1. Chapter 4 Deriving the Zero-Coupon Yield Curve FIXED-INCOME SECURITIES

2. Outline General Principle Spot Rates Recovering the Term Structure Direct Methods Interpolation Indirect Methods Splines Term Structure of Credit Spreads

3. Last Time The current price of a bond (P 0 ) paying cash-flows F t is given by: Now, do we expect to get the same rate when borrowing/lending for a year versus 10 years? Not necessarily Term structure of interest rates

4. General Principle General formula –R(0,t) is the discount rate –B(0,t) is the discount factor (present value of $1 received at date t) –Discount factor more convenient: no need to specify frequency What exactly does that equation mean? –Q1: Where do we get the B(0,t) or R(0,t) from? –Q2: Do we use the equation to obtain bond prices or implied discount factors/discount rates? –Q3: Can we deviate from this simple rule? Why?

5. Spot Rates Q1: Where do we get the B(0,t) or R(0,t) from? –Any relevant information concerning how to price a security should be obtained from market sources –More specifically, B(t,T) is the price at date t of a unit pure discount bond paying $1 at date T Discount factor B(0,t) is the price of a T-Bond with unit face value and maturity t Spot rate R 0,t is the annualized rate on a pure discount bond: Bad news is no such abundance of zero-coupon bonds exists in the real world Good news is we might still be able to compute the spot rate

6. Bond Pricing Answer to the “chicken-and-egg'' second question (Q2) is –It depends on the situation –Roughly speaking, one would like to use the price of primitive securities as given, and derive implied discount factors or discount rates from them –Then, one may use that information (more specifically the term structure of discount rates) to price any other security –This is known as relative pricing Answer to the third question (Q3) is –Any deviation from the pricing rules would imply arbitrage opportunities –Practical illustrations of that concept shall be presented in what follows –Everything we cover in this Chapter can be regarded as some form of perspective on these issues

7. Spot Rates Example of spot rate: –Consider a two-year pure discount bond that trades at $92 –The two-year spot rate R 0,2 is: The collection of all spot rates for all maturities is: –The Term Structure of Interest Rates

8. Recovering the Term Structure Direct Methods - Principle Consider two securities (nominal $100): –One year pure discount bond selling at $95 –Two year 8% bond selling at $99 One-year spot rate: Two-year spot rate:

9. Recovering the Term Structure Direct Methods - Principle We may “construct” a two year pure discount bond Two components: –Buy the two year bond –Shortsell the first $8 coupon Cost:

10. Recovering the Term Structure Direct Methods - Principle Schedule of payments: Today1 year 2 years -91.40108 This is like a two-year pure discount bond Two-year rate is again:

11. Recovering the Term Structure Direct Methods - Example If you can find different bonds with same anniversary date, then you can directly get the spot rates : Coupon Maturity (year)Price Bond 151101 Bond 25.52101.5 Bond 35399 Bond 464100 Solve the following system 101 = 105 B(0,1) 101.5 = 5.5 B(0,1) + 105.5 B(0,2) 99 = 5 B(0,1) + 5 B(0,2) + 105 B(0,3) 100 = 6 B(0,1) + 6 B(0,2) + 6 B(0,3) + 106 B(0,4) And obtain B(0,1)=0.9619, B(0,2)=0.9114, B(0,3)=0.85363, B(0,4)= 0.7890 R(0,1)=3.96%, R(0,2)=4.717%, R(0,3)=5.417%, R(0,4)=6.103%

12. 1 year and 2 months rate x=5.41% 1 year and 9 months rate y= 5.69% 2 year rate z= 5.69% Recovering the Term Structure Bootstrap: Practical Way of Implementing Direct Method MaturityZC Overnight4.40% 1 month4.50% 2 months4.60% 3 months4.70% 6 months4.90% 9 months5.00% 1 year5.10% CouponMaturity (years)Price Bond 15%1 y and 2 m103.7 Bond 26%1 y and 9 m102 Bond 35.50%2 y99.5

13. Recovering the Term Structure Interpolation - Linear Interpolation –Term structure is a mapping  -> R(t,  ) for all possible  –Need to interpolate Linear interpolation –We know discount rates for maturities t 1 et t 2 –We are looking for the rate with maturity t such that t 1 < t <t 2 Example: R(0,3) =5.5% and R(0,4)=6%

14. Recovering the Term Structure Interpolation – Piecewise Polynomial Cubic interpolation for different segments of the term structure –Define the first segment: maturities ranging from t 1 to t 4 (say 1 to 2 years) –We know R(0, t 1 ), R(0, t 2 ), R(0, t 3 ), R(0, t 4 ) The discount rate R(0, t) is defined by Impose the constraint that R(0, t 1 ), R(0, t 2 ), R(0, t 3 ), R(0, t 4 ) are on the curve

15. Recovering the Term Structure Piecewise Polynomial - Example We have computed the following rates –R(0,1) = 3% –R(0,2) = 5% –R(0,3) = 5.5% –R(0,4) = 6% Compute the 2.5 year rate R(0,2.5) = a x 2.5 3 + b x 2.5 2 + c x 2.5 1 + d = 5.34375% with

16. Recovering the Term Structure Piecewise Polynomial versus Piecewise Linear

17. Rather than obtain a few points by boostrapping techniques, and then extrapolate, it usually is more robust to use a model for the yield curve So-called indirect methods involve the following steps –Step 1: select a set of K bonds with prices P j paying cash-flows F j (t i ) at dates t i >t –Step 2: select a model for the functional form of the discount factors B(t,t i ;ß), or the discount rates R(t,t i ;ß), where ß is a vector of unknown parameters, and generate prices Recovering the Term Structure Indirect Methods –Step 3: estimate the parameters ß as the ones making the theoretical prices as close as possible to market prices

18. Nelson and Siegel have introduced a popular model for pure discount rates Recovering the Term Structure Indirect Methods – Nelson Siegel R(0,  ) : pure discount rate with maturity   0 : level parameter - the long-term rate  1 : slope parameter – the spread sort/long-term  2 : curvature parameter  1 : scale parameter

19. Recovering the Term Structure Inspection of Nelson Siegel Functional

20. Recovering the Term Structure Slope and Curvature Parameters To investigate the influence of slope and curvature parameters in Nelson and Siegel, we perform the following experiment –Start with a set of base case parameter values  0 = 7%  1 = -2%  2 = 1%  = 3.33 –Then adjust the slope and curvature parameters  1 = between –6% and 6%  2 = between –6% and 6%

21. Recovering the Term Structure Initial Curve

22. Recovering the Term Structure Impact of Changes in the Slope Parameter

23. Recovering the Term Structure Impact of Changes in the Curvature Parameter

24. Recovering the Term Structure Possible Shapes for the Yield Curve

25. Recovering the Term Structure Evolution of Parameters of the Nelson and Siegel Model on the French Market - 1999-2000 Beta(0) oscillates between 5% and 7% and may be regarded as the very long term rate Beta(1) is the short to long term spread. It varies between -2% and -4% in 1999, and then decreases in absolute value to almost 0% at the end of 2000 Beta(2), the curvature parameter, is the more volatile parameter which varies from -5% to 0.7%.

26. An augmented form exists Recovering the Term Structure Indirect Methods – Augmented Nelson Siegel R(0,  ) : pure discount rate with maturity   3 : level parameter  2 : scale parameter Allows for more flexibility in the short end of the curve

27. Recovering the Term Structure Augmented Nelson Siegel - Illustration Augmented Nelson-Siegel.035.037.039.041.043.045.047.049 0510152025      

28. These models are heavily used in practice One key advantage is they are parsimonious –Do not involve many parameters –This induces robustness and stability –Very important in the context of hedging One drawback is their lack of flexibility –Can not account for all possible shapes of the TS we see in practice Alternative approach: spline models –More flexible –Better for pricing –Less parsimonious Spline models come in different shapes –Cubic splines –Exponential splines –B-splines Recovering the Term Structure Parsimonious Models – Pros and Cons

29. Discount factors as polynomial splines Polynomial Splines Impose smooth-pasting constraints Cut down the number of parameters from 12 to 5

30. Example (French Market)

31. 4.70% 4.80% 4.90% 5.00% 5.10% 5.20% 5.30% 02468101214 Confidence interval Yield curve on 09/01/00 Example