditions:
• 3 expectations,
• 3 standard deviations,
• 3 correlations,
• 3 equalities for risk-neutral probabilities.
Here we have too many conditions, and with our experience we can arrive at the conclusion that we should consider more branches in the tree, for instance a quadrinomial model. However, this becomes very complicated and bearing in mind that we are considering just the first step, this does not seem a viable path for further development. As will be seen in a later vol-ume in this series, continuous-time interest rate models, though in principle more sophisticated, provide a more tractable approach to this problem.
Multi-step models
We have seen that introducing more complex models leads to complica-tions and does not really solve the problem. We have to make simplifying assumptions in order to build some viable examples.
We would like to follow the general approach of the CRR model where the returns determine the prices and the returns in all steps are the same. Here we can apply this only partially and some extra care will be needed.
The general pattern is this: at step 1 we build binomial trees for various bonds (with various maturities). These trees have the following form
Bu(1, n) B(0, n)
Bd(1, n)
where n = 2, . . . , N. At each step the time to maturity decreases by one for each bond. We would like to apply the same scheme in each subse-quent step, using appropriately structured changes related to the distance to maturity.
So, for instance, with N = 4 we have three trees related to the bonds of maturities 2,3,4 and their times to maturity are reduced to 1,2,3, re-spectively. We have at our disposal three schemes, which could be denoted by 2to1, 3to2, 4to3. Now at time one, at the up state, we have the prices Bu(1, 4), Bu(1, 3), Bu(1, 2). The last bond requires no action since it will become the unit at the next time. For the design of trees for the first two bonds we now use the 3to2 and 2to1 schemes. In other words, the structure of the tree
Buu(2, 3) Bu(1, 3)
Bud(2, 3) will be the same as the structure of the tree
Bu(1, 2) B(0, 2)
Bd(1, 2)
Our task is to identify this structure. Among the many calibrating methods available we choose the one based on taking all risk-neutral probabilities to be 1/2. This implies the following structure of returns
U = R + ε, D= R − ε,
where R is the risk-free return. Let us explore this with a numerical example.
Example 6.17
We will continue Example 6.15 and we recognise the form of the structure related to the 2to1 tree. The risk-free return for step 1 is
R(0)= 1
B(0, 1)− 1 = 0.3412%.
For the 2to1 reduction we use the bond prices computed before Bu(1, 2) = 0.99690136,
Bd(1, 2) = 0.99627544, which are related to returns at time 0 for the 2-bond
U2to1= Bu(1, 2)
B(0, 2) − 1 = 0.3727%, D2to1= Bd(1, 2)
B(0, 2) − 1 = 0.3096%, and clearly
U2to1= R(0) + 0.0315%, D2to1= R(0) − 0.0315%, hence
ε2to1= 0.0315%.
Similarly, employing B(0, 3) and Bu(1, 3), Bd(1, 3) we can find ε3to2= 0.0856%.
Now we will design the prices B(2, 3) at time 2 (there are four of them).
Recall that earlier we found the values
Bu(1, 3) = 0.99408791, Bd(1, 3) = 0.99206501.
Consider two cases:
(i) ‘Up’ move at step 1. First note that we have the risk-free return Ru(1)= 1
Bu(1, 2)− 1 = 0.3108%.
We need two prices Buu(2, 3) and Bud(2, 3) and we build them by applying the above structure and requiring that
Buu(2, 3)
Bu(1, 3) − 1 = Ru(1)+ ε2to1, Bud(2, 3)
Bu(1, 3) − 1 = Ru(1)− ε2to1, so that
Buu(2, 3) = Bu(1, 3)(1 + Ru(1)+ ε2to1)= 0.997326, Bud(2, 3) = Bu(1, 3)(1 + Ru(1)− ε2to1)= 0.996699.
(ii) ‘Down’ move at step 1. The risk-free return is Rd(1)= 1
Bd(1, 2)− 1 = 0.3738%,
and working in the same manner (using the same perturbations
±ε2to1) we get
Bdu(2, 3) = Bd(1, 3)(1 + Rd(1)+ ε2to1)= 0.996252, Bdd(2, 3) = Bd(1, 3)(1 + Rd(1)− ε2to1)= 0.995626.
To summarise, here is the complete tree:
Buu(2, 3) = 0.9973 Bu(1, 2) = 0.9969
Bu(1, 3) = 0.9939
B(0, 1) = 0.9966 Bud(2, 3) = 0.9967
B(0, 2) = 0.9932
B(0, 3) = 0.9897 Bdu(2, 3) = 0.9963
Bd(1, 2) = 0.9963 Bd(1, 3) = 0.9922
Bdd(2, 3) = 0.9956
In general, at the first step, assuming q= 1/2, we have the perturbations εnto(n−1), n ≤ N and then at time k, in the state ωk we design single-step trees for all bonds using the same risk-free returns, and appropriate pertur-bations, the same for each step, depending solely on distance to maturity
Bωku(k+ 1, n) = Bωk(k, n)(1 + Rωk(k)+ ε(n−k)to(n−k−1)), Bωkd(k+ 1, n) = Bωk(k, n)(1 + Rωk(k)− ε(n−k)to(n−k−1)).
One important final comment is related to the No Arbitrage Principle.
Using q = 1/2 seems to guarantee lack of arbitrage but one has to be careful, since in general the bond prices might increase beyond 1. Should this happen in the model, we must find appropropriate adjustments to deal with it.
Remark 6.18
If we adopt the method of using the first step structure at the subsequent stages, the first step is crucial. An alternative to assuming a value for the risk-neutral probability arbitrarily, as we have done, is to take the historical variances (or standard deviations) related to bonds with various maturities.
Such a sequence of standard deviationsσ(0, k), k ≤ N, is called the initial term structure of volatilities. Clearly their sizes are related to the corre-sponding perturbationsεkto(k−1). The binomial distribution gives a simple formula for the standard deviation and if U = R + ε, D = R − ε then ε is the standard deviation of the returns but computed with respect to the probability q= 1/2: σ =
q(1− q)(U − D) = ε. If we have the historical standard deviations we have to remember that they correspond to physical probabilities, not the risk-neutral ones.
Exercise 6.18 Build the initial tree for a bond maturing at time 4 with initial price B(0, 4) = 0.9859, such that the price of 100 000 puts with strike K = 0.9898 is 80 and find the perturbation ε4to3. Build the complete tree of prices B(k, 4) using the perturbations found in the previous example.