The Science of Term Structure Models
After completing this reading, you should be able to: Calculate the expected discounted... Read More
After completing this reading, you should be able to:
In the previous chapters, the model of the short-term rate involved any of the following:
In this chapter, we introduce a model that has time-dependent volatility.
Time-dependent drift can be used to fit many bond or swap rates. In the same way, a time-dependent volatility function can be used to fit many option prices.
A simple model with a time-dependent volatility function might be written as follows:
$$ \text{dr}=\lambda(\text t)\text{dt}+\sigma( \text t)\text{dw} $$
A closer look at the function above reveals that this model augments Model 1 and the Ho-Lee model.
If the function \(\sigma(\text t)\) were such that \(\sigma(1)\) = 1.20% and \(\sigma(2)\) 1.30%, then the volatility of the short rate in one year is 120 basis points per year while the volatility of short rate in two years is 130 basis points per year. The relationships between volatility in various periods can take on an almost limitless number of combinations.
A special case of time–dependent volatility (which we call Model 3) assumes time- can be represented as follows:
$$ \text{dr}=\lambda (\text t)\text{dt}+\sigma {\text e}^{-\alpha \text{t}} \text{dw} $$
The volatility of the short rate starts at the constant \(\sigma\) and then exponentially declines to zero.
Assume a current short-term rate, \(\text r_0\), of 3%, a drift, \(\lambda\),of 0.24%, and, instead of constant volatility, include time-dependent volatility of \(\text e^{-0.3{\text t}}\) (with initial \(\sigma\) = 1.30%). Further, we assume the dw realization drawn with mean = 0 and standard deviation = \(\sqrt{\frac {1}{12}}\) takes on a value of 0.2. Determine the change in the short-term rate after one month.
$$ \begin{align*} dr & =\lambda(\text t)\text{dt}+\sigma {\text e}^{-\alpha {\text t}} {\text {dw}} \\ & =0.24\%×1/12+1.3\%×\text e^{-0.3×{\frac {1}{12}} }×0.2 \\ & =0.02\%+0.25\%=0.27\% \\ \end{align*} $$
After 1 month, therefore, the expected short-term rate is 3.27%.
Similarly, in the fifth year (month = 60), assume that the prior rate is 3.35% (rate in month 59). If the simulated random normal, dw = -0.35, then the rate change is given by:
$$ \begin{align*} dr & =\lambda(\text t)\text{dt}+\sigma {\text e}^{-\alpha {\text t}} {\text {dw}} \\ & =0.24×1/12+1.3×\text e^{-0.3×(5)}×-0.35 \\ & =0.02\%-0.10=-0.08 \\ & \quad \quad \quad \text{and} \\ r_5 & = 3.35\% – 0.08\% = 3.27\% \\ \end{align*} $$
As noted, the volatility of the short rate starts at sigma, but over time, declines exponentially toward zero. We can illustrate this using data from the above example,
At \(\text t = 0, \text e^{-0.3×(0)}=1.0\); volatility term \(=\sigma \text e^{-\alpha {\text t}} \text {dw}=1.3\%×1.0×\text{dw} \).
At \(\text t = 5, \text e^{-0.3×(0)}=0.223\); volatility term \(=\sigma \text e^{-\alpha {\text t}} \text {dw}=1.3\%×0.223×\text{dw} \).
At \(\text t = 10, \text e^{-0.3×(0)}=0.0498\); volatility term \(=\sigma \text e^{-\alpha {\text t}} \text {dw}=1.3\%×0.0498×\text{dw} \).
Time-dependent volatility models provide a useful tool for pricing fixed income options in situations where a precise market price is not easily observable. The models provide a means of interpolating from known to unknown option prices.
However, if the purpose of the model is to value and hedge fixed income securities, including options, then a model with mean reversion might be preferred for two reasons:
Time-dependent volatility models are also useful for pricing multi-period derivatives such as caplets and floorlets. A caplet also, called interest rate cap, is a type of derivative in which case, a buyer receives payments at the end of each period if the interest rate exceeds the agreed strike price. For example, the agreement can be such that a monthly payment is received if the monthly LIBOR rate exceeds 1.5%. The payment is equal to the difference between the interest rate and the strike and is usually made every month, three months, or six months. A floorlet works in the opposite direction – the holder collects payments for every period the interest rate is below the predetermined strike. The payoff to each caplet or floorlet is based on the strike rate and the current short-term rate over the next period. As such, their pricing will depend critically on the forecast of \(\sigma(\text t)\) at several future dates.
Time-dependent volatility models are also criticized because they forecast volatility far out into the future, which calls their long- to medium-term reliability into question.
The aforementioned models assume that the basis-point volatility of the short rate is independent of the level of the short rate. However, this is unlikely to be true at extreme levels of the short rate.
In periods with high inflation, short-term interest rates are usually high and inherently unstable. Consequently, the basis-point volatility of the short rate tends to be high.
When the short-term rate is very low, basis-point volatility is limited by the constraint that interest rates cannot decline much below zero. This has prompted calls to specify the basis-point volatility of the short rate as an increasing function of the short rate. The Cox-Ingersoll-Ross (CIR) model helps us to achieve just that.
The risk-neutral dynamics of the Cox-Ingersoll-Ross (CIR) model are
$$ \text{dr}=\text k(\theta-{\text r}){\text{dt}}+\sigma \sqrt{\text{r}} \text{dw} $$
In this equation, the first term on the right-hand side is not a random variable, and the standard deviation of dw equals \(\sqrt{\text{dt}}\). It follows that the annualized standard deviation of dr (i.e., the basis-point volatility) is proportional to the square root of the rate. Put differently, in the CIR model the parameter \(\sigma\) is constant, but basis-point volatility is not: annualized basis-point volatility equals \(\sigma \sqrt{\text r}\) and increases with the level of the short rate.
In essence, the CIR model exhibits mean reversion (i.e., the tendency of the short-term rate to move toward the equilibrium rate denoted by theta, \(\theta\)), just like the Vasicek model. However, the (CIR) model multiplies volatility by the square root of the level of the interest rate. Unlike the Vasicek model, the CIR model does not Allow for negative interest rates because of the square root component.
Assume a current short-term rate, \(\text r_0\), of 6%, a long-run value of the short-term rate, \(\theta\), of 20%, spread of mean reversion parameter, k, of 0.05 and a volatility, \(\sigma\), of 1.3%. Further, we assume the dw realization drawn with mean = 0 and standard deviation = \(\sqrt{\frac {1}{12}}\) takes on a value of 0.2. Using the CIR model, determine the change in the short-term rate after one month.
$$ \begin{align*} \text{dr} & =\text k(\theta-{\text r}){\text{dt}}+\sigma \sqrt{\text{r}} \text{dw} \\ & =0.05(20\%-6\%)\left(\frac {1}{12} \right)+1.3\% \sqrt{6}\% ×0.2 \\ & =0.0583\%+0.06369\%=0.122\% \end{align*} $$
Therefore, the expected short-term rate after one month is 6.122% (6% plus 0.122%).
Given the above parameters, the short-term rate would evolve as follows:
$$ \textbf{Evolution of Short-term Rates under the CIR Model } $$
$$ \begin{array}{l|c|c|c|c|c|c|c} \textbf{Month} & \textbf{(t)} & \textbf{Mean} & \textbf{dw} & \textbf{SQRT(r)} & \textbf{Volatility} & \textbf{dr} & \textbf{r} \\ \textbf{} & \textbf{} & \textbf{Reverting} & \textbf{} & \textbf{} & \textbf{term} & \textbf{} & \textbf{} \\ \textbf{} & \textbf{} & \textbf{Drift} & \textbf{} & \textbf{} & \bf{\sigma \sqrt{\text r}} & \textbf{} & \textbf{} \\ \textbf{} & \textbf{} & \textbf{term} & \textbf{} & \textbf{} & \textbf{dw} & \textbf{} & \textbf{} \\ \textbf{} & \textbf{} & \bf{\text k(\theta-\text r){\text dt}} & \textbf{} & \textbf{} & \textbf{} & \textbf{} & \textbf{} \\ \hline \bf{0} & {} & {} & {} & {} & {} & {} & \bf{6.00\%} \\ \hline {1} & {0.08} & {0.0583\%} & {0.20} & {24.29\%} & {0.0632\%} & {0.1215\%} & {6.1215\%} \\ \hline {2} & {0.17} & {0.0578\%} & {0.5145} & {24.74\%} & {0.1655\%} & {0.2233\%} & {6.3448\%} \\ \hline {3} & {0.25} & {0.0569\%} & {0.1235} & {25.19\%} & {0.0404\%} & {0.0973\%} & {6.4421\%} \\ \hline {4} & {0.33} & {0.0565\%} & {-0.4872} & {25.38\%} & {-0.1607\%} & {-0.1402\%} & {6.3379\%} \\ \hline {5} & {0.42} & {0.0569\%} & {-0.4522} & {25.18\%} & {-0.1480\%} & {-0.0911\%} & {6.2467\%} \\ \hline {6} & {0.58} & {0.0573\%} & {0.5531} & {24.99\%} & {0.1797\%} & {0.2370\%} & {6.4837\%} \\ \hline {7} & {0.83} & {0.0563\%} & {0.0568} & {25.46\%} & {0.0188\%} & {0.0751\%} & {6.5588\%} \\ \end{array} $$
The change in the short-term rate after 4 months, for example, is calculated as follows:
$$ \begin{align*} \text{dr} & =\text k(\theta-{\text r}){\text{dt}}+\sigma \sqrt{\text{r}} \text{dw} \\ & =0.05(20\%-6.4421\%)\left(\frac {1}{12} \right)+1.3\% \sqrt{6.} 4421\%×-0.4872 \\ & =0.0565\%-0.1607\%=-0.1042\% \\ \end{align*} $$
The short rate after 4 months is thus 6.3379% (= 6.4421% -0.1042%)
In the long-run, the interest rate reverts to the long-run value of 20%.
Note: The values of dw in the table are simulated random normals. Each step accepts a different random normal.
We also have other models where the basis-point volatility is proportional to the rate. In this case, the parameter \(\sigma\) is often called yield volatility.
$$ \text{dr} =\text k(\theta-{\text r}){\text{dt}}+\sigma \sqrt{\text{r}} \text{dw} $$
$$ \text{dr}=\text{ardt}+\sigma \text{rdw} $$
In these two specifications, yield volatility is constant but basis-point volatility equals \(\sigma {\text r}\) and increases with the level of the rate.
Let us now look at two versions of the lognormal model: one with deterministic drift and the other with mean reversion.
A lognormal model with a deterministic drift can be applied to a model created by Salomon Brothers in the ‘80s. The dynamics of the model are as follows:
$$ \text d[\text{ln(r)} ]=\text a(\text t)\text{dt}+\sigma \text{dw} $$
This equation states that the natural logarithm of the short interest rate is normally distributed. Remember that by definition, a random variable has a lognormal distribution if its natural logarithm has a normal distribution. Therefore, we can infer that the short rate has a lognormal distribution.
The lognormal model with deterministic drift has a few parallels with the Ho-Lee model. In the latter, drift can vary from period to period.
$$ \textbf{The Ho-Lee Model Interest Rate Tree} $$
$$ \textbf{Ho-Lee Model} $$
$$ \begin{array} \hline {} & {} & {} & {\scriptsize {1/2} } & { \text r }_{ 0 }+(\lambda_1 + \lambda_2) \text{dt}+2\sigma \sqrt { \text {dt} } \\ {} & {} & { \text r }_{ 0 }+\lambda_1 \text{dt}+\sigma \sqrt { \text { dt} } & {\Huge \begin{matrix} \nearrow \\ \searrow \end{matrix} } & {} \\ { \text r }_{ 0 } & {\begin{matrix} \scriptsize {1/2} \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \nearrow \\ \end{matrix} \\ \quad \quad \quad \Huge \searrow \end{matrix} \\ \scriptsize {1/2} \end{matrix} } & {} & {\scriptsize \begin{matrix} \begin{matrix} {1/2} \\\scriptsize \begin{matrix} \\ \end{matrix} \end{matrix} \\ \begin{matrix} \\ \end{matrix} \\ {1/2} \end{matrix} } & { \text r }_{ 0 }+(\lambda_1 + \lambda_2) \text{dt} \\ {} & {} & { \text r }_{ 0 }+\lambda_1 \text{dt}-\sigma \sqrt { \text {dt} } & {\Huge \begin{matrix} \nearrow \\ \searrow \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize {1/2}} & { \text r }_{ 0 }+(\lambda_1 + \lambda_2) \text{dt}-2\sigma \sqrt { \text {dt} } \\ \end{array} $$
We construct a binomial tree for the natural logarithm of the rate as follows:
$$ \textbf{Binomial Tree for the Natural Logarithm of the Rate} $$
$$ \begin{array} \hline {} & {} & {} & {\scriptsize {0.5} } & { \text {ln r} }_{ 0 }+(\text a_1 + \text a_2) \text{dt}+2\sigma \sqrt { \text {dt} } \\ {} & {} & { \text {ln r} }_{ 0 }+{\text a}_1 \text{dt}+\sigma \sqrt { \text { dt} } & {\Huge \begin{matrix} \nearrow \\ \searrow \end{matrix} } & {} \\ { \text {ln r} }_{ 0 } & {\begin{matrix} \scriptsize {0.5} \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \nearrow \\ \end{matrix} \\ \quad \quad \quad \Huge \searrow \end{matrix} \\ \scriptsize {0.5} \end{matrix} } & {} & {\scriptsize \begin{matrix} \begin{matrix} {0.5} \\\scriptsize \begin{matrix} \\ \end{matrix} \end{matrix} \\ \begin{matrix} \\ \end{matrix} \\ {0.5} \end{matrix} } & { \text {ln r} }_{ 0 }+(\text a_1 + \text a_2) \text{dt} \\ {} & {} & { \text {ln r} }_{ 0 }+\text a_1 \text{dt}-\sigma \sqrt { \text {dt} } & {\Huge \begin{matrix} \nearrow \\ \searrow \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize {0.5}} & { \text {ln r} }_{ 0 }+(\text a_1 + \text a_2) \text{dt}-2\sigma \sqrt { \text {dt} } \\ \end{array} $$
The binomial tree for the rate can be derived from the natural logarithm of the rate, by taking the exponential of each node. For example, the adjusted period 2 upper node would be calculated as:
$$ {\text e}^{\left[\text{ln r}_0+(\text a_1+\text a_2 )\text{dt}+2\sigma {\sqrt{\text{dt}}} \right] }=\text r_0 {\text e}^{\left[(\text a_1+\text a_2 )\text{dt}+2\sigma \sqrt{\text{dt}} \right] } $$
[Recall that \(\text e^{\text{ln(x)}} =\text x\)]
$$ \textbf{Lognormal Model Rates at Each Node } $$
$$ \begin{array} \hline {} & {} & {} & {\scriptsize {0.5} } & { \text {r} }_{ 0 } {\text e}^{\left[(\text a_1 + \text a_2) \text{dt}+2\sigma \sqrt { \text {dt} }\right]} \\ {} & {} & { \text {r} }_{ 0 } {\text e}^{\left[{\text a}_1 \text{dt}+\sigma \sqrt { \text { dt} }\right]} & {\Huge \begin{matrix} \nearrow \\ \searrow \end{matrix} } & {} \\ { \text {r} }_{ 0 } & {\begin{matrix} \scriptsize {0.5} \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \nearrow \\ \end{matrix} \\ \quad \quad \quad \Huge \searrow \end{matrix} \\ \scriptsize {0.5} \end{matrix} } & {} & {\scriptsize \begin{matrix} \begin{matrix} {0.5} \\\scriptsize \begin{matrix} \\ \end{matrix} \end{matrix} \\ \begin{matrix} \\ \end{matrix} \\ {0.5} \end{matrix} } & { \text {r} }_{ 0 } {\text e}^{\left[(\text a_1 + \text a_2) \text{dt}\right] } \\ {} & {} & { \text {r} }_{ 0 } {\text e}^{\left[\text a_1 \text{dt}-\sigma \sqrt { \text {dt} } \right]} & {\Huge \begin{matrix} \nearrow \\ \searrow \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize {0.5}} & { \text {r} }_{ 0 } e^{\left[ (\text a_1 + \text a_2) \text{dt}-2\sigma \sqrt { \text {dt} } \right]} \\ \end{array} $$
From these trees, we can see that the future movements of the short rate in a lognormal model are multiplicative instead of being additive as in normal models (such as the Ho-Lee model). Since \(\text e^{\text x}\) can only be positive for any value of x, the short rate in this model cannot turn negative as long as \(\text r_0\) is positive in every node of the lognormal tree.
For small values of x, we know that \(\text e^{\text x} \approx 1 + \text x\).
If we assume \(\alpha = 0\) and dt = 1, then:
$$ {\text r}_0 {\text e}^{ \sigma }\approx {\text r}_0 (1+\sigma) $$
Hence, we see that volatility is a percentage of the rate. For example if \(\sigma\) = 8.5%, then the short rate in the up state is 8.5% above the initial short rate.
The constants which explain the drift, (i.e., \(\text a_1\) and \(\text a_2\),) can be used to match market bond prices just like in the Ho-Lee model.
The lognormal distribution combined with a mean-reverting process is known as the Black-Karasinski model. This model is very flexible since volatility, mean reversion, and the central tendency of the short rate are all allowed to depend on time. The model is given as:
$$ \text d[\text {ln(r)} ]=\text k(\text t)[\text {ln} \theta (\text t)-\text {ln(r)} ]\text {dt}+\sigma (\text t)\text {dw} $$
What the above equation implies is that the natural logarithm of the short rate is normally distributed. It reverts to \(\text {ln} \theta (\text t)\) at a speed of k(t) with a volatility of \(\sigma (\text t)\).
As in the previous section where we looked at the lognormal model with deterministic drift, the binomial tree of this model can be represented in terms of the natural logarithm of the rate. The process over the first date takes the following shape:
$$ \textbf{Interest Rate Tree with Lognormal Model (Mean Revision)} $$
$$ \begin{array} \hline {} & {\scriptsize 0.5 } & { \text r }_{ 0 }{ \text e }^{\text k\left( 1 \right) \left( \text {ln } \theta \left( 1 \right) -\text {ln } { \text r }_{ 0 } \right) \text {dt}+\sigma \left( 1 \right) \sqrt { \text {dt} } }\equiv {\text r }_{ 1 }{\text e }^{ \sigma \left( 1 \right) \sqrt { \text {dt} } } \\ {\text r }_{ 0 } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {\scriptsize 0.5 } & {\text r }_{ 0 }{\text e }^{\text k\left( 1 \right) \left( \text {ln } \theta \left( 1 \right) -\text {ln }{ \text r }_{ 0 } \right) \text {dt}-\sigma \left( 1 \right) \sqrt {\text {dt} } }\equiv {\text r }_{ 1 }{ \text e }^{- {\sigma}\left( 1 \right) \sqrt { \text {dt} } } \\ \end{array} $$
The notation \(\text r_1\) is used to condense the exposition. As such,
$$ \text{In (upper node)} =\text{ln } \text r_1+\sigma (1) \sqrt{\text{dt}} $$
and
$$ \text{ln (lower node)} =\text{ln r}_1-\sigma (1) \sqrt{\text{dt}} $$
The tree will recombine in the second period only if:
$$ \text k(2)=\cfrac {\sigma (1)-\sigma (2)}{\sigma (1)\text {dt} } $$
As earlier illustrated in the mean-reverting model, it is possible to shift from a non-recombining tree into a combining tree. This is done by tweaking the probabilities in the second period. A similar adjustment can be made in this model. This time, however, we manipulate the time (dt) variable. After choosing a value for \(\text{dt}_1\), we determine a value for \(\text{dt}_2\) using the following equation:
$$ \text k(2)=\cfrac {1}{{\text {dt}}_2 } \left[ 1- \cfrac {\sigma (2) \sqrt{\text{dt}_2 }}{\sigma (1) \sqrt{\text{dt}_1 }} \right] $$
Question 1
You have been asked to employ the Cox-Ingersoll-Ross (CIR) model for the short-term rate process and you have been given the following figures:
\(\sigma=0.02\), \(r = 0.011\), \(\theta = 0.053\), \(k = 0.44\)
An initial interest rate of 1% is assumed.
If for the first month, \(dw = 0.12\), then, what is the short-rate in the first month under this CIR process, \(r\left( { 1 }/{ 12 } \right) \)?
- 1.23%.
- 0.23%.
- 0.16%.
- 1.16%.
The correct answer is A.
Recall that from the CIR model, we have:
$$ dr=k\left( \theta -r \right) dt+\sigma \sqrt { r } dw $$
From the data provided in the question:
\(\sigma = 0.02\), \(r = 0.011\), \(\theta = 0.053\), \(k = 0.44\), \(dw = 0.12\)
Therefore:
$$ dr=0.44\left( 0.053-0.011 \right) \times \frac { 1 }{ 12 } +0.02\sqrt { 0.011 } \times 0.12 =0.0023 = 0.23\% $$
The short rate in the first month under this CIR model is:
$$ 1\% + 0.23\% = 1.23\% $$
Question 2
A binomial tree is constructed by an analyst according to a lognormal model. The analyst makes the following assumptions:
- The time-step is monthly, \(dt={ 1 }/{ 12 }\).
- Today’s initial rate, \(r\left( 0 \right) =3.16\%\).
- Annual drift is constant at 46 basis points.
- The annual basis point volatility, \(\sigma = 8.32\%\).
What is the rate at node \(\left[ 1,1 \right] \left( X \right) \) if:
$$ \begin{array} \hline {} & {} & {} & {\scriptsize 0.5 } \\ {} & {} & { _{ \_ }{ X }_{ \_ } } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } \\ 3.160 & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & \scriptsize 0.5 \\ {} & {} & 3.086 & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } \\ {} & {} & {} & {\scriptsize 0.5} \\ \end{array} $$
- 3.238%.
- 3.160%.
- 3.458%.
- 3.065%.
The correct answer is A.
Remember that node \(\left[ 1,1 \right]\) is given by:
$$ r_{ 0 }{ e }^{ { a }_{ 1 }dt+\sigma \sqrt { dt } } $$
From the question,we have:
\(r_{ 0 }=0.0316\), \(dt={ 1 }/{ 12 }\), \(\sigma=0.0832\), \({ a }_{ 1 }=0.0046\)
Therefore at node \(\left[ 1,1 \right]\):
$$\begin{align*}X&=0.0316\times { e }^{ 0.0046\times { 1 }/{ 12+0.0832\times \sqrt { { 1 }/{ 12 } } } }\\&=0.03238=3.238\% \end{align*}$$