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Chapter 4 Deriving the Zero-Coupon Yield Curve

FIXED-INCOME SECURITIES. Chapter 4 Deriving the Zero-Coupon Yield Curve. Outline. General Principle Spot Rates Recovering the Term Structure Direct Methods Interpolation Indirect Methods Splines Term Structure of Credit Spreads. Last Time.

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Chapter 4 Deriving the Zero-Coupon Yield Curve

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

  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 (P0) paying cash-flows Ft 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 R0,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 R0,2 is: • The collection of all spot rates for all maturities is: • The Term Structure of Interest Rates

  8. Recovering the Term StructureDirect 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 StructureDirect 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. -91.4 0 108 2 years Today 1 year Recovering the Term StructureDirect Methods - Principle • Schedule of payments: • This is like a two-year pure discount bond • Two-year rate is again:

  11. Coupon Maturity (year) Price Bond 1 5 1 101 Bond 2 5.5 2 101.5 Bond 3 5 3 99 Bond 4 6 4 100 Recovering the Term StructureDirect Methods - Example • If you can find different bonds with same anniversary date, then you can directly get the spot rates : • 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. Maturity ZC Coupon Maturity (years) Price Overnight 4.40% Bond 1 5% 1 y and 2 m 103.7 1 month 4.50% Bond 2 6% 1 y and 9 m 102 2 months 4.60% Bond 3 5.50% 2 y 99.5 3 months 4.70% 6 months 4.90% 9 months 5.00% 1 year 5.10% Recovering the Term StructureBootstrap: Practical Way of Implementing Direct Method • 1 year and 2 months rate x=5.41% • 1 year and 9 months rate y= 5.69% • 2 year rate z= 5.69%

  13. Recovering the Term StructureInterpolation - Linear • Interpolation • Term structure is a mapping  -> R(t, ) for all possible  • Need to interpolate • Linear interpolation • We know discount rates for maturities t1 et t2 • We are looking for the rate with maturity t such that t1< t <t2 • Example: • R(0,3) =5.5% and R(0,4)=6%

  14. Recovering the Term StructureInterpolation – Piecewise Polynomial • Cubic interpolation for different segments of the term structure • Define the first segment: maturities ranging from t1 to t4 (say 1 to 2 years) • We know R(0, t1), R(0, t2), R(0, t3), R(0, t4) • The discount rate R(0, t) is defined by • Impose the constraint that R(0, t1), R(0, t2), R(0, t3),R(0, t4) are on the curve

  15. Recovering the Term StructurePiecewise 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 2.53 + b x 2.52 + c x 2.51 + d = 5.34375% with

  16. 6.50% Linear Cubic 6.00% 5.50% 5.00% Rate 4.50% 4.00% 3.50% 3.00% 1 1.5 2 2.5 3 3.5 4 Maturity Recovering the Term StructurePiecewise Polynomial versus Piecewise Linear

  17. Recovering the Term StructureIndirect Methods • 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 Pj paying cash-flows Fj(ti) at dates ti>t • Step 2: select a model for the functional form of the discount factors B(t,ti;ß), or the discount rates R(t,ti;ß), where ß is a vector of unknown parameters, and generate prices • Step 3: estimate the parameters ß as the ones making the theoretical prices as close as possible to market prices

  18. Recovering the Term StructureIndirect Methods – Nelson Siegel • Nelson and Siegel have introduced a popular model for pure discount rates 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. Inspection of Nelson-Siegel Functional Form 3.5% b Long-term limit 0: 3.0% 2.5% 2.0% Term structure 1.5% 1.0% 0.5% b q t t q q t ((1-exp(- / )). / -exp(- / )) medium-term component 2 0.0% 0 5 10 15 20 25 -0.5% b q t t q (1-exp(- / )). / : short-term component 1 -1.0% -1.5% Recovering the Term StructureInspection of Nelson Siegel Functional

  20. Recovering the Term StructureSlope 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 StructureInitial Curve

  22. Recovering the Term StructureImpact of Changes in the Slope Parameter

  23. Recovering the Term StructureImpact of Changes in the Curvature Parameter

  24. Recovering the Term StructurePossible Shapes for the Yield Curve

  25. Recovering the Term StructureEvolution 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. Recovering the Term StructureIndirect Methods – Augmented Nelson Siegel • An augmented form exists 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 StructureAugmented Nelson Siegel - Illustration Augmented Nelson-Siegel .049 .047 .045 b 0 3 > .043 .041 .039 b 0 3 < .037 .035 0 5 10 15 20 25

  28. Recovering the Term StructureParsimonious Models – Pros and Cons • 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

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

  30. Bond Maturity Coupon Market price Model Price Residuals % error BTAN 12/03/2001 5.75 103.113 103.095 0.018 0.02% BTAN 12/07/2001 3 98.724 98.729 -0.005 0.00% BTAN 12/10/2001 5.5 105.332 105.310 0.022 0.02% BTAN 12/01/2002 4 101.106 101.102 0.005 0.00% BTAN 12/03/2002 4.75 101.710 101.706 0.004 0.00% BTAN 12/07/2002 4.5 99.526 99.540 -0.015 -0.01% OAT 25/11/2002 8.5 113.454 113.440 0.013 0.01% BTAN 12/01/2003 5 102.843 102.874 -0.031 -0.03% OAT 25/04/2003 8.5 111.054 111.105 -0.051 -0.05% OAT 25/10/2003 6.75 110.295 110.314 -0.019 -0.02% BTAN 07/12/2003 4.5 101.603 101.405 0.198 0.20% OAT 27/02/2004 8.25 113.785 113.858 -0.074 -0.06% BTAN 12/07/2004 3.5 94.778 94.812 -0.035 -0.04% OAT 25/10/2004 6.75 111.515 111.532 -0.017 -0.02% OAT 25/04/2005 7.5 111.952 112.003 -0.051 -0.05% BTAN 12/07/2005 5 99.910 99.918 -0.009 -0.01% OAT 25/10/2005 7.75 117.958 117.900 0.058 0.05% OAT 25/04/2006 7.25 112.312 112.309 0.003 0.00% OAT 25/10/2006 6.5 112.170 112.085 0.084 0.08% OAT 25/04/2007 5.5 103.239 103.312 -0.073 -0.07% OAT 25/10/2007 5.5 106.037 106.032 0.004 0.00% OAT 25/04/2008 5.25 101.559 101.574 -0.015 -0.02% OAT 25/10/2008 8.5 127.936 127.942 -0.006 0.00% OAT 25/04/2009 4 92.297 92.326 -0.029 -0.03% OAT 25/10/2009 4 93.832 93.817 0.015 0.02% OAT 25/04/2010 5.5 103.049 102.983 0.066 0.06% OAT 25/04/2011 6.5 111.331 111.380 -0.049 -0.04% Example (French Market)

  31. 5.30% Yield curve on 09/01/00 5.20% 5.10% Confidence interval 5.00% 4.90% 4.80% 4.70% 0 2 4 6 8 10 12 14 Example

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