 ### The Evolution of Short Rates and the Shape of the Term Structure

This chapter will help explain the role of interest rate expectation in determining the shape of the term structure. It shows how spot or forward rates are determined by expectations of future short-term rates, the volatility of short-term rates and an interest rate premium.

Unlike with the arbitrage-free models (those that take the initial term structure as given), a different approach starts with an assumption about the interest rate process and the risk premium demanded by the market for bearing interest rate risk. Then, deriving the risk-neutral process can also be used.These models are called equilibrium models.

Note: Equilibrium models, in the context of their assumption, which does not include market prices for the initial term structure, are also arbitrage-free.

# Expectations

Expectations refer to uncertainty. In determining the shape of the term structure, the role of interest rate forecasting can be highlighted by the following consideration: that one-year interest rate is currently 10% and all investor forecast this will be the case the next year and the year after. In such a case, investors will discount cash flows using forward rates of 10%. The price of one, two and three-year zero-coupon bonds per dollar face value will be shown in the following example, where the term structure of spot rate is flat at 10%:

$${ P }^{ 1 }=\frac { 1 }{ 10 }$$

$${ P }^{ 2 }=\frac { 1 }{ \left( 1.10 \right) \left( 1.10 \right) } =\frac { 1 }{ { 1.10 }^{ 2 } }$$

$${ P }^{ 3 }=\frac { 1 }{ \left( 1.10 \right) \left( 1.10 \right) \left( 1.10 \right) } =\frac { 1 }{ { 1.10 }^{ 3 } }$$

Assuming that the one-year rate is still 10% but all investors forecast the one-year rate next year to be 12% and 14% the year after, then the one-year spot rate will still be 10%, but the two-year spot rate $$\rho \left( 2 \right)$$ will be given as:

$${ \rho }^{ 2 }=\frac { 1 }{ \left( 1.10 \right) \left( 1.12 \right) } =\frac { 1 }{ { \left( 1+\rho \left( 2 \right) \right) }^{ 2 } }$$

$$\Rightarrow \rho \left( 2 \right) =10.995\%$$

And the three-year spot rate is such that:

$${ \rho }^{ 3 }=\frac { 1 }{ \left( 1.10 \right) \left( 1.12 \right) \left( 1.14 \right) } =\frac { 1 }{ { \left( 1+\rho \left( 3 \right) \right) }^{ 3 } }$$

From this example, the evolution of the one-year rate from 10%, to 12% to 14% generates an upward sloping term structure of spot rates. In such cases, investors require rates above 10% when locking up their money for two or three years.

On the other hand, assume that a one-year rate is 10% but investors forecast it to fall to 8% and 6% in one and two years respectively. In such a case, it can be shown that the term structure of spot rates will be downward-sloping.

From this example, it is clear that expectations can cause the term structure to take on different shapes. Over short horizons, the financial community can have a very specific view of future short-term rates. Over the long-term horizon, expectations cannot be so granular. It is therefore very clear that forecasts can be useful in describing the shape and level of the term structure over a short horizon and the level of the term rates at very long horizons.

# Volatility and Convexity

The assumption that investors are risk-neutral in pricing securities using expected discounted values is important if we are to isolate the implications of volatility on the shape of the term structure. Unlike in the previous section with no volatility around expectations, a flat expectation of 10% implies a flat term structure of spot rates.

Consider the following interest rate tree:

$$\begin{array} {} & {} & {} & {\scriptsize { 1 }/{ 2 } } & 14\% \\ {} & {} & 12\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ 10\% & {\begin{matrix} \scriptsize { 1 }/{ 2 } \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize { 1 }/{ 2 } \end{matrix} } & {} & {\scriptsize \begin{matrix} \begin{matrix} { 1 }/{ 2 } \\\scriptsize \begin{matrix} \\ \end{matrix} \end{matrix} \\ \begin{matrix} \\ \end{matrix} \\ { 1 }/{ 2 } \end{matrix} } & 10\% \\ {} & {} & 8\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize { 1 }/{ 2 }} & 6\% \\ \end{array}$$

From the above, unlike with flat term structure, the price of a one year zero is 0.09091 or 1/1.10. Under the assumption of risk-neutrality, the price of a two-year zero may be calculated by discounting the terminal cash flow. The two-year spot rate is such that

$$0.82672={ \left( 1+p\left( 2 \right) \right) }^{ -2 }$$ implying that $$p\left( 2 \right) =9.982\%$$.

The 1.8 basis point difference between the spot rate that we would obtain in the absence of uncertainty(10%) and the spot rate in the presence of volatility (9.982%) is the effect of convexity on the spot rate. This effect arises from the mathematical fact, a special case of Jensen’s Inequality, that:

$$E\left[ \frac { 1 }{ 1+r } \right] >\frac { 1 }{ E\left[ 1+r \right] } =\frac { 1 }{ 1+E\left[ r \right] }$$

From the example in this section, a three-year spot such that $$.752309={ \left( 1+p\left( 3 \right) \right) }^{ -3 }$$ is $$9.952\%$$. The value of convexity in this spot rate is $$10\% – 9.952\%$$ or $$4.8$$ basis points. Note, the value of convexity in the two-year spot rate was only 1.8 basis points. It is thus clear that all else equal, the value of convexity increases with maturity.

From this, securities with greater convexity perform better when yields change a lot and perform worse when yields do not change by much. Given that yield are determined by forecast, with an introduction of volatility, the yield is reduced by the value of convexity which arises from volatility.

Risk premium effect on the term structure can be illustrated by taking into consideration an interest rate tree with a 400 basis points volatility per annum. Risk-neutral investors would price a two-year zero by the following calculation:

$$0.827541=\frac { 0.5\left[ \frac { 1 }{ 1.14 } +\frac { 1 }{ 1.06 } \right] }{ 1.10 }$$

$$=\frac { 0.5\left[ 0.877193+0.943396 \right] }{ 1.10 }$$

Over the next year, the expected return from owning the 2-year zero is 10% by discounting the expected future price by 10%. Compared to an investment having a risky return averaging at 10%, investors will go for an investment with a certain return of 10% if they have the choice. The implication here is that investors require compensation for bearing interest rate risk.

Over the next year, a demand of higher than 10% for the 2-year zero is what risk-averse investors will put forward. One year later, the pricing of a zero-coupon bond at less than the prices $$\frac { 1 }{ 1.14 }$$ and $$\frac { 1 }{ 1.06 }$$ is a way to effect the return. Future cash flows could be discounted at rates higher than the possible rates of 14% and 6%.

Consider a situation in which 20 basis points are added to each of the rates; it can be proved that this is equivalent to an assumption that for each year of duration risk, an extra 20 basis points are demanded by investors. To compute the price of the two-year zero, it is assumed that the above situation is truly the fair market risk premium and is calculated as follows:

$$0.826035=\frac { 0.5\left[ \frac { 1 }{ 1.142 } +\frac { 1 }{ 1.062 } \right] }{ 1.10 }$$

From this, the price obtained is lower than previously obtained,and the assumption is that investors are risk-neutral; the increase in the expected return of the two-year zero is due to an increase in the discounting rates. In one year, if the interest rate is 14%, then the price of a one-year zero will be 0.877193. If the rate is 6%, then the price will be 0.943396. Hence, the expected return of the two-year zero priced at 0.826035 is:

$$\frac { 0.5\left[ 0.877193+0.943396 \right] -0.826035 }{ 0.826035 } =10.20\%$$

Still, under the same assumption of 20 basis points required by investors in each duration risk year, a return of 40 basis points should be offered by the three-year zero.

Pricing the three-year zero assuming the following years rates are above their true values by 20 basis points and 40 basis points above their true values in the consecutive year affects the returns.

In summary, actual or true interest rate processes are summarized by the following tree ($$a$$) while in tree ($$b$$) investors are provided with a 20 basis points risk premium for each of the duration week,and tree ($$c$$) will help compute the price of three-year zero.

$$\begin{array} \hline {} & {} & {} & {\scriptsize 0.5 } & 18\% \\ {} & {} & 14\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ 10\% & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & \scriptsize 0.5 & 10\% \\ {} & {} & 6\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize 0.5} & 2\% \\ \end{array}$$

(a)

$$\begin{array} \hline {} & {} & {} & {} & 18.4\% \\ {} & {} & 14.2\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ 10\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} & {} & 10.4\% \\ {}& {} & 6.2\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {}& {} & {} & {} & 2.4\% \\ \end{array}$$

(b)

$$\begin{array} \hline {} & {} & {} & & {} & {\scriptsize 0.5 } & 1 \\ {} & {} & {} & {\scriptsize 0.5 } & 0.844595 & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } \\ {} & {} & 0.766371 & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} & {\scriptsize 0.5 } & 1\\ 0.751184 & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & \scriptsize 0.5 & 0.905797 & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } \\ {} & {} & 0.86234 & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} & {\scriptsize 0.5 } & 1 \\ {} & {} & {} & {\scriptsize 0.5} & 0.976563 & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } \\ {} & {} & {} & {} & {} & {\scriptsize 0.5 } & 1 \\ \end{array}$$

(c)

Above a one-year zero’s certain of 10 %, the three-year zero return is 40 basis points in order to compensate investors for two years of duration risk.

With the assumption of 400 basis-point volatility, graphing the term structure of spot rates for three cases: no risk premium, a risk premium of 2 basis points and a risk premium of 40 basis point, the following observations are made:

1. In the case of no risk premium the term structure of spot rates is downward-sloping due to convexity
2. A risk premium of 20 basis points pushes up spot rates while convexity pulls them down, in the short end, the risk premium effect dominates,and the term structure is mildly upward-sloping
3. A risk premium of 40 basis points dominates the convexity effect,and the term structure of spot rates is upward-sloping. The convexity effect is still evident.

This section assumes that bonds with interest rate risk earn a premium. An argument supporting this assumption is that interest ratesfall when inflation is correlated with good times.

# Practice Questions

1) Investors value the current one-year interest rate at 8.30%. If they also forecast that for the following year, the one-year interest rate will be 9.43%, then the two-year spot rate, $$\rho \left( 2 \right)$$, is closest to:

1. 8.86%
2. 9.43%
3. 18.51%
4. 9.26%

The correct answer is A.

The two- year spot rate $$\rho \left( 2 \right)$$ is such that:

$${ P }^{ 2 }=\frac { 1 }{ \left( 1.0830 \right) \left( 1.0943 \right) } =\frac { 1 }{ { \left( 1+\rho \left( 2 \right) \right) }^{ 2 } }$$

$$\Rightarrow { \left( 1+\rho \left( 2 \right) \right) }^{ 2 }=\left( 1.0830 \right) \left( 1.0943 \right)$$

$$\rho \left( 2 \right) =0.886=8.86\%$$

2) Assume that the following tree gives the true process for the one-year rate.

$$\begin{array} \hline {} & {} & {} & {\scriptsize 0.5 } & 22\% \\ {} & {} & 18.5\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ 15\% & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & \scriptsize 0.5 & 15\% \\ {} & {} & 11.5\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize 0.5} & 8\% \\ \end{array}$$

Compute the expected interest rate for date 1 and 2, respectively.

1. Both are 15%
2. 12% and 19%
3. 18.5% and 11.5%
4. 8% and 22%

The correct answer is A.

The expected interest rate on $$date \quad 1$$ is:

$$0.5\times 11.5+0.5\times 18.5=15\%$$

The expected interest rate for $$date \quad 2$$ is:

$$0.25\times 22+0.5\times 15+0.25\times 8=15\%$$

3) Using the data provided in the following tree, apply the Jensen’s inequality for estimation of connectivity to show why the 1-year spot rate is less than 15%.

$$\begin{array} \hline {} & {} & {} & {\scriptsize 0.5 } & 22\% \\ {} & {} & 18.5\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ 15\% & {\begin{matrix} \scriptsize 0.5 \\ \begin{matrix} \begin{matrix} \quad \quad \quad \Huge \diagup \\ \end{matrix} \\ \quad \quad \quad \Huge \diagdown \end{matrix} \\ \scriptsize 0.5 \end{matrix} } & {} & \scriptsize 0.5 & 15\% \\ {} & {} & 11.5\% & {\Huge \begin{matrix} \diagup \\ \diagdown \end{matrix} } & {} \\ {} & {} & {} & {\scriptsize 0.5} & 8\% \\ \end{array}$$

1. 0.853 >0.657
2. 0.798 >0.713
3. 0.758 >0.756
4. 0.658 >0.456

The correct answer is C.

According to the Jensen’s inequality:

$$E\left[ \frac { 1 }{ 1+r } \right] >\frac { 1 }{ E\left[ 1+r \right] } =\frac { 1 }{ 1+E\left[ r \right] }$$

$$\Rightarrow 0.5\times \frac { 1 }{ 1.185 } +0.5\times \frac { 1 }{ 1.115 } >\frac { 1 }{ 0.5\times 1.185+0.5\times 1.115 } =\frac { 1 }{ 1.15 }$$

Dividing both sides by 1.15:

$$\Rightarrow \frac { 1 }{ 1.15 } \left[ 0.5\times \frac { 1 }{ 1.185 } +0.5\times \frac { 1 }{ 1.115 } \right] >\frac { 1 }{ { 1.15 }^{ 2 } }$$

$$\Rightarrow 0.758>0.756$$