Models for the short rate

The Ho and Lee tree approach

The binomial tree of the previous chapter was very useful in order for one to price contigent claims in the one-period setting. We will now start with a generalization of the binomial tree model. The 50-50 rule assumption will be maintained, implying that

$$c_{0} = q_{u}c_{u}+q_{d}c_{d}$$

$$q_{u} = q_{d}=\frac{0.5}{1+r/2}$$

Example 22 (Interest rate tree)   Consider the state variable to be the interest rate, which can move up or down by $100$ basis points [=$1\%$] over a six months period. In addition, assume that today's short rate is $5\%$. Such a specification would lead into an interest rate tree which looks like

					$9\%$
				$8\%$
			$7\%$		$7\%$
		$6\%$		$6\%$
	$5\%$		$5\%$		$5\%$
		$4\%$		$4\%$
			$3\%$		$3\%$
				$2\%$
					$1\%$
Years	0	0.5	1	1.5	2

To be concrete, we could denote those movements of the short rate using the recursive relation

$$r_{t+1}=r_{t}+\mu _{t+1}+\varepsilon _{t+1}$$,

where

$$\varepsilon _{t+1}=\left\{ \begin{tabular}{ll} $+\sigma $ & , wi... ...{2}$ \\ $-\sigma $ & , with probability $\frac{1}{2}$%% \end{tabular}\right.$$   .

Here, we have used $r_{0}=5\%$, $\mu _{t+n}=0$ for all $n$, and $\sigma =1\%$. The probabilities mentioned above are the risk-neutral probabilities of the previous chapter, which are a result of the 50-50 rule. The above model is the Ho and Lee model, named after the people that developed it. There are two important observations that one can make:

• the parameters of the model are chosen in a way to make the branches recombine. The reason for that is to just make the calculations later simpler; it is not something that has to hold; and

• since the short rate is different across different nodes of the tree, we would expect the state prices to be different too.

Example 23 (Bond pricing)   Let us now consider how one can price a $6\%$ three period [one-and-a-half year] bond. We can start at the maturity where the cash flows are known: $%% \pounds 100$ the face value plus $\pounds 3$ the coupon, or $\pounds 103$. From these we can calculate the column before last

			$r=7\%$ $c=\pounds 99.52$		$c=\pounds 103$
		$r=6\%$ $c=-$
	$r=5\%$ $c=-$		$r=5\%$ $c=\pounds 100.49$		$c=\pounds 103$
		$r=4\%$ $c=-$
			$r=3\%$ $c=\pounds 101.48$		$c=\pounds 103$
Years	0	0.5	1		1.5

We can now continue our recursive computations using the pricing equations, for example the top element of the second column will be equal to

$$\frac{0.5}{1+0.06/2}\left[ \left( 99.52+3\right) +\left( 100.49+3\right) \right] =100.00$$.

The other element of the second column is going to be equal to $\pounds 101.94$, and the root of the tree --the price of this particular bond today, is going to be equal to

$$\frac{0.5}{1+0.05/2}\left[ \left( 100.00+3\right) +\left( 101.94+3\right) \right] =101.44$$.

The complete tree is going to be

			$\pounds 99.52$		$c=\pounds 103$
		$\pounds 100.00$
	$\pounds 101.44$		$\pounds 100.49$		$c=\pounds 103$
		$\pounds 101.94$
			$\pounds 101.48$		$c=\pounds 103$
Years	0	0.5	1		1.5

Obviously, this is a lot of work to be done when the time horizon starts to increase and one has to compute the tree again and again to value different kinds of bonds. At this point, the state prices come to our rescue: if one could compute the state price tree, one could use the same tree in order to value all fixed income securities. Instead of moving backwards in time, state prices are computed moving forwards. One pound today is worth exactly one pound today, setting the state price of the root of the tree equal to one. From there, the one period state prices are computed as in the previous chapter

		-
	$0.4878$
$1.0000$		-
	$0.4878$
		-

Example 24 (State price tree)   There is one way of moving to the first element of the third column, and that is up. The state price will be equal to

$$0.4878\times \frac{0.5}{1+0.06/2}=0.2368$$;

whereas there are two ways of moving towards the middle element. Its state price is

$$0.4878\times \frac{0.5}{1+0.06/2}+0.4878\times \frac{0.5}{1+0.04/2}=0.4759$$

and so on, giving the state price tree

				$.055$
			$.114$
		$.237$		$.223$
	$.488$		$.347$
$1.00$		$.476$		$.340$
	$.488$		$.350$
		$.239$		$.230$
			$.118$
				$.058$

What is the value of the above coupon bearing bond in this case? It will be equal to

$$\begin{multline*} \left( .1144+.3466+.3499+.1178\right) \times 103 \\ +\left( .... ... \times 3 \\ +\left( .4878+.4878\right) \times 3=101.44\text{.} \end{multline*}$$

The asset price is equal to the sum of all individual claims, namely

$$c_{0}=\sum_{\text{time}}\left[ \sum_{\text{states}}q\times c\right] \text{.}$$

Choosing the parameters

There are two important points that one has to make. The first is the 50-50 split that we employed. It might be wrong but it has been followed traditionally without questioning it. The second point has to do with the choice of the parameters, namely the values of $\mu _{t}$ for all $t$ and the value of $\sigma$. The value of $\sigma$ is chosen arbitrarily^5.1 while the values of $\mu _{t}$ are chosen as to generate the observed yield curve. This is not difficult, you just need a good computer program that will match the spot rates.

Example 25 (Tree parameters)   Suppose that you observe the following LIBOR yield curve information

$$\begin{tabular}{cc} \textbf{Maturity [1/2y]} & \textbf{Discount f... ...3 & 0.9175 \\ 4 & 0.8931 \\ 5 & 0.8644 \\ 6 & 0.8378 \hline \end{tabular}$$

In addition suppose that $\sigma =1\%$ is a decent value for the volatility parameter. The parameters that reproduce this yield curve are found step-by-step. Therefore the spot rate has to solve

$$97.07\times \left( 1+\frac{r}{2}\right) =100\Rightarrow r=6.036\%$$.

This interest rate will dictate the values of the state prices after the first branching, given by

$$q_{u}=q_{d}=\frac{1}{2}\frac{1}{1+r/2}=0.48535$$.

The value of $\mu _{t+1}$ has to be chosen in a way that the state prices after the second branching sum up to the discounting factor. The tree is

		-- $q_{0}$
	$7.036\%+\mu _{t+1}$ $.4854$
$r=6.036\%$ $q=1.000$		-- $q_{1}$
	$5.036\%+\mu _{t+1}$ $.4854$
		-- $q_{2}$

and the state prices are

$$q_{0} = \frac{1}{2}\frac{0.4854}{1+\frac{7.036\%+\mu _{t+1}}{2}}$$

$$q_{1} = \frac{1}{2}\frac{0.4854}{1+\frac{7.036\%+\mu _{t+1}}{2}}+\frac{1}{2}%% \frac{0.4854}{1+\frac{5.036\%+\mu _{t+1}}{2}}$$

$$q_{2} = \frac{1}{2}\frac{0.4854}{1+\frac{5.036\%+\mu _{t+1}}{2}}$$

$$q_{0}+q_{1}+q_{2} = 0.9443$$

The solution is $\mu _{t+1}=-0.418\%$. This gives the state prices

		-- $0.2349$
	$6.618\%$ $.4854$
$r=6.036\%$ $q=1.000$		-- $0.4722$
	$4.618\%$ $.4854$
		-- $0.2372$

Example 26 (cont. Tree parameters)   The value of $\mu _{t+2}$ can be computed in a similar fashion. The tree will be

			-- $q_{0}$
		$7.618\%+\mu _{t+2}$ $.2349$
	$6.618\%$ $.4854$		-- $q_{1}$
$\cdots$		$5.618\%+\mu _{t+2}$ $.4722$
	$4.618\%$ $.4854$		-- $q_{2}$
		$3.618\%+\mu _{t+2}$ $.2372$
			-- $q_{3}$

which gives $\mu _{t+2}=0.2386\%$. The state prices are

			-- $0.1130$
		$7.857\%$ $.2349$
	$6.618\%$ $.4854$		-- $0.3424$
$\cdots$		$5.857\%$ $.4722$
	$4.618\%$ $.4854$		-- $0.3457$
		$3.857\%$ $.2372$
			-- $0.1164$

Example 27 (cont. Tree parameters)   By repeating the above procedure once more we find $\mu _{t+3}=-0.3636\%$ and $\mu _{t+4}=0.7793\%$. The interest rate tree becomes

				$10.272\%$
			$8.493\%$
		$7.857\%$		$8.272\%$
	$6.618\%$		$6.493\%$
$6.036\%$		$5.857\%$		$6.272\%$
	$4.618\%$		$4.493\%$
		$3.857\%$		$4.272\%$
			$2.493\%$
				$2.272\%$

and the state price tree is

				$.0541$
			$.1130$
		$.2349$		$.2196$
	$.4854$		$.3424$
$1.000$		$.4722$		$.3342$
	$.4854$		$.3457$
		$.2372$		$.2261$
			$.1164$
				$.0573$

It is sometimes intuitive to visualize the resulting tree as in diagram 5.1

In the same fashion as in the examples above, one could compute all unknown parameters, and afterwards all the state prices associated with the interest rate model. Having computed the state prices it is straightforward to value more complex instruments such as coupon bearing bonds, options on bonds, etc. Now we turn to some examples to illustrate the use of the above analysis.

Claim pricing

Example 28 (Digital option)   Consider a bet on the future level of the interest rates: you win $\pounds 10$ if the short rate is greater than $7\%$ four periods [two years] from now, and nothing otherwise. This is not a very common instrument, it reminds us of the bookies, but nevertheless it has some similarity to the digital options. The bet generates state contigent cash flows that look like this

				$10$
			0
		0		$10$
	0		0
0		0		0
	0		0
		0		0
			0
				0

What is this bet worth? According to our model the price of the bet is the sum of the product of the state prices times the payoffs, or

$$\left( 0.0541+0.2196\right) \times 10=2.737$$.

Example 29 (Price path of a zero)   Consider a 30m zero, and check the different price paths that it can follow through its life. At the last nodes --just before maturity, the short rate is known and the price of the bond can be easily calculated as $100\times \frac{1}{1+r/2}$. For the rest one can work backwards using the relationships

$$= q_{u}c_{u}+q_{d}c_{d}$$ Asset price

$$q_{u} = q_{d}=\frac{0.5}{1+r_{t}/2}$$

The tree will be

				$95.12$
			$91.68$
		$89.08$		$96.03$
	$87.50$		$93.48$
$86.62$		$91.71$		$96.99$
	$90.97$		$95.31$
		$94.42$		$97.91$
			$97.18$
				$98.88$

Example 30 (Option on a zero)   Consider the zero of the previous example, and suppose that one wants to find the price of an one year European call option on that zero, with strike price $\pounds 92$. This is the right to buy the bond on the second period for the strike price. The option will be exercised of course only if the price of the bond after one year is higher than the exercise price. The payoff of the option is illustrated using the tree

		0
	0
0		0
	0
		$\pounds 2.42$

Today's price will be equal to the payoffs times the corresponding state prices, namely

$$C=0.2372\times 2.42=\pounds 0.5740$$.

One can calculate the whole price path of the option, as

		0
	0
$57.40p$		0
	$\pounds 1.18$
		$\pounds 2.42$

Example 31 (American put on a zero)   Suppose now that the option was an American put instead of a European call, with the same exercise price $\pounds 92$. Now the payoffs are not as easy as before, since one has the option to exercise in the first period too. The result is obtained going backwards. Now the right is to sell, so the terminal payoffs are

		$\pounds 2.92$
	--
--		$29p$
	--
		0

On the first period the option will be priced as it if were European, since it expires in the next period. Therefore the price of the option given that one will not exercise it will be^5.2

		$\pounds 2.92$
	$\pounds 1.80$
--		$29p$
	$14p$
		0

Consider the upper node of this tree: The bond price at this point is $%% \pounds 87.50$; the option if not exercised is worth $\pounds 1.55$, while if the investor exercises then the payoff will be $\pounds %% \left( 92-87.50\right) =\pounds 4.50$. Therefore the option will be exercised on this node, and the payoff will be $\pounds 4.50$. Now turn to the lower node of the tree: There the bond price is $\pounds 90.97$, if the option is exercised --buying at $\pounds 90.97$ and selling at $%% \pounds 92$-- the payoff will be $\pounds \left( 92-90.97\right) =\pounds %% 1.03$. The value of the option if it is not exercised is $14p<\pounds %% 1.03$, and exercising is optimal. The price of the American put will therefore evolve as

		$\pounds 2.92$
	$\pounds 4.50$
--		$29p$
	$\pounds 1.03$
		0

The value of the option at the starting node if it is not exercised is given by

$$\frac{0.5}{1+6.036\%/2}\left( 4.50+1.03\right) =\pounds 2.69$$.

while the value if it is exercised is $\pounds \left( 92-86.62\right) =\pounds 5.38$. Exercising the option is optimal at time zero, therefore its price will be equal to its payoffs, i.e. $\pounds 5.38$.

Other models

The model discussed above has the disadvantage that it allows the asset price --the interest rate is the above examples, to become negative. A simple solution is to consider the logarithm of the asset price to be driven by the tree model. In this spirit the interest rate tree would be generated by

$$\ln r_{t+1}=\ln r_{t}+\mu _{t+1}+\varepsilon _{t+1}$$,

where $\varepsilon _{t+1}$ is defined in the same fashion as above

$$\varepsilon _{t+1}=\left\{ \begin{tabular}{ll} $+\sigma $ & , wi... ...{2}$ \\ $-\sigma $ & , with probability $\frac{1}{2}$%% \end{tabular}\right.$$   .

The changes are multiplicative rather than additive, and in terms of the interest rates themselves the model says that

$$r_{t+1}=r_{t}e^{\mu _{t+1}+\varepsilon _{t+1}}$$.

The very popular among practitioners Black-Derman-Toy paradigm takes this one step further and sets the volatility changing with time, in order to replicate not only the term structure of the interest rates but the observed patterns of volatilities across time --the so called volatility term structure. In their model

$$\varepsilon _{t+1}=\left\{ \begin{tabular}{ll} $+\sigma _{t+1}$ ... ...\\ $-\sigma _{t+1}$ & , with probability $\frac{1}{2}$%% \end{tabular}\right.$$   .

Obviously in this specification one has too many unknowns to solve for when the term structure is replicated, especially the values of the future volatilities which have to be set explicitly. Apart from using estimated historical volatilities practitioners and academics alike investigate the use of volatility values which are implied from option prices^5.3

The binomial tree framework discussed above is widely used due to its simplicity, but of course has a number of problems.

• Mean reversion: When the interest rates peaked in the 80s, everyone was expecting them to start decreasing sooner or later. Interest rates are different from asset prices because they exhibit mean reversion. This is a feature that cannot be incorporated into the binomial tree framework.

• Multiple factors: In this setting it has been the short rate that drives the bond prices of all maturities. More complex models extend this setting to include multiple factors.

• Calibration: The choice of some parameters is arbitrary, therefore the ability of the model to explain future movements has an element of artistry. The choice will depend on the data available and on the application.

The binomial framework of Cox, Ross and Rubinstein

Cox, Ross and Rubinstein [CRR] developed a simplified approach in order to value derivative securities, employing the binomial framework that was discussed in the previous chapter. In fact they generalize the valuation of contigent claims from the one period to the $\tau$ period setting. One difference between the model of CRR and the HL model is the fact that the CRR has a multiplicative evolution while the Ho-Lee model is additive. In that respect one can say that the CRR model is more related to the Heath-Jarrow-Morton specification.

In the risk neutral world that we had developed in the previous chapter for the one period case, the discounted values of all assets are martingales. Denote $S_{t}$ the price of the asset at any time $t$ and suppose that the price can move up --state $u$, reaching $S_{t+1}=S_{t}u$, or it can drop --state $d$, down to $S_{t+1}=S_{t}d$. Suppose also that these movements can take place during an interval $\Delta$ which we consider to be small --small enough to make quantities of order $o\left( \Delta ^{2}\right)$ and higher negligible. The risk neutral probability measure for the one period case will have elements $\left\{ Q_{u},Q_{d}\right\} \equiv \left\{ Q_{0},Q_{1}\right\}$. For the $\tau$ period case it will have elements of the form

$$\left\{ Q_{0},Q_{1},\ldots ,Q_{\tau }\right\}$$   ,

the subscripts denoting the number of upward movements in this time interval. In the above formulation

$$Q_{k}=\binom{\tau }{k}Q_{u}^{k}Q_{d}^{\tau -k}=\binom{\tau }{k}%% Q_{u}^{k}\left( 1-Q_{u}\right) ^{\tau -k}$$.

Such a tree structure always recombines: the sequences $\left\{ u,d\right\}$ or $\left\{ d,u\right\}$ will result into the same asset price, since $%% S_{t}ud=S_{t}du$.

Now fix the interest rate at $r$, and consider compounding to be continuous. The starting price of the asset is known, and let it be $S_{0}$. Then, there are two possible values

	$S_0u$
$S_0$
	$S_0d$

According to the risk neutral valuation principle, the risk neutral probabilities will ensure that the discounted expected prices form martingales. This implies the relationship

$$e^{-r\Delta }\mathbf{E}^{Q}\left[ S_{1}\right] =S_{0}$$,

or equivalently $e^{-r\Delta }\left( Q_{u}u+Q_{d}d\right) =1$, hence

$$\left\{ \begin{tabular}{l} $Q_{u}=\frac{e^{r\Delta }-d}{u-d}$ \\ $Q_{d}=1-Q_{u}$%% \end{tabular}\right.$$   .

One can easily verify that the discounted asset prices under this measure form martingales for any time horizon:

$$e^{-r\tau \Delta }\mathbf{E}^{Q}\left[ S_{\tau }\right] = e^{-r\tau \Delta }S_{0}\sum_{k=0}^{\tau }u^{k}d^{\tau -k}Q_{k}$$

$$= e^{-r\tau \Delta }S_{0}\sum_{k=0}^{\tau }u^{k}d^{\tau -k}\binom{\tau }{k}%% Q_{u}^{k}Q_{d}^{\tau -k}$$

$$= e^{-r\tau \Delta }S_{0}\sum_{k=0}^{\tau }\binom{\tau }{k}\left[ uQ_{u}%% \right] ^{k}\left[ dQ_{d}\right] ^{\tau -k}$$

$$= e^{-r\tau \Delta }S_{0}\left[ uQ_{u}+dQ_{d}\right] ^{\tau }$$

$$= e^{-r\tau \Delta }S_{0}\left[ e^{-r\Delta }\right] ^{\tau }$$

$$= S_{0}$$

The above tree specification allows one to price all derivative contracts, for instance futures, European options, American options, etc., as illustrated in the following examples

Example 32 (European put)   Suppose that the real asset price follows a binomial tree, with $\Delta$ being one year, the risk free rate $r=5\%$, and $u=1+20\%$. Say that the current asset price is £50, and consider a two year European put written on this asset, with strike price £52. The price tree is shown in the following table

		$\pounds 72$
	$\pounds 60$
$\pounds 50$		$\pounds 48$
	$\pounds 40$
		$\pounds 32$

The value of the risk neutral probability is

$$Q_{u}=\frac{\exp \left[ 1\times 5\%\right] -0.8}{1.2-0.8}=0.628$$.

The payoffs of this put option after two years are given by $\max \left[ K-S_{2},0\right]$, or

		0
	--
--		$\pounds 4$
	--
		$\pounds 20$

Using the risk neutral probabilities, one can work her way backwards through the tree, computing the discounted cash flows to get the price path of the option

		0
	$e^{-0.05} \times$ $\left[ 0.628\times 0+0.372\times 4\right]$ $=\pounds 1.42$
$e^{-0.05} \times$ $\left[ 0.628\times 1.415+0.372\times 9.47\right]$ $=\pounds 4.19$		$\pounds 4$
	$e^{-0.05} \times$ $\left[ 0.6282\times 4+0.372\times 20\right]$ $=\pounds 9.47$
		$\pounds 20$

Example 33 (American put)   Now suppose that the put were of the American type. At every node of the tree, the value of the contract would be the maximum of: 1. the payoff of the option were it exercised immediately, and 2. the discounted cash flows under risk neutrality. This would generate the tree

		0
	$\max \left[ \pounds 1.415,-\pounds 8\right] =\pounds 1.415$
$\max \left[ \pounds 9.47,\pounds 12\right] =\pounds 12$		$\pounds 4$
	$\max \left[ \pounds 9.47,\pounds 12\right] =\pounds 12$
		$\pounds 20$

In practice, one if faced with the real market process, where the asset price does not grow with the interest rate, but according to some constant $%% \mu$. Therefore, in order to match the real trend one has to choose values for $u$, $d$ and $P_{u}$ that match the expected stock price

$$e^{\mu \Delta }=P_{u}u+\left( 1-P_{u}\right) d$$.

The second step involves matching the volatility of the stock return, or

$$\sigma ^{2}\Delta =P_{u}u^{2}+\left( 1-P_{u}\right) d^{2}-\left[ P_{u}u+\left( 1-P_{u}\right) d\right] ^{2}$$.

The above will give

$$e^{\mu \Delta }\left( u+d\right) -ud-e^{2\mu \Delta }=\sigma ^{2}\Delta$$   .

One solution of the above equation is taking the symmetric values proposed in CRR

$$u = e^{\sigma \sqrt{\Delta }}$$

$$d = e^{-\sigma \sqrt{\Delta }}$$

In the risk neutral world the growth is equal to $r$ and the probability of moving upwards is as we saw before

$$Q_{u}=\frac{e^{r\Delta }-d}{u-d}$$.

One can verify that under risk neutrality the volatility of the asset return remains

$$Q_{u}u^{2}+\left( 1-Q_{u}\right) d^{2}-\left[ Q_{u}u+\left( 1-Q_{u}\right) d\right] ^{2}=\sigma ^{2}\Delta$$

for the aforementioned values of $u$ and $d$. CRR show that their model gives $\tau$ period European call option prices with strike price $K$ according to

$$S_{0}\mathcal{W}\left( a;\tau ,Q_{u}^{\prime }\right) +e^{-r\tau \Delta }K\mathcal{W}\left( a;\tau ,Q_{u}\right)$$   ,

where

$$\mathcal{W}\left( a;\tau ,Q_{u}\right) =\sum_{k=a}^{\tau }\binom{\tau }{k}%% Q_{u}^{k}\left( 1-Q_{u}\right) ^{\tau -k}$$,

the complementary binomial distribution function, $Q_{u}^{\prime }=e^{-r\Delta }uQ_{u}$, and $a$ is the smallest non-negative integer greater than $\ln \left( K/Sd^{\tau }\right) /\ln \left( u/d\right)$. They also present the convergence of their model to the Black and Scholes formula, as the interval $\Delta$ becomes finer.