In this chapter, we look at the most common building blocks of short-term rate models. Since the art of term structure modeling is the purpose at hand, this chapter seeks to create suitable models by the selection and rearrangement of the mentioned building blocks.

# Model I: Normally Distributed Models and No Drift

The following equation depicts the evolution of the instantaneous rate \({ r }_{ t }\).

$$ dr=\sigma dw $$

Where \(dr\) is the small time interval changes, \(dt\), in rate and is measured in years, \(\sigma\) is the changes in the rate of volatilities for annual basis-point, and \(dw\) is a random variable which is normally distributed whose standard deviation is \(\sqrt { dt } \) and mean \(0\).

The expected value of \(dw\)(mean) zero implies a drift of zero,and the standard deviation of the rate is \(\sigma dw\) as the standard deviation of \(dw\) is \(\sqrt { dt } \). This process can be approximated by a rate tree where, in the up-state, the interest rate change is \(\sigma dt\), and, – \(\sigma dt\) in the down state. The expected rate change, using probabilities provided in the rate tree, will be:

$$ E\left[ dr \right] =0.5\times \sigma \sqrt { dt } +0.5\times -\sigma \sqrt { dt } =0 $$

To compute the variance of the rate, \(V\left[ dr \right]\), between dates 0 and 1, we apply the following equation:

$$ V\left[ dr \right] =E\left[ d{ r }^{ 2 } \right] -{ \left\{ e\left[ dr \right] \right\} }^{ 2 } $$

$$ =0.5\times { \left( \sigma \sqrt { dt } \right) }^{ 2 }+0.5\times { \left( -\sigma \sqrt { dt } \right) }^{ 2 }-0 $$

$$ ={ \sigma }^{ 2 }dt $$

It is worth noting that as opposed to a single step in the tree leading to only two possible values, the normally distributed random variable in the process can take on any value; hence the slight difference in the process and the tree.

The model is a normal or Gaussian model since there is a normal distribution in the terminal distribution of interest rates. Its limitation is the possibility of negativity in the short-term rate hence causing a lack of willingness by investors to lend money at the negative rates while holding cash that can earn them zero rates instead.

# Model II: Drift and Risk Premium

As pointed out in chapter 12, the term structure tends to have an upward slope, a tendency explained by the existence of a risk premium. An addition of a drift to model 1, by this model, aims to obtain an economically coherent rich model with the following dynamics:

$$ dr=\lambda dt+\sigma dw $$

The risk-neutral process drift is a combination of the true expected change in the interest rate and of a risk premium. The model can also be approximated by a tree:

$$ \begin{array} \hline {} & {} & {} & {\scriptsize 0.5 } & { r }_{ 0 }+2\lambda dt+2\sigma \sqrt { dt } \\ {} & {} & { r }_{ 0 }+\lambda dt+\sigma \sqrt { dt } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ { r }_{ 0 } & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & \scriptsize 0.5 & { r }_{ 0 }+2\lambda dt \\ {} & {} & { r }_{ 0 }+\lambda dt-\sigma \sqrt { dt } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize 0.5} & { r }_{ 0 }+2\lambda dt-2\sigma \sqrt { dt } \\ \end{array} $$

The calibration of the \({ r }_{ 0 }\) and \(\lambda \) values, is such that they match the 2- and 10-year par swap rates and \(\sigma \) is such that it is the average implied volatility of the 2- and 10-year swap par rates. Due to lack of sufficient flexibility to describe the term structure’s shape in a way that is meaningful to the economy, in the long run, the interpretation of \(\lambda \) as a combination of true drift and risk premium is not easy.

# The Ho-Lee Model: Time-Dependent Drift

The Ho-Lee model has its dynamics written as:

$$ dr={ \lambda }_{ t }dt+\sigma dw $$

In this model the drift varies from date to date, hence is time-dependent and represents some combination in the short-term rate some combination of the risk premium and expected changes over each period of time. The model can also be represented by a tree.

$$ \begin{array} \hline {} & {} & {} & {\scriptsize 0.5 } & { r }_{ 0 }+\left( { \lambda }_{ 1 }+{ \lambda }_{ 2 } \right) dt+2\sigma \sqrt { dt } \\ {} & {} & { r }_{ 0 }+{ \lambda }_{ 1 }dt+\sigma \sqrt { dt } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ { r }_{ 0 } & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & {\scriptsize \begin{matrix} \begin{matrix} 0.5 \\\scriptsize \begin{matrix} \\ \end{matrix} \end{matrix} \\ \begin{matrix} \\ \end{matrix} \\ 0.5 \end{matrix} } & { r }_{ 0 }+\left( { \lambda }_{ 1 }+{ \lambda }_{ 2 } \right) dt \\ {} & {} & { r }_{ 0 }+{ \lambda }_{ 1 }dt-\sigma \sqrt { dt } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize 0.5} & { r }_{ 0 }+\left( { \lambda }_{ 1 }+{ \lambda }_{ 2 } \right) dt-2\sigma \sqrt { dt } \\ \end{array} $$

Securities prices can be matched using free parameters,\({ \lambda }_{ 1 }\) and \({ \lambda }_{ 2 }\), with fixed cash flows. This can be achieved by setting \(dt = 0.5\),\(r_{ o }=1\) -month rate, \({ \lambda }_{ 1 }\): a two-month spot rate produced by the model equal the the one in the market, and, \({ \lambda }_{ 2 }\): a three-month spot rate produced by the model equal the the one in the market. This trend is continued until the end of the tree.

# Desirability of Fitting to the Term Structure

When deciding between arbitrage-free and equilibrium models, the desirability of fitting to the term structure is a crucial factor, whose choice is dependent on the purpose of building the model.

Based on the prices of liquid securities, arbitrage-free models can quote prices of securities not actively traded. Odd maturity-swap can be priced as ameans of interpolating between observed values using an arbitrage-free model.

Furthermore, in order to make markets or proprietary trading, valuation and hedging of derivative securities can be achieved by means of arbitrage-free models. Thus, many practitioners will assume that some set of underlying securities is fairly priced.

Recognizing that a risk-neutral process’ drift arises only from expectation and risk premium is a good way to express the problem of fitting drift to the term structure.

Practically, starkly contrasted arbitrage-free models and equilibrium modelshould not have a clear distinction between them. The art of term structure modeling takes the blending of the two approaches.

# The Vasicek Model: Mean Diversion

The Vasicek model has its risk-neutral dynamics written as:

$$ dr=k\left( \theta -r \right) dt+\sigma dt\quad \quad \quad \quad \left( a \right) $$

Where constant \(\theta\) is the central tendency of the short term of the short term rate in the risk-neutral process while \(k\) is the speed of mean reversion. Great difference between \(r\) and \(\theta\) implies a greater expected change in the short-term rate toward \(\theta\).

Given that mean reversion to a long-term value \({ r }_{ \infty }\) exhibited by the true interest rate process and as a constant drift the risk premium enters into the risk-neutral process, then the Vasicek model becomes:

$$ dr=k\left( { r }_{ \infty }-r \right) dt+\lambda dt+\sigma dw $$

$$ =k\left( \left[ { r }_{ \infty }+\frac { \lambda }{ k } \right] -r \right) dt+\sigma dw\quad \quad \quad \quad \left( b \right) $$

Process (\(a\)) and (\(b\)) are identical under the following condition:

$$ \theta \equiv { r }_{ \infty }+\frac { \lambda }{ k } $$

At the same market prices, lots of \({ r }_{ \infty }\) and \(\lambda\) combinations will give the same \(\theta\), the process (\(b\)).

Assuming that we take \(k=0.025\), \(\sigma =126\) basis points per year, \({ r }_{ \infty }=6.179\% \) and \(\lambda=0.229\) then \(\theta=15.339\%\). Then according to equation (\(a\)), the expected change in the short rate is:

$$ 0.025\left( 15.339\%-5.121\% \right) \frac { 1 }{ 12 } =0.0213\%\quad or\quad 2.13\quad basis\quad points $$

Over the next month, the volatility is:

$$ 126\times \sqrt { \frac { 1 }{ 12 } } =36.4\quad basis\quad points $$

Hence, the following trees will represent the above process:

$$ \begin{array} {} & {} & {\scriptsize 0.5 } & 5.8902\% \\ {} & 5.5060\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ 5.121\% \begin{matrix} & {\scriptsize 0.5 } & \\ &\Huge \diagup & \\ &\Huge \diagdown & \\ & {\scriptsize 0.5 } & \end{matrix} & {} & \begin{matrix} {\scriptsize 0.5 } \\ \\ {\scriptsize 0.5 } \end{matrix} & \begin{matrix} 5.1628\% \\ \\ 5.1643\% \end{matrix} \\ {} & 4.7786\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {} & {\scriptsize 0.5 } & 4.4369\% \\ \end{array} $$

For \(date 2\), the expected value after the first time step is:

$$ 5.121\%+0.025\left( 15.339\%-5.121\% \right) \frac { 1 }{ 12 } =5.1423\% $$

Then after the second step, the expected value will be:

$$ 5.1423\%+0.025\left( 15.339\%-5.1423\% \right) \frac { 1 }{ 12 } =5.1635\% $$

From 5.5060%, the standard deviation of the rate and the expected rate are:

$$ 5.5060\%+0.025\left( 15.339\%-5.5060\% \right) \frac { 1 }{ 12 } =5.5265\% $$

and

$$ 1.26\%\sqrt { \frac { 1 }{ 12 } } =0.3637\% $$

To create a recombining tree matching this expectation and standard deviation, then:

$$ p\times r^{ uu }+\left( 1-p \right) \times 5.1635\%=5.5265 $$

and the standard deviation is:

$$ \sqrt { p{ \left( { r }^{ uu }-5.5265\% \right) }^{ 2 }+\left( 1-p \right) { \left( 5.51635\%-5.5265\% \right) }^{ 2 } } =0.3637\% $$

Solving the two problems simultaneously, then \({ r }^{ uu }=5.8909\%\) and \(p=0.4990\).

The same procedure can be used to compute \(q\) and \({ r }^{ dd }\) from 4.7786% to get \(q\) and \({ r }^{ dd }\) simultaneous to be 0.5011 and 4.4361% consecutively.

Where \({ r }^{ uu }\) and \({ r }^{ dd }\) are the up- and down-states consecutively. As described in the tree building for date 1 and 2, the process can continue until the desired period is achieved.

The Vasicek model’s expectation of the rate after \(T\) years is:

$$ { r }_{ 0 }e^{ -kT }+\theta \left( 1-{ e }^{ -kT } \right) $$

The mean-reverting parameter determines the rate of the exponential decay of the weight of the current short rate whereby expectation is the current short rate and its long-run value. However, the factor’s half-life is considered a more intuitive way to determine the time taken by the factor to progress half the distance towards its goal.

In the Vasicek model, the terminal distribution’s short rate after \(T\) years has the following standard deviation:

$$ \sqrt { \frac { { \sigma }^{ 2 }\left( 1-{ e }^{ -2kT } \right) }{ 2k } } $$

It is observed that a model with mean reversion and one without will result in different volatility term structures. Also, impact rates are lowered by mean reversion just as they lower volatilities of longer-term par rates.

The assumption that short rates tend towards a long-term goal is reinterpreted to understand better the mean reversion implication for the term structure for factor shape and volatility.

In this regard, if for many future years economic news will change the economy’s market view, then the news is said to be long-lived. If however, it changes the market’s view in the near future rather than far, it is said to be short-lived.

# Practice Questions

1) Assuming that we are provided with a set of data whose current short-term rate of 3.02% has a volatility of 103 basis points per year. We are also given a constant \(\lambda \) whose value is 0.225%. Using the model of drift and risk premium, compute the change in rate if the monthly realization of the random variable \(dw\) is 0.24.

- 0.0050%
- 0.2660%
- 0.0212%
- 0.0027%

The correct answer is **D**.

From model 2 (drift and risk premium model), we have:

$$ dr=\lambda dt+\sigma dw $$

We are given:

\(\lambda=0.225\%\)

\({ r }_{ 0 }=3.02\%\)

\(\sigma =1.03\%\) and

\(dw=0.24\)

and that the time interval under consideration is one month or \(\frac { 1 }{ 12 } \) years.

Hence:

$$ dr=0.225\%\times \frac { 1 }{ 12 } +1.03\%\times 0.24 $$

$$ \Rightarrow dr=0.0027\% $$

2) We are given the following for the Vasicek model:

$$ dr=k\left( { r }_{ \infty }-r \right) dt+\lambda dt+\sigma dw $$

$$ =k\left( \left[ { r }_{ \infty }+\frac { \lambda }{ k } \right] -r \right) dt+\sigma dw $$

The speed of mean reversion is 0.63, and the true interest rate process exhibits mean reversion to a long-term value \({ r }_{ \infty }\) of \(6.3\%\). We are also provided that the current short term rate is \(5.2\%\) and \(\sigma=3.1\%\). Using the model provided, determine the expected change in the short rate and the volatility over the next month consecutively if \(\lambda=0.3\%\).

- 69.3 basis points
- 9.2 basis points
- 8.3 basis points
- 5.8 basis points

The correct answer is **C**.

From the above information, we have that:

\(k=0.63\),

\({ r }_{ \infty }=6.3\%\),

\( \sigma=3.1\%\),

\({ r }_{ 0 }=5.2\%\) and

\(\lambda=0.256\%\).

Thus,the expected change in the short rate is:

$$ dr=k\left( \theta -r \right) dt+\sigma dt $$

$$ =k\left( \left[ { r }_{ \infty }+\frac { \lambda }{ k } \right] -r \right) dt+\sigma dw $$

Recall that:

$$ \theta \equiv { r }_{ \infty }+\frac { \lambda }{ k } $$

$$ \therefore \theta \equiv 6.3\%+\frac { 0.3\% }{ 0.63 } =6.78\% $$

$$ =0.63\times \left( 6.78\%-5.2\% \right) \frac { 1 }{ 12 } =0.0008295\quad or\quad 8.295\quad basis\quad points $$

3) A dataset has a mean-reverting parameter of 0.22 and a volatility of 103 basis points. If the short-term rate is normally distributed, then use the Vasicek model to determine the standard deviation of the terminal distribution of the short rate after 2 years.

- 0.0119
- 0.0001
- 0.0168
- 0.0093

The correct answer is **A**.

Recall that the standard deviation of the terminal distribution of the short rate after \(T\) years is:

$$ \sqrt { \frac { { \sigma }^{ 2 }\left( 1-{ e }^{ -2kT } \right) }{ 2k } } $$

Thus:

$$ \sqrt { \frac { { 0.0103 }^{ 2 }\left( 1-{ e }^{ -2\times 0.22\times 2 } \right) }{ 2\times 0.22 } } =0.0119 $$