Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the term structure of interest rates equation, also known as the expectations hypothesis theory of interest rates,

###\left(1+r_\text{0-2yrs, eff 6mth}\right)^4 = \left(1+r_\text{0-1yrs, eff 6mth}\right)^2\left(1+r_\text{1-2yrs, eff 6mth}\right)^2 ### ###\left(1+\frac{r_\text{0-2yrs, apr 6mth}}{2}\right)^4 = \left(1+\frac{r_\text{0-1yrs, apr 6mth}}{2}\right)^2\left(1+\frac{r_\text{1-2yrs, apr 6mth}}{2}\right)^2 ### ###\left(1+\frac{0.1}{2}\right)^4 = \left(1+\frac{0.08}{2}\right)^2\left(1+\frac{r_\text{1-2yrs, apr 6mth}}{2}\right)^2 ### ###\left(1+\frac{r_\text{1-2yrs, apr 6mth}}{2}\right)^2 = \frac{\left(1+\frac{0.1}{2}\right)^4}{\left(1+\frac{0.08}{2}\right)^2} ### ###\begin{aligned} r_\text{1-2yrs, apr 6mth} &= \left( \left( \frac{\left(1+\frac{0.1}{2}\right)^4}{\left(1+\frac{0.08}{2}\right)^2} \right)^{1/2} - 1 \right) \times 2\\ &= 0.1202 \\ \end{aligned} ###

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the term structure of interest rates equation, also known as the expectations hypothesis theory of interest rates,

###\left(1+r_{\text{0}\rightarrow\text{4yr, eff yrly}}\right)^4 = \left(1+r_{\text{0}\rightarrow\text{3yr, eff yrly}}\right)^3 \left(1+r_{\text{3}\rightarrow\text{4yr, eff yrly}}\right)^1 ### ###\left(1+0.065\right)^4 = \left(1+0.06 \right)^3 \left(1+r_{\text{3}\rightarrow\text{4yr, eff yrly}}\right)^1 ### ###1+r_{\text{3}\rightarrow\text{4yr, eff yrly}} = \frac{\left(1+0.065\right)^4}{\left(1+0.06 \right)^3} ### ###\begin{aligned} r_{\text{3}\rightarrow\text{4yr, eff yrly}} &= \frac{\left(1+0.065\right)^4}{\left(1+0.06 \right)^3} - 1\\ &= 0.080141955 \\ \end{aligned} ###

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the term structure of interest rates equation, also known as the expectations hypothesis theory of interest rates,

###(1+r_{0-3})^3 = (1+r_{0-2})^2(1+r_{2-3}) ### ###(1+0.1)^3 = (1+0.08)^2(1+r_{2-3}) ### ###1+r_{2-3} = \frac{(1+0.1)^3}{(1+0.08)^2} ### ###\begin{aligned} r_{2-3} =& \frac{(1+0.1)^3}{(1+0.08)^2} - 1 \\ =& 0.14111797 \\ \end{aligned} ###

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the term structure of interest rates equation, also known as the expectations hypothesis theory of interest rates,

###(1+r_{0-2})^2 = (1+r_{0-1})(1+r_{1-2}) ### ###(1+0.1)^2 = (1+0.08)(1+r_{1-2}) ### ###1+r_{1-2} = \frac{(1+0.1)^2}{1+0.08} ### ###\begin{aligned} r_{1-2} =& \frac{(1+0.1)^2}{1+0.08} - 1 \\ =& 0.12037037 \\ \end{aligned} ###

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the term structure of interest rates equation, also known as the expectations hypothesis theory of interest rates,

###\left(1+r_\text{0-2yrs, eff 6mth}\right)^4 = \left(1+r_\text{0-1yrs, eff 6mth}\right)^2\left(1+r_\text{1-2yrs, eff 6mth}\right)^2 ### ###\left(1+\frac{r_\text{0-2yrs, apr 6mth}}{2}\right)^4 = \left(1+\frac{r_\text{0-1yrs, apr 6mth}}{2}\right)^2\left(1+\frac{r_\text{1-2yrs, apr 6mth}}{2}\right)^2 ### ###\left(1+\frac{0.08}{2}\right)^4 = \left(1+\frac{0.1}{2}\right)^2\left(1+\frac{r_\text{1-2yrs, apr 6mth}}{2}\right)^2 ### ###\left(1+\frac{r_\text{1-2yrs, apr 6mth}}{2}\right)^2 = \frac{\left(1+\frac{0.08}{2}\right)^4}{\left(1+\frac{0.1}{2}\right)^2} ### ###\begin{aligned} r_\text{1-2yrs, apr 6mth} &= \left( \left( \frac{\left(1+\frac{0.08}{2}\right)^4}{\left(1+\frac{0.1}{2}\right)^2} \right)^{1/2} - 1 \right) \times 2\\ &= 0.0602 \\ \end{aligned} ###

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the term structure of interest rates equation, also known as the expectations hypothesis theory of interest rates,

###\left(1+\frac{r_{\text{0}\rightarrow\text{12mth, apr 6mth}}}{2}\right)^2 = \left(1+\frac{r_{\text{0}\rightarrow\text{6mth, apr 6mth}}}{2}\right)^1 \left(1+\frac{r_{\text{6}\rightarrow\text{12mth apr 6mth}}}{2}\right)^1 ### ###\left(1+\frac{0.07}{2}\right)^2 = \left(1+\frac{0.06}{2}\right)^1 \left(1+\frac{r_{\text{6}\rightarrow\text{12mth apr 6mth}}}{2}\right)^1 ### ###\left(1+\frac{r_{\text{6}\rightarrow\text{12mth, apr 6mth}}}{2}\right)^1 = \frac{\left(1+\frac{0.07}{2}\right)^2}{\left(1+\frac{0.06}{2}\right)^1} ### ###\begin{aligned} r_{\text{6}\rightarrow\text{12mth, apr 6mth}} &= \left( \frac{\left(1+\frac{0.07}{2}\right)^2}{\left(1+\frac{0.06}{2}\right)^1} - 1 \right) \times 2 \\ &= 0.0800 \\ \end{aligned} ###

Answer: Good choice. You earned $10. Poor choice. You lost $10.

All statements are true except b. This is because ##r_{0-1}## is the one year spot rate, not the forward rate. Spot interest rates are apply straight away 'on the spot', so they begin at time zero. Borrowers can borrow at the spot rate straight away.

Forward rates start at a future time, so they can't be borrowed at straight away. Forward rates can be locked-in using forward rate agreements (FRA's) which allow you to borrow in the future at an interest rate agreed to now, usually with a bank.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Statement b is false. If the liquidity premium theory is true, then the forward rates are higher than the expected future spot rates due to the liquidity premium, not lower.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Let ##z## be the spot or forward 'zero coupon yield' and ##r## be the spot yield to maturity (YTM). All rates will be quoted as annualised percentage rates (APR's) compounding every 6 months. So, for example, when you see ##z_{0 \rightarrow 0.5}## which is the zero coupon spot rate between now and 6 months (0.5 a year) away, it really means ##z_{0 \rightarrow 0.5 \text{ APR comp 6mth}}## since it's an APR compounding semi-annually. For brevity, the extra 'APR compounding every 6 months' description in ##z## and ##r## will be omitted in the working below.

Precise method

Use the 0.5 year bond to find ##z_{0 \rightarrow 0.5}##

The 0.5 year bond pays only a single coupon at maturity, so effectively there's only one cash flow: the coupon and principal paid at maturity. Therefore its yield to maturity (YTM) is equal to the 0.5 year zero coupon spot yield per annum compounding semi-annually, which is 3% pa.

###z_{0 \rightarrow 0.5} = r_{0 \rightarrow 0.5} = 0.03 = 3 \% ###

Use the 1 year bond to bootstrap and find ##z_{0 \rightarrow 1}##

The 1 year bond pays two coupons, one at 6 months and another at 1 year, together with the principal. We already know the 6 month spot zero coupon yield. We can use boot-strapping to find the 1-year spot zero-coupon yield. Since the bond price is given ##P_{0, \text{ 1yr bond}} = 100##, we don't need to use the YTM ##r_{0 \rightarrow 1} = 0.04##. But if we did, this first bond-pricing step would look like:

###\begin{aligned} P_{0, \text{ 1yr bond}} =& \dfrac{C_{0.5, 1}}{r_{0 \rightarrow 1}/2}\left(1 - \frac{1}{(1+r_{0 \rightarrow 1}/2)^{1 \times 2}} \right) + \frac{F_1}{(1+r_{0 \rightarrow 1}/2)^{1 \times 2}} \\ =& \dfrac{0.04/2 \times 100}{0.04/2}\left(1 - \frac{1}{(1+0.04/2)^{1 \times 2}} \right) + \frac{100}{(1+0.04/2)^{1 \times 2}} \\ =& 100 \\ \end{aligned} ###

To boot-strap and solve for the 1-year spot zero-coupon yield ##z_{0 \rightarrow 1}##, price the bond again but discount each coupon or principal by the appropriate zero coupon yield:

###P_{0, \text{ 1yr bond}} = \dfrac{C_{0.5}}{(1 + z_{0 \rightarrow 0.5}/2)^{0.5 \times 2}} + \dfrac{C_{1} + F_1}{(1 + z_{0 \rightarrow 1}/2)^{1 \times 2}} ### ###100 = \dfrac{0.04/2 \times 100}{(1 + 0.03/2)^{0.5 \times 2}} + \dfrac{0.04/2 \times 100 + 100}{(1 + z_{0 \rightarrow 1}/2)^{1 \times 2}} ### ###100 = 1.97044335 + \dfrac{102}{(1 + z_{0 \rightarrow 1}/2)^2} ### ###(1 + z_{0 \rightarrow 1}/2)^2 = \dfrac{102}{98.02955665} ### ###1 + z_{0 \rightarrow 1}/2 = \left( \dfrac{102}{98.02955665} \right)^{1/2}### ###\begin{aligned} z_{0 \rightarrow 1} &= \left( \left( \dfrac{102}{98.02955665} \right)^{1/2} - 1 \right) \times 2 \\ &= 0.0401005 \\ &\approx 4.0101 \% \\ \end{aligned} ###

Use the 1.5 year bond to bootstrap and find ##z_{0 \rightarrow 1.5}##

The 1.5 year bond pays three coupons, one at 6 months, another at 1 year and the last at 1.5 years together with the principal. We already know the 6 month and 1 year spot zero coupon yields. We can use boot-strapping to find the 1.5-year spot zero-coupon yield. Since the bond price is given ##P_{0, \text{ 1.5yr bond}} = 98.572##, we don't need to use the YTM ##r_{0 \rightarrow 1.5} = 0.05##. But if we did, this first bond-pricing step would look like:

###\begin{aligned} P_{0, \text{ 1.5yr bond}} =& \dfrac{C_{0.5, 1, 1.5}}{r_{0 \rightarrow 1.5}/2}\left(1 - \frac{1}{(1+r_{0 \rightarrow 1.5}/2)^{1.5 \times 2}} \right) + \frac{F_{1.5}}{(1+r_{0 \rightarrow 1.5}/2)^{1.5 \times 2}} \\ =& \dfrac{0.04/2 \times 100}{0.05/2}\left(1 - \frac{1}{(1+0.05/2)^{1.5 \times 2}} \right) + \frac{100}{(1+0.04/2)^{1.5 \times 2}} \\ =& 98.572 \\ \end{aligned} ###

To boot-strap and solve for the 1.5-year spot zero-coupon yield ##z_{0 \rightarrow 1.5}##, price the bond again but discount each coupon or principal by the appropriate zero coupon yield:

###P_{0, \text{ 1.5yr bond}} = \dfrac{C_{0.5}}{(1 + z_{0 \rightarrow 0.5}/2)^{0.5 \times 2}} + \dfrac{C_{1}}{(1 + z_{0 \rightarrow 1}/2)^{1 \times 2}} + \dfrac{C_{1.5} + F_{1.5}}{(1 + z_{0 \rightarrow 1.5}/2)^{1.5 \times 2}} ### ###98.572 = \dfrac{0.04/2 \times 100}{(1 + 0.03/2)^{0.5 \times 2}} + \dfrac{0.04/2 \times 100}{(1 + 0.0401005/2)^{1 \times 2}} + \dfrac{0.04/2 \times 100 + 100}{(1 + z_{0 \rightarrow 1.5}/2)^{1.5 \times 2}} ### ###98.572 = 1.97044335 + 1.92214817 + \dfrac{102}{(1 + z_{0 \rightarrow 1.5}/2)^3} ### ###\begin{aligned} z_{0 \rightarrow 1.5} &= \left( \left( \dfrac{102}{98.572 - 1.97044335 - 1.92214817} \right)^{1/3} - 1 \right) \times 2 \\ &= 0.050272369 \\ &\approx 0.050272 \% \\ \end{aligned} ###

Use the expectations hypothesis and spot zero yields to find the forward zero yield ##z_{0.5 \rightarrow 1}##

###(1+z_{0 \rightarrow 1}/2)^{1 \times 2} = (1+z_{0 \rightarrow 0.5}/2)^{0.5 \times 2} (1+z_{0.5 \rightarrow 1}/2)^{0.5 \times 2}### ###(1+0.0401005/2)^{2} = (1+0.03/2) (1+z_{0.5 \rightarrow 1}/2)### ###1+z_{0.5 \rightarrow 1}/2 = \dfrac{(1+0.0401005/2)^2}{(1+0.03/2)} ### ###\begin{aligned} z_{0.5 \rightarrow 1} &= \left( \left( \dfrac{(1+0.0401005/2)^2}{(1+0.03/2)} \right) - 1 \right) \times 2 \\ &= 0.050251256 \\ &\approx 5.0251 \% \\ \end{aligned}###

Use the expectations hypothesis and spot zero yields to find the forward zero yield ##z_{1 \rightarrow 1.5}##

###(1+z_{0 \rightarrow 1.5}/2)^{1.5 \times 2} = (1+z_{0 \rightarrow 0.5}/2)^{0.5 \times 2} (1+z_{0.5 \rightarrow 1}/2)^{0.5 \times 2} (1+z_{1 \rightarrow 1.5}/2)^{0.5 \times 2}### ###(1+0.050272369/2)^{3} = (1+0.03/2) (1+0.050251256/2) (1+z_{1 \rightarrow 1.5}/2)### ###1+z_{1 \rightarrow 1.5}/2 = \dfrac{(1+0.050272369/2)^3}{(1+0.03/2) (1+0.050251256/2)} ### ###\begin{aligned} z_{1 \rightarrow 1.5} &= \left( \left( \dfrac{(1+0.050272369/2)^3}{(1+0.03/2) (1+0.050251256/2)} \right) - 1 \right) \times 2 \\ &= 0.070768511 \\ &\approx 7.0769 \% \\ \end{aligned}###

Quick but imprecise method

Use the 0.5 year bond to find ##z_{0 \rightarrow 0.5}##

The first TYM (##r##) is always equal to the zero coupon spot rate (##z##), otherwise you can't begin the bootstrapping process.

###z_{0 \rightarrow 0.5} = r_{0 \rightarrow 0.5} = 0.03 = 3 \% ###

Use the 1 year bond to find an approximation for ##z_{0 \rightarrow 1}##

The 1 year YTM (##r_{0 \rightarrow 1}##) is approximately equal to a weighted average of the zero coupon yields (##z_{0 \rightarrow 0.5}## and ##z_{0 \rightarrow 1}##), weighted by the cash flows (using the present value of cash flows is more accurate, but harder to calculate):

###r_{0 \rightarrow 1} \approx \dfrac{C_{0.5}}{C_{0.5} + C_{1} + F_{1}} \times z_{0 \rightarrow 0.5} + \dfrac{C_1 + F_1}{C_{0.5} + C_{1} + F_{1}} \times z_{0 \rightarrow 1} ### ###0.04 \approx \dfrac{2}{104} \times 0.03 + \dfrac{102}{104} \times z_{0 \rightarrow 1} ### ###\begin{aligned} z_{0 \rightarrow 1} &\approx \dfrac{0.04 - \dfrac{2}{104} \times 0.03}{\left( \dfrac{102}{104} \right)} \\ &\approx 0.040196078 \end{aligned}###

This is pretty close to the exact answer of 0.0401005! Of course these weighted average calculations were a lot of effort too, but after getting a feel for weighted averages, you will quickly notice that to make an overall average of 4%, where one of the two things being averaged is 3%, then the second thing, with the big weight, must be a little higher than the 4% average. So 4.01% seems about right.

Use the 1.5 year bond to find an approximation for ##z_{0 \rightarrow 1.5}##

Similarly to before, the YTM (##r##) is approximately the weighted average of the zero rates (##z##):

###r_{0 \rightarrow 1.5} \approx \dfrac{C_{0.5}}{C_{0.5} + C_{1} + C_{1.5} + F_{1.5}} \times z_{0 \rightarrow 0.5} + \dfrac{C_{1}}{C_{0.5} + C_{1} + C_{1.5} + F_{1.5}} \times z_{0 \rightarrow 1} + \dfrac{C_{1.5} + F_{1.5}}{C_{0.5} + C_{1} + C_{1.5} + F_{1.5}} \times z_{0 \rightarrow 1.5} ### ###0.05 \approx \dfrac{2}{106} \times 0.03 + \dfrac{2}{106} \times 0.0401 + \dfrac{102}{106} \times z_{0 \rightarrow 1.5} ###

Without actually solving for ##z_{0 \rightarrow 1.5}##, 5.0272% looks about right, because the 5% YTM would be the weighted average of the 3%, 4.01% and 5.02% zero rates, considering the huge $100 principal weight on that 5.02%.

Use the expectations hypothesis and spot zero yields to find an approximation for the forward zero yield ##z_{0.5 \rightarrow 1}##

The expectations hypothesis is a geometric weighted average, weighted by time. Geometric averages can be approximated by arithmetic averages.

Exact geometric average of gross returns:

###(1+z_{0 \rightarrow 1}/2) = \left( (1+z_{0 \rightarrow 0.5}/2)^{1} (1+z_{0.5 \rightarrow 1}/2)^{1} \right)^{1/2}###

Approximation using the arithmetic average of net returns:

###z_{0 \rightarrow 1} \approx \dfrac{z_{0 \rightarrow 0.5} \times 1 + z_{0.5 \rightarrow 1} \times 1 }{2}### ###0.0401005 \approx \dfrac{0.03 + z_{0.5 \rightarrow 1}}{2}### ###\begin{aligned} z_{0.5 \rightarrow 1} &\approx 0.0401005 \times 2 - 0.03 \\ &\approx 0.050201 \\ \end{aligned}###

This approximation is close to the exact answer of 0.050251256. In this working the exact numbers were used as inputs for ##z_{0 \rightarrow 1}## and ##z_{0 \rightarrow 0.5}##. Approximations for ##z_{0 \rightarrow 1}## could have been used instead, but errors in that input would make the error in this approximation for ##z_{0 \rightarrow 1}## even worse.

Use the expectations hypothesis and spot zero yields to find an approximation for the forward zero yield ##z_{1 \rightarrow 1.5}##

The expectations hypothesis is a geometric weighted average, weighted by time. Geometric averages can be approximated by arithmetic averages.

Exact geometric average of gross returns:

###(1+z_{0 \rightarrow 1.5}/2) = \left( (1+z_{0 \rightarrow 0.5}/2)^{1} (1+z_{0.5 \rightarrow 1}/2)^{1} (1+z_{1 \rightarrow 1.5}/2)^{1} \right)^{1/3}###

Approximation using the arithmetic average of net returns:

###z_{0 \rightarrow 1.5} = \dfrac{z_{0 \rightarrow 0.5} \times 1 + z_{0.5 \rightarrow 1} \times 1 + z_{1 \rightarrow 1.5} \times 1}{3}### ###0.050272369 = \dfrac{0.03 + 0.050251256 + z_{1 \rightarrow 1.5}}{3}### ###\begin{aligned} z_{1 \rightarrow 1.5} &= 0.050272369 \times 3 - 0.03 - 0.050251256 \\ &= 0.070565851 \\ \end{aligned}###

This approximation is close to the exact answer of 0.070768511. Again, in this working the exact numbers were used as inputs for ##z_{0 \rightarrow 1.5}##, ##z_{0 \rightarrow 0.5}## and ##z_{0.5 \rightarrow 1}##. Approximations for ##z_{0 \rightarrow 1.5}## and ##z_{0.5 \rightarrow 1}## could have been used instead, but errors in those inputs would make the error in this approximation for ##z_{1 \rightarrow 1.5}## even worse.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Let ##z## be the spot or forward 'zero coupon yield' and ##r## be the spot yield to maturity (YTM). All rates will be quoted as annualised percentage rates (APR's) compounding annually. So, for example, when you see ##z_{0 \rightarrow 1}## which is the zero coupon spot rate between now and 1 year away, it really means ##z_{0 \rightarrow 1 \text{ APR comp yearly}}## since it's an APR compounding annually. For brevity, the extra 'APR compounding every year' description in ##z## and ##r## will be omitted in the working below.

Precise method

Use the 1 year bond to find ##z_{0 \rightarrow 1}##

The 1 year bond pays only a single coupon at maturity, so effectively there's only one cash flow: the coupon and principal paid at maturity. Therefore its yield to maturity (YTM) is equal to the 1 year zero coupon spot yield per annum compounding annually, which is 0% pa.

###z_{0 \rightarrow 1} = r_{0 \rightarrow 1} = 0 = 0 \% ###

Use the 2 year bond to bootstrap and find ##z_{0 \rightarrow 2}##

The 2 year bond pays two coupons, the first at 1 year and the next at maturity in 2 years, together with the principal. We already know the 1 year spot zero coupon yield. We can use boot-strapping to find the 2-year spot zero-coupon yield. Since the bond price is given ##P_{0, \text{ 2yr bond}} = 101.9703951##, we don't need to use the YTM ##r_{0 \rightarrow 2} = 0.01##. But if we did, this first bond-pricing step would look like:

###\begin{aligned} P_{0, \text{ 2yr bond}} =& \dfrac{C_{1}}{r_{0 \rightarrow 2}}\left(1 - \frac{1}{(1+r_{0 \rightarrow 2})^{2}} \right) + \frac{F_1}{(1+r_{0 \rightarrow 2})^{2}} \\ =& \dfrac{0.02 \times 100}{0.01}\left(1 - \frac{1}{(1+0.01)^{2}} \right) + \frac{100}{(1+0.01)^{2}} \\ =& 101.9703951 \\ \end{aligned} ###

To boot-strap and solve for the 2-year spot zero-coupon yield ##z_{0 \rightarrow 2}##, price the bond again but discount each coupon or principal by the appropriate zero coupon yield:

###P_{0, \text{ 2yr bond}} = \dfrac{C_{1}}{(1 + z_{0 \rightarrow 1})^{1}} + \dfrac{C_{2} + F_2}{(1 + z_{0 \rightarrow 2})^{2}} ### ###101.9703951 = \dfrac{0.02 \times 100}{(1 + 0)^{1}} + \dfrac{0.02 \times 100 + 100}{(1 + z_{0 \rightarrow 2})^{2}} ### ###101.9703951 = 2 + \dfrac{102}{(1 + z_{0 \rightarrow 2})^2} ### ###(1 + z_{0 \rightarrow 2})^2 = \dfrac{102}{101.9703951 - 2} ### ###1 + z_{0 \rightarrow 2} = \left( \dfrac{102}{101.9703951 - 2} \right)^{1/2}### ###\begin{aligned} z_{0 \rightarrow 2} &= \left( \dfrac{102}{101.9703951 - 2} \right)^{1/2} - 1 \\ &= 0.010100025 \\ &=1.010100025 \% \\ \end{aligned} ###

Use the 3 year bond to bootstrap and find ##z_{0 \rightarrow 3}##

The 2 year bond pays three coupons, the first at 1 year, another at 2 years and the last at 3 years together with the principal. We already know the 1 year and 2 year spot zero coupon yields. We can use boot-strapping to find the 3 year spot zero-coupon yield. Since the bond price is given ##P_{0, \text{ 3yr bond}} = 100##, we don't need to use the YTM ##r_{0 \rightarrow 3} = 0.02##. But if we did, this first bond-pricing step would look like:

###\begin{aligned} P_{0, \text{ 3yr bond}} =& \dfrac{C_{1}}{r_{0 \rightarrow 3}}\left(1 - \frac{1}{(1+r_{0 \rightarrow 3})^{3}} \right) + \frac{F_{3}}{(1+r_{0 \rightarrow 3})^{3}} \\ =& \dfrac{0.02 \times 100}{0.02}\left(1 - \frac{1}{(1+0.02)^{3}} \right) + \frac{100}{(1+0.02)^{3}} \\ =& 100 \\ \end{aligned} ###

To boot-strap and solve for the 3-year spot zero-coupon yield ##z_{0 \rightarrow 3}##, price the bond again but discount each coupon or principal by the appropriate zero coupon yield:

###P_{0, \text{ 3yr bond}} = \dfrac{C_{1}}{(1 + z_{0 \rightarrow 1})^{1}} + \dfrac{C_{2}}{(1 + z_{0 \rightarrow 2})^{2}} + \dfrac{C_{3} + F_{3}}{(1 + z_{0 \rightarrow 3})^{3}} ### ###100 = \dfrac{0.02 \times 100}{(1 + 0)^{1}} + \dfrac{0.02 \times 100}{(1 + 0.010100025)^{2}} + \dfrac{0.02 \times 100 + 100}{(1 + z_{0 \rightarrow 3})^{3}} ### ###100 = 2 + 1.960203825 + \dfrac{102}{(1 + z_{0 \rightarrow 3})^3} ### ###\begin{aligned} z_{0 \rightarrow 3} &= \left( \dfrac{102}{100 - 2 - 1.960203825} \right)^{1/3} - 1 \\ &= 0.020272812 \\ &=2.0272812 \% \\ \end{aligned} ###

Use the expectations hypothesis and spot zero yields to find the forward zero yield ##z_{1 \rightarrow 2}##

###(1+z_{0 \rightarrow 2})^{2} = (1+z_{0 \rightarrow 1})^{1} (1+z_{2 \rightarrow 3})^{1}### ###(1+0.010100025)^{2} = (1+0) (1+z_{1 \rightarrow 2})### ###1+z_{1 \rightarrow 2} = \dfrac{(1+0.010100025)^2}{1} ### ###\begin{aligned} z_{1 \rightarrow 2} &= (1+ 0.010100025)^2 - 1 \\ &= 0.02030206 \\ & 2.030206\% \\ \end{aligned}###

Use the expectations hypothesis and spot zero yields to find the forward zero yield ##z_{2 \rightarrow 3}##

###(1+z_{0 \rightarrow 3})^{3} = (1+z_{0 \rightarrow 1})^{1} (1+z_{1 \rightarrow 2})^{1} (1+z_{2 \rightarrow 3})^{1}### ###(1+0.020272812)^{3} = (1+0) (1+0.02030206) (1+z_{2 \rightarrow 3})### ###1+z_{2 \rightarrow 3} = \dfrac{(1+0.020272812)^3}{(1+0) (1+0.02030206)} ### ###\begin{aligned} z_{2 \rightarrow 3} &= \dfrac{(1+0.020272812)^3}{(1+0) (1+0.02030206)} - 1 \\ &= 0.040926772 \\ &= 4.0926772\% \\ \end{aligned}###