 ### The Art of Term Structure Models: Volatility and Distribution

The main point of interest in this chapter is the volatility of interest rates and the models in which rates lack normal distribution.

# Time-Dependent Volatility: Model 3

One of the roles of time-dependent volatility function is fitting many option prices in a similar manner time-dependent drift fits the swap rates or many bonds. The volatility function applies to the following model (3):

$$dr=\lambda \left( t \right) dt+\sigma \left( t \right) dw\quad \quad \quad \quad \quad \left( I \right)$$

In the equation, the short rate volatility is time-dependent. The following is a special case of equation ($$I$$) that should be considered when illustrating the features of time-dependent volatility:

$$dr=\lambda \left( t \right) dt+\sigma { e }^{ -\alpha t }dw\quad \quad \quad \quad \quad \left( II \right)$$

Starting at constant $$\sigma$$, the short rate volatility declines to zero in an exponential manner but could have been designed to decline to another constant rather than zero. Assuming that $$\sigma=126$$ basis points and $$\alpha=0.025$$. At various horizons graph, the standard deviation’s short rate terminal distribution would show at first a rapid rise of the standard distribution which then would slow down.

The volatility function chose for equation ($$II$$) determines the shape of the curve with the number of shapes dependent on the specifications of the general volatility.

The impact of mean reversion on the horizon standard deviation is resembled by the behavior of standard deviation as a horizon function. However, the models remain different in other ways apart from some parameterizations which give equivalent terminal distributions.

Model 3 is a parallel shift model whose term structure of volatility at levels changing over time is flat since its volatility changes over time.

# The Cox-Ingersoll-Ross (CIR) and Lognormal Models: Volatility as a Function of the Short Rate

At extreme levels of the short rate, the assumption, by the models seen so far, that the short rate level and short rate’s basis-point volatility are independent is almost untrue. The basis-point of short rates tends to be high due to inherent instability in high inflation periods and short-term interest rates that are high.

As a result of the inability by interest rates to decline below zero, basic volatility is limited when the rates are very low.

The following are the risk-neutral dynamics of the CIR model:

$$dr=k\left( \theta -r \right) dt+\sigma \sqrt { r } dw$$

There exist a proportionality in the square root of the rate and the annulated standard deviation of $$dr$$ due to the fact that $$\sqrt { dt }$$ is the standard deviation of $$dw$$ and $$k$$ not being a random variable.

Also proportional to rate is the basis point volatility whereby the parameter $$\sigma$$ is popularly called yield volatility whose examples are Courtadon and Lognormal models, respectively presented below.

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

and

$$dr=ardt+\sigma rdw$$

While basis point volatility equals $$\sigma r$$ and increases in the level of the rate yield volatility is constant in the above two specifications.

At a short rate of 8%, $$\sigma$$ is set such that basis-point volatility equals to 100 for the purpose of comparison. Mathematically:

$${ \sigma }^{ bp }=0.1$$

$${ \sigma }^{ CIR }\times \sqrt { 8\% } =1\%\Rightarrow { \sigma }^{ CIR }=0.0354$$

$${ \sigma }^{ y }\times 8\%=1\%\Rightarrow { \sigma }^{ y }=12.5$$

An increase with rate at different speed of the basis point volatility of $$CIR$$ and proportional volatility specification is visible in their graphs with both models having zero rates and zero basis point volatility. The short rate is always positive due to this property combined with the condition that thezero rate implies positive drift.

# Tree for the Original Salomon Brothers Models

The following are the dynamics of this model:

$$dr=\tilde { a } \left( t \right) rdt+\sigma rdw\quad \quad \quad \quad \left( III \right)$$

And:

$$d\left[ ln\left( r \right) \right] =\frac { dr }{ r } -\frac { 1 }{ 2 } { \sigma }^{ 2 }dt\quad \quad \quad \quad \left( IV \right)$$

Substituting ($$III$$) into ($$IV$$):

$$d\left[ \tilde { a } \left( t \right) -\frac { 1 }{ 2 } { \sigma }^{ 2 } \right] dt+\sigma dw\quad \quad \quad \quad \left( V \right)$$

In the time-dependent drift notation, letting:

$$a\left( t \right) =\tilde { a } \left( t \right) -\frac { 1 }{ 2 } { \sigma }^{ 2 }$$

equation ($$V$$) becomes:

$$d\left[ ln\left( r \right) \right] =a\left( t \right) dt+\sigma dw\quad \quad \quad \quad \left( VI \right)$$

The implication in equation ($$VI$$) is that for the short rate, the natural logarithm is normally distributed. If the natural logarithm of the random variable is normally distributed, then its distribution is lognormal as implied by equation ($$VI$$) whose distribution is lognormal.

The tree for the first three dates adopting the Ho-Lee model tree is:

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

As opposed to the natural logarithm of rate, the tree can be in rate by exponentiation each node as follows:

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

In normal models, perturbations are additive, a situation which is different from the above tree depicting short rate lognormal model perturbations as multiplicative, thus the reason why short rates in the model are always positive. The positive rate in every node of the lognormal tree is due to the fact that $${ e }^{ x }$$ is nonnegative for any $$x$$ value as long as $${ r }_{ 0 }$$ is also nonnegative

From the tree, it is also clear why there is a percentage of the rate expression of lognormal model volatility.

To match the market bond prices, constants that determine drift can be applied just the same way as in Ho-Lee model.

# The Black-Karasinski Model: A Lognormal model with Mean Reversion

The model is firmly placed in the arbitrage-free class due to the fact that it allows volatility, mean reversion and the short rate’s central tendency to be time-dependent at the same time allowing a user to remove as much time dependence as they wish.

The model’s dynamics are written as:

$$dr=k\left( t \right) \left( ln\tilde { \theta } \left( t \right) -lnr \right) dt+\sigma rdw$$

Or as:

$$d\left[ ln\quad r \right] =k\left( t \right) \left( ln\theta \left( t \right) -ln r \right) dt+\sigma \left( t \right) dw$$

The implication here is that the short rate is normally distributed. The version of the Vasicek model followed here by the short rate’s natural logarithm is dependent on time.

The corresponding tree can be written in terms of the rate and in terms of the natural logarithm as depicted below:

$$\begin{array} \hline {} & {\scriptsize 0.5 } & { r }_{ 0 }{ e }^{ k\left( 1 \right) \left( ln \theta \left( 1 \right) -ln{ r }_{ 0 } \right) dt+\sigma \left( 1 \right) \sqrt { dt } }={ r }_{ 1 }{ e }^{ v\left( 1 \right) \sqrt { dt } } \\ { r }_{ 0 } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {\scriptsize 0.5 } & { r }_{ 0 }{ e }^{ k\left( 1 \right) \left( ln \theta \left( 1 \right) -ln{ r }_{ 0 } \right) dt-\sigma \left( 1 \right) \sqrt { dt } }={ r }_{ 1 }{ e }^{- v\left( 1 \right) \sqrt { dt } } \\ \end{array}$$

In the up- and down-stares, the natural logarithms of the rates are shown below respectively,and $${ r }_{ 1 }$$ is introduced for readability.

$$ln{ r }_{ 1 }+\sigma \left( 1 \right) \sqrt { dt }$$

And:

$$ln{ r }_{ 1 }-\sigma \left( 1 \right) \sqrt { dt }$$

The following rate is required by the step down from the up-state:

$${ r }_{ 1 }{ e }^{ \sigma \left( 1 \right) \sqrt { dt } }{ e }^{ k\left( 2 \right) -\left[ ln\theta \left( 2 \right) – \left\{ ln{ r }_{ 1 }+\sigma \left( 1 \right) \sqrt { dt } \right\} \right] dt-\sigma \left( 2 \right) \sqrt { dt } }\quad \quad \quad \quad \left( VII \right)$$

Whereas the following rate is required by the step up from the down-state:

$${ r }_{ 1 }{ e }^{- \sigma \left( 1 \right) \sqrt { dt } }{ e }^{ k\left( 2 \right) -\left[ ln\theta \left( 2 \right) – \left\{ ln{ r }_{ 1 }-\sigma \left( 1 \right) \sqrt { dt } \right\} \right] dt+\sigma \left( 2 \right) \sqrt { dt } }\quad \quad \quad \quad \left( VIII \right)$$

Only under the following condition will the tree recombine:

$$k\left( 2 \right) =\frac { \sigma \left( 1 \right) -\sigma \left( 2 \right) }{ \sigma \left( 1 \right) dt }$$

The following up-down and down-up rates are obtained by applying steps labeled $${ dt }_{ 1 }$$ and $${ dt }_{ 2 }$$ to equations ($$VII$$) and ($$VIII$$) :

Therefore the tree will recombine under the following condition

$$k\left( 2 \right) =\frac { 1 }{ d{ t }_{ 2 } } \left[ 1-\frac { \sigma \left( 2 \right) \sqrt { d{ t }_{ 2 } } }{ \sigma \left( 1 \right) \sqrt { d{ t }_{ 1 } } } \right] \quad \quad \quad \quad \quad \left( IX \right)$$

To allow freedom in independently choosing the mean reversion and volatility functions, equation ($$IX$$) is satisfied by the length of the second time step as opposed to the length of the first time step which can be arbitrarily set.

# Closed Form Solution for Spot Rates

The following are spot rates for the continuously compounded spot rates for terms $$T$$ and $$\hat { r } \left( t \right)$$:

Model 1:

$$\hat { r } \left( t \right) ={ r }_{ 0 }-\frac { { \sigma }^{ 2 }{ T }^{ 2 } }{ 6 }$$

Model 2:

$$\hat { r } \left( t \right) ={ r }_{ 0 }+\frac { \lambda T }{ 2 } -\frac { { \sigma }^{ 2 }{ T }^{ 2 } }{ 6 }$$

Vasicek:

$$\hat { r } \left( t \right) =\theta +\frac { 1-{ e }^{ -kT } }{ kT } \left( { r }_{ 0 }-\theta \right) -\frac { { \sigma }^{ 2 } }{ 2{ k }^{ 2 } } \left( 1+\frac { 1-{ e }^{ -kT } }{ kT } -2\frac { 1-{ e }^{ -kT } }{ kT } \right)$$

Model 3 with $$\lambda \left( t \right) =\lambda$$:

$$\hat { r } \left( t \right) ={ r }_{ 0 }+\frac { \lambda T }{ 2 } -{ \sigma }^{ 2 }\frac { 2{ \alpha }^{ 2 }{ T }^{ 2 }-2\alpha T+1-e^{ -kT } }{ 8{ \alpha }^{ 3 }{ T } }$$

Letting $$P\left( T \right)$$ be the price of a zero-coupon bond maturing at time $$T$$, then:

$$P\left( T \right) =A{ e }^{ -B\left( T \right) { r }_{ 0 } }$$

$$A\left( T \right) ={ \left[ \frac { 2he^{ \left( k+h \right) { T }/{ 2 } } }{ 2h=\left( k+h \right) \left( { e }^{ hT }-1 \right) } \right] }^{ { 2k\theta }/{ { \sigma }^{ 2 } } }$$

$$B\left( T \right) =\frac { 2\left( { e }^{ hT }-1 \right) }{ 2h+\left( k+h \right) \left( { e }^{ hT }-1 \right) }$$

$$h=\sqrt { { k }^{ 2 }+2{ \sigma }^{ 2 } }$$

# Practice Questions

1) You have been asked to employ the Cox-Ingersoll-Ross (CIR) model for the short-term rate process and 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. 1.23%
2. 0.23%
3. 0.16%
4. 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:

$$dr=0.44\left( 0.053-0.011 \right) \times \frac { 1 }{ 12 } +0.02\sqrt { 0.011 } \times 0.12$$

$$\Rightarrow dr=0.0023 = 0.23\%$$

The short rate in the first month under this CIR model is:

$$1\% + 0.23\% = 1.23\%$$

2) A binomial tree is constructed by an analyst according to a lognormal model. The analyst makes the following assumptions:

1. The time-step is monthly, $$dt={ 1 }/{ 12 }$$
2. Today’s initial rate, $$r\left( 0 \right) =3.16\%$$
3. Annual drift is constant at 46 basis points
4. The annual basis point volatility, $$\sigma = 8.32\%$$

What is the rate at node $$\left[ 1,1 \right] \left( X \right)$$ if:

$$\begin{array} \hline {} & {} & {} & {\scriptsize 0.5 } \\ {} & {} & { _{ \_ }{ X }_{ \_ } } & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } \\ 3.160 & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & \scriptsize 0.5 \\ {} & {} & 3.086 & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } \\ {} & {} & {} & {\scriptsize 0.5} \\ \end{array}$$

1. 3.238%
2. 3.160%
3. 3.458%
4. 3.065%

The correct answer is A.

Recall 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 $$\left[ 1,1 \right]$$:

$$X=0.0316\times { e }^{ 0.0046\times { 1 }/{ 12+0.0832\times \sqrt { { 1 }/{ 12 } } } }$$

$$X=0.03238=3.238\%$$