Student Solution Manual. Fixed Income Securities: Valuation, Risk, and Risk Management

Transcription

1 Student Solution Manual to accompany the textbook Fixed Income Securities: Valuation, Risk, and Risk Management by Pietro Veronesi Authors: Francisco Javier Madrid Anna Cieslak Francesco Benintende Version 1.3 Date: March 02, 2011 This student solution manual provides solutions to selected exercises at the end of each chapter of the book Fixed Income Securities. Solutions to all of the exercises are instead available only to instructors. 1

2 Solutions are provided to the following exercises: Chapter 2 Exercises 2, 4, and 5; Chapter3 Exercises1,3,4,7,8,and9; Chapter 4 Exercises 1, 4, and 7; Chapter5 Exercises2,5,7,9,and11; Chapter6 Exercises1,3,6,7,and10; Chapter 7 Exercises 2 and 5; Chapter 8 Exercises 1, 2, and 7; Chapter 9 Exercises 1 and 5; Chapter 10 Exercises 1 and 3; Chapter 11 Exercises 3, 5, and 7; Chapter 12 Exercises 2, 3, and 5; Chapter 13 Exercises 2, 3, and 4; Chapter 14 Exercises 2, 4, 5, and 7; Chapter 15 Exercises 1, 2, 3, 6, 7, and 9; Chapter 16 Exercises 1, 4, and 7; Chapter 17 Exercises 1, 3, and 6; Chapter 18 Exercises 3, and 5; Chapter 19 Exercises 1, and 4; Chapter 20 Exercises 1, 4, and 5; Chapter 21 Exercises 2, 4, 6, and 8; Chapter 22 Exercises 1, 2, and 3.

3 Solutions to Chapter 2 Exercise 2. Compute the quoted price P of the T-bill as: [ P = n ] 360 d, (1) using the discount rate given, d. The simple (bond equivalent) yield measures your annualized return as: BEY = 100 P P 365 n. (2) Let τ = n 365 be the time to maturity expressed as fraction of a year, and let T denote the maturity date of a given T-bill. The continuously compounded yield follows as: r(t, T )= 1 τ ln P 100. (3) Finally, to obtain the semi-annually compounded yield for the 1-year T-bill, use: ( ) 1 r 2 (0, 1) = 2 (P/100) 1 (4) 1/2 Semi-annual n T t Discount, d Price, P BEY yield Date a % % 3.53% 12/12/2005 b % % 0.13% 11/6/2008 c % % 5.03% 7/10/2006 d % % 4.86% 5/8/2007 e % % 0.49% 11/4/2008 f % % 4.84% 4/21/2006 g % % 4.87% 6/6/2007 h % % 0.90% 11/11/2008 i % % 1.77% 9/30/2008 j % % 1.21% 11/5/2008 For bond i. and j. the continuously compounded yields are 1.75% and 1.20% respectively. 2

4 Exercise 4. Using Table 2.4, obtain the discount factor Z(t, T )foreachmaturity T t from 0.25 to 7.5 years: Z(t, T )= ( 1 1+ r2(t,t ) 2 ) 2(T t). (5) Use Z to price each bond: a. P z (0, 5) = 100 Z(0, 5) = b. P c=15%,n=2 (0, 7) = i=1 Z(0,i/2) Z(0, 7) = c. P c=7%,n=4 (0, 4) = i=1 Z(0,i/4) Z(0, 4) = d. P c=9%,n=2 (0, 3.25) = i=1 Z(0,i/2 0.25)+100 Z(0, 3.25) = e. 100 (see Fact 2.11) f. P FR,n=1,s=0 = Z(0, 0.5) 100 ( % ) = , where we assume 1 that r 1 (0) = 6.8% g. P FR,n=4,s=0.35% (0, 5.5) = i=1 Z(0,i/4) = h. P FR,n=2,s=0.40% = Z(0, 0.25) 100 ( % )+ 2 i=1 Z(0,i/2 0.25) = 104, where we assume that r 2 (0) = 6.4% Exercise 5. a. When coupon c is equal to the yield to maturity y the bond trades at par; when coupon is below (above) the yield to maturity the bond trades above (below) par. Obtain bond prices given yield and the coupon using: P c (0,T)= 20 i=1 c/2 100 (1 + y/2) i (1 + y/2) 20 (6) 3

5 It follows: c y P 5% 6% % 6% 100 7% 6% b. Figure?? plots bond prices implied by different yields to maturity. 150 Bond price vs YTM Bond price Yield Figure 1: Bond price as function of yield to maturity 4

6 Solutions to Chapter 3 Exercise 1. a. 3; equal to the maturity of the zero bond b ; the duration of the coupon bond is the weighted average of the coupon payment times c d. 0.5; equal to the time left to the next coupon payment e ; obtain the price P FR of the floating rate bond (see Chapter 2, equation (2.39)). In analogy to a coupon bond, the duration is computed as: D FR = s 3 t=0.5 Z(0,t) t. (7) P FR P FR f ; proceed as in point e. above but recognize that the valuation is outside the reset date. Assume that the coupon applying to the next reset date has been set at r 2 (0) = 6.4%. Exercise 3. Obtain yield to maturity y for each security. Compute modified and Macaulay duration accodring to equation (3.19) and (3.20) in the book. Yield Duration Modified Macaulay a. 6.95% b. 6.28% c. 6.66% d. 0.00% e. 6.82% f. 6.76%

7 Exercise 4. Compute the duration of each asset and use the fact that the dollar duration is the bond price times its duration. Exercise 7. Price Duration $ Duration a. $ $ b. $ ($473.39) c. $ $ d. $ $50.00 e. $ ($25.00) f. $ ($28.38) a. $10 mn b. Compute the dollar duration of the cash flows in each bond, and then the dollar duration of the portfolio: Security $ (mn) Price N $D $D N 6yr 20% - fl quart , yr fl 45bps semi yr zero (30.00) Portfolio $10.00 mn Note: negative values denote short positions. Exercise 8. a. The price of the 5% semi bond is $ You want the duration of the hedged portfolio to be zero. You need to short units of the 3-year bond, i.e. the short position is -$5.69. b. The total value of the portfolio is: $4.31 mn. 6

8 Exercise 9. Compute the new value of the portfolio assuming the term structure of interest rates as of May 15, a. $8.97 mn Original Now Δ value Unhedged port. $10.00 $8.97 ($1.03) Hedge ($5.69) ($5.44) $0.25 Total $4.31 $3.53 ($0.78) b. $3.53 mn c. The immunization covered part of the loss. The change in the value of the portfolio is both due to (i) the passage of time (coupon) and (ii) the increase in interest rates. 7

9 Solutions to Chapter 4 Exercise 1. a. 16 b. 4.7 (see Fact 4.3) c d = 0.25 (due to floater with zero spread) (due to the spread) e = 0.06 (due to floater with zero spread) (due to the spread). Assume r 2 (0) = 6.40%. Exercise 4. Follow the steps in Example 4.3 using a 2-year bond instead of a 10-year bond. Duration Hedging Spot Curve Change in Shift P c (0, 10) P z (0, 2) Position Portfolio Value Initial Values dr =.1% dr = 1% dr = 2% dr =.1% dr = 1% dr = 2% The hedge performs better in that for any scenario the change in the value of the portfolio is positive. 8

10 Exercise 7. You need factor sensitivities for yields with maturities from 3 months to 4.25 years with a 0.25-year spacing. To obtain the 0.25-year grid of sensitivities, interpolate linearly the β s between available maturities. To compute factor durations, use Fact 4.5. Security Price Level D Slope D Curvature D a. 4 yr zero b % semi c yr float 0 bps d yr float 35 bps

11 Solutions to Chapter 5 Exercise 2. Use the facts that: F (0,T Δ,T) = e f(0,t Δ,T )Δ (8) Z(0,T i ) = Z(0,T i 1 ) F (0,T i 1,T i ). (9) (t, T) f(t, T Δ,T) F (t, T Δ,T) yield Z(t, T) % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %

12 Exercise 6. a. P fwd = 100 F (0, 0.5, 2) = $ b. M = $50 mn P fwd =0.554 mn Exercise 7. a. Payoff = M (P (0.5, 2) P fwd )=$0.64 mn b. You make money on the long forward. Exercise 9. Use fact 5.11 to obtain the swap rate c(t, T ): (t, T ) yield Z L (t, T ) c(t, T ),n= % % % % % % % % Exercise 11. On that day, the net spread was strongly negative: SS = c ytm = 0.33% LRS = LIBOR repo =1.71% SS LRS = 2.04% You could envision the following strategy to exploit the large negative spread: 1. Buy Treasury through a repo transaction: get coupon, pay repo 2. Enter the floating-for-fixed swap: pay fixed, get LIBOR 11

13 Solutions to Chapter 6 Exercise 1. From semi-annually compounded yields obtain (i) the discount factors, and (ii) the forward discount factors. Use Fact 5.9 in Chapter 5 to compute the forward price of a coupon bond. a b. 0 c. Compute the time-t value of the forward as: V (t) =Z(t, T ) [ Pc fwd (t, T, T ) K ],K= Pc fwd (0,T,T ) (10) d. See Table below e. Obtain the daily overnight discount factor Z(o/n), then: TotalP/L(t i )=P/L(t i )+ TotalP/L(t i 1) Z o/n (t i 1 ) (11) 2/15/94 2/16/94 2/17/94 2/18/94 2/22/94 2/23/94 t 0 t 1 t 2 t 3 t 4 t 5 Pc fwd V (t) P/L r o/n 3.54% 3.78% 3.90% 4.44% 2.88% Z(o/n) Tot. P/L

14 Exercise 3. a. Yes, the rates did converge: Libor equals the futures rate at expiry. b. Total profit from futures: = 0.25 ( ) c. Total profit from forward: = 100 ( ) d.,e.,f. See Figure?? Cumulative P&L P&L fut P&L fwd Sep07 Nov07 Dec07 Feb08 Apr08 May08 Figure 2: P&L from the futures and forward contracts Exercise 6. a.i. The empirical probability that the firm has enough cash for the lawsuit is 79% (see Figure??, a.i). a.ii The firm would have enough cash in 54% of scenarios (see Figure??, a.ii). α β s.e r(0) % b. Ex post the company will not have enough cash: ( %/4) =

15 300 a. Histogram of Δr a.i. Histogram of cash available (empirical) a.ii. Histogram of cash available (parametric) Figure 3: Distribution of possible cash flows for the firm Exercise 7. Price P&L Total 3mo Libor After 90 days Final ($ mn) a. $ % b. $ % Under scenario in b. (from Table 6.12), the firm is short on the cash they need, receiving out of 100 required. Exercise 10. a. The securities are not correctly priced, as the put-call parity is violated: Put $ Call $ P fwd (0, 0.5, 0.75) K Z(0, 0.5) (P fwd K) Call from P/C parity ($0.0546) 14

16 b. Strategy long call, short put, short forward gives a positive cash-flow of $ at no risk. 15

17 Solutions to Chapter 7 Exercise 2. a. Using Equation 7.28, minimize squared distance between the observed and the model-based prices (computed from equation 7.27) to obtain the parameters in the table below. Using the parameters compute real yields and the discount factors (see Figure??). Parameter Value θ θ θ θ κ κ b. Obtain the analytical expression for the short interest rate by taking the limit of r(0,t) implied by the Nelson-Siegel model: Use the following facts: lim T 0 r(0,t)=θ 0 + θ 1 =7.58% (12) 1 e T/κ1 lim T 0 T/κ 1 = 1 (13) lim T 0 e T/κ1 = 1. (14) c.,d. Using the NS yield curve, the TIPS with maturity years (maturing 4/14/2012) is priced at $ This is $2.755 lower relative to the traded price. 16

18 0.08 NS real yield curve, r(t, T ) Maturity, T t NS discount curve, Z(t, T ) Maturity, T t Figure 4: NS yields and discount factors. Vertical lines indicate the maturities of bonds used in estimation. Exercise 5. b. If there is no inflation, the security will pay real coupon , as the index ratio will not change. If there is a deflation, and should the index ratio for the security go below 1, it will be automatically reset to 1. In this case, the security will pay an amount equal to the real coupon. c. The bond is more valuable to investors. Compared to the other securities, the bond offers a deflation floor, and this attractive feature will be priced in. d. The price of the security using the NS real curve is (You have to compute dirty prices of bonds first.) e. The squared pricing error for this security is ( ˆP NS Dirty ˆP ) 2 =( ) 2 = , (15) and is much larger than on all other securities. 17

19 f. The model-implied clean price is: ( ) = The squared error on this price is: ( ) 2 = g. Yes. 18

20 Solutions to Chapter 8 Exercise 1. The results are reported in the following Table. T t Y ield Z(t, T ) T t Yield Z(t, T ) T t Y ield Z(t, T )

21 Exercise 2. Follow the lines of Example 8.2 and Table 6.3. Plug in the NS parameters to compute the discount curve. a. The price is $ mn, and is above the $300 mn par value of the security. b. Under the (unrealistic) assumption of constant PSA, you can apply the definition of duration in Chapter 3, and compute it in a standard way (see Section 3.2.3). The duration is c.,d. Compute the prices of the pass through under the different scenarios taking into account the parallel shift in the curve and the change in the PSA. Use definitions 8.1 and 8.2 to obtain the effective duration and convexity, respectively. Prices under the three scenarios: P(dr = 0bps, PSA=150%) P(dr = +50bps, PSA=120%) P(dr = 50bps, PSA=200%) Effective Duration 5.44 Effective Convexity The effective duration is lower than the one obtained under the assumption that the change in rates does not affect the PSA. Standard duration overstates the sensitivity of the MBS price to changes in interest rates. In contrast to the Treasury bonds, the convexity of an MBS is negative (i.e. the value profile is concave with respect to interest rate changes). Therefore, convexity presents a source of risk to investors. This risk is associated with the prepayment option that homeowners have. Effectively, an MBS investor is short an American call option to the homeowners. 20

22 Exercise 7. Results are reported in the following table Tranche A B C E G H Face a. 10/1/1993, PSA = 450% Price Duration b. 4/4/1994, PSA = 450% Price Duration c. 4/4/1994, PSA = 200% Price Duration (b) i. The interest rates have increased (in a nonparallel fashion). The behavior of the G+H portfolio is shown in the following Table. a. b.ii. c.iv. PSA 450% 450% 200% Date 10/1/93 4/4/94 4/4/94 Tranche Inv. Weight Price Price Price G H Port ΔPort (c) i. Yes, the term structure has increased, so the PSA has declined (lower prepayment speed). ii. No. Since all tranches are PO, the portfolio is a bet on decreasing interest rates (it wins when interest rates decline). Between October and April interest rates moved up, therefore the portfolio is losing money. The computed duration confirms this intuition. 21

23 Solutions to Chapter 9 Exercise 1. a. The expected return is equal to 2.5%. b. The forward rate (continously compounded) is equal to %. This is higher than the expected rate computed in Part (a). If we observe high forward rates it may be because of two posibilities: either market participants expect higher future interest rates; or they are strongly averse to risk, and thus the price of long term bonds is low today. c. The market price of risk equals: λ = The high (negative) market price of risk, means that market participants have high risk aversion, which may explain the price of long term bonds today. d. The risk neutral probability equals: p = The interpretation is the same as in Part (c). Exercise 5. a. The value of r 0 is: 5% b. The following three pairs of values for (r 1,u, r 1,d ) are consistent with the two bond prices: (7.010, 3.005), (7.000, 3.015) and (6.000, 4.000). In fact the relationship between both interest rate scenarios is linear and can be summarized by the following equation: r 1,d = r 1,u (16) 22

24 Note that this is consistent with the idea of risk neutrality. In order to compensate a decrease (increase) in the up state, the down side must increase (decrease); given that the risk neutral probabilities are fixed. c. The option gives an information on actual prices, so we can infer the market price of risk and therefore can pin down the actual values for (r 1,u, r 1,d ). These values should be consistent in order to value the option at the given price. The values for (r 1,u, r 1,d ) are: (7.000, 3.015). d. If you didn t know p you would also need the value of the interest rate in the up and down states or the prices of the bonds in these states. 23

25 Solutions to Chapter 10 Exercise 1. a. The bond evolves in the following way: i= b. The market price of risk is: λ = c. The anwers are the following: i. The market price of risk is: λ = Which is the same as the one computed from the bond, this is because the market price of risk is the same across securities since it measures the willingness of agents to hold risky assets. In other words it is a measure on the economic agents and not on the instruments themselves. ii. The price of the option is iii. The result holds the same. d. The price of the option is e. The option has the following structure over time: i=

26 The positions on the short-term bond (N 1 ) and on the three-period bond (N 2 ) vary over time, in order to replicate the option structure, in the following way: N 1 N 2 i=0 1 i= Exercise 3. The following results vary greatly with the dataset used. The results presented use values from September 1, Which are the following: LIBOR 3-month I % I % I % II % II % II % III % III % III % IV % IV % IV % I % I % I % II % II % II % III % III % III % IV % IV % IV % I % I % I % II % II % II % III % III % III % IV % IV % I % I % II % II % III % III % IV % IV % Swaps and Eurodollar Futures Maturity Swap rate Maturity Eurodollar Futures 1 year 0.61% 3 months year 1.29% 6 months year 1.91% 9 months year 2.37% 12 months year 2.72% 7 year 3.20% 25

27 a. The current 3-month interest rate is %, when converted to continous compounding we get: %. The regression gives the following parameters (t-stats in parenthesis): α = ( ) and β = ( ). The predicted value for next quarter is m t+i = %. b. We estimate σ = The tree for the first 3 (i = 3) periods looks like this: i = % 0.74% 1.12% 1.50% 0.13% 0.52% 0.90% -0.09% 0.29% -0.31% c. The zero coupon yield curve can be summarized with the following discounts: Z(0,T) Value Z(0, 0.25) Z(0, 0.50) Z(0, 0.75) Z(0, 1.00) Z(0, 1.25) Z(0, 1.50) Z(0, 1.75) Z(0, 2.00) Z(0, 2.25) Z(0, 2.50) Z(0, 2.75) Z(0, 3.00) These values were computed with the Eurodollar futures and swap data. d. With the emprical σ you get probabilites outside of the [0,1] range. If you increase σ enough you can go into the range. In this specific case, volatility had to be increased by 10 percentage points, to get the first eight periods to be within the range. e. When the probabilities are working as they should, we get that the expected risk neutral rate is higher than the one predicted by the regression. 26

28 This makes sense in order to make economic agents risk neutral, they must be compensated with enough upside. Recall the discussion at the end of chapter 9 (section 9.4.4). 27

29 Solutions to Chapter 11 Exercise 3. a. The price of the zero coupon bonds is: b. The swap rate c(3) is: 4.89%. Z(0,T) Value Z(0, 1) Z(0, 2) Z(0, 3) c. For the option described in the exercise: i. The value of the option is: ii. In order to hedge this security: 1. Choose two securities to hedge this one, for example the 1-year zero coupon bond and the 2-year zero coupon bond. 2. Compute the values for N 1 and N 2 which give the amount of each security that we need to buy. In this case the values are N 1 = and N 2 = Verify that this strategy actually gives the value of the bond: [N 1 Z(0, 1)] + [[N 2 Z(0, 2)] = You can also verify that this replicates the up and down states, by using the inputs for i =1: 28

30 Up State - When interest rates go from 4% to 7% we have that: [N 1 Z up (1, 1)] + [[N 2 Z up (1, 2)] = Down State - When interest rates go from 4% to 3% we have that: [N 1 Z dn (1, 1)] + [[N 2 Z dn (1, 2)] = Which matches exactly the payoffs of the option. d. For the Procter & Gamble leveraged swap we have: i. The value of the security is: ii. Intuitively the value of r that makes the swap value zero should be higher than c(3). P&G is paying more now, through the states in which it has to pay the spread, which means that in order to be compensated it should expect a higher rate from the fixed rate that Bankers Trust pays. iii. Using MS Excel Solver we find the value of this rate to be: 14.94%. Exercise 5. a. The zeros from the LIBOR curve are the following: t P z (0,t) t P z (0,t) t P z (0,t)

31 b. TheHo-Leemodelinthefirst2years(i =8): i = t = % 2.82% 2.96% 3.42% 4.08% 4.67% 5.28% 5.87% 6.67% 2.04% 2.17% 2.63% 3.30% 3.89% 4.50% 5.08% 5.89% 1.39% 1.85% 2.51% 3.11% 3.72% 4.30% 5.11% 1.07% 1.73% 2.33% 2.93% 3.52% 4.32% 0.95% 1.54% 2.15% 2.73% 3.54% 0.76% 1.37% 1.95% 2.76% 0.59% 1.17% 1.98% 0.39% 1.19% 0.41% c. The following table compares the risk neutral expected future interest rate with the forward rate. σ = σ =0.05 σ =0.001 t E [r t] f(t-1,t) E [r t] f(t-1,t) E [r t] f(t-1,t) % 2.849% 2.849% 2.849% 2.849% 2.849% % 2.430% 2.438% 2.430% 2.430% 2.430% % 2.173% 2.204% 2.173% 2.173% 2.173% % 2.241% 2.311% 2.241% 2.241% 2.241% % 2.511% 2.636% 2.511% 2.511% 2.511% % 2.712% 2.908% 2.712% 2.712% 2.712% % 2.928% 3.209% 2.928% 2.928% 2.928% % 3.117% 3.499% 3.117% 3.117% 3.117% % 3.530% 4.029% 3.530% 3.530% 3.530% % 4.132% 4.764% 4.132% 4.132% 4.132% % 4.534% 5.315% 4.534% 4.534% 4.534% % 4.614% 5.558% 4.614% 4.614% 4.614% Note that as volatility increases, so does the difference between the rates. d. The value of the 1-year cap is , the value of the 2-year cap is , and the value of the 3-year cap is In general the model underestimates the price of the securities. e. The value of the swap is zero, as expected. f. The value of the swaption is:

32 Exercise 7. a. The following table reports the first values of the LIBOR curve, in discount factors: b. The implied volatilities are: c. The forward volatilities are: t P z (0,t) t σ implied t t σ forward t d. The price of the corridor note is

33 Solutions to Chapter 12 Exercise 2. a. The value of the American swaption is b. The value of the callable bond is c. In order to only hedge the prepayment risk you hedge the underlying option, intead of the callable bond itself. The hedging strategy is the following: i = 0 1 Short-term bond (N1) American swaption (N2) Note that there is no hedging strategy for node (1,1) since the option is retired at that point. d. To hedge the interest rate risk (which includes prepayment risk) of the callable bond the investor can use the following strategy: i = 0 1 Short-term bond (N1) American swaption (N2) As in the previous part, there is no hedging strategy for node (1,1). Exercise 3. (a) The results on the mortgage are as follows: i. The value of the coupon is C =11, The stream of scheduled interest payments, principal payments, and the remaining principal is: 32

34 Interest Principal Remaining i t paid payments principal , ,000 9,133 90, ,817 9,315 81, ,631 9,502 72, ,441 9,692 62, ,247 9,885 52, ,049 10,083 42, ,285 32, ,491 21, ,700 10, ,914 0 ii. The interest rate tree at semi-annual frequency, for the first seven periods, is: i = % 2.87% 4.69% 6.39% 8.72% 10.67% 12.17% 13.72% 2.17% 3.53% 4.82% 6.58% 8.04% 9.17% 10.34% 2.66% 3.63% 4.96% 6.06% 6.91% 7.79% 2.73% 3.73% 4.57% 5.21% 5.87% 2.81% 3.44% 3.92% 4.43% 2.60% 2.96% 3.34% 2.23% 2.51% 1.90% iii. The value of the mortgage without the prepayment option is: i = ,359 88,791 77,636 67,110 57,045 47,527 38,283 29,049 91,416 80,252 69,578 59,266 49,384 39,716 30,054 82,305 71,519 61,015 50,845 40,840 30,839 73,031 62,377 51,981 41,713 31,447 63,430 52,859 42,387 31,915 53,533 42,903 32,273 43,297 32,546 32,753 33

35 iv. The value of the American option implicit in the mortgage is: i = v. The option-adjusted value of the mortgage is: 100,071. A. The prepayment option is going to be exercised when the value of the mortgage exceeds the outstanding principal. This occurs in the following nodes: i = H H H H H H H H 1 X H H H H H H 2 R H H H H H 3 R H H H H 4 R X H H 5 R R X 6 R R 7 R where H stands for Hold, X for Exercise, and R for Retired. B. Yes. Any path leading to nodes (7,5) or (9,6). C. It is fairly priced in the sense that it takes into account the prepayment option, assuming that agents act optimally. But this may not be the case, people may forget to exercise at the right time or they may be other factors affecting their decision that are not taken into account in the model. D. If the homeowner doesn t refinance then the value of the option is higher than before. Refinancing is done for the homeowner s 34

36 benefit, if he forgets to do so, it is the holder of the security that gets an extra amount of cash. (b) The results for the mortgage backed securities are as follows: i. The value of the pass-through is 99,334. The spot rate duration is: ii. The value of the PO strip is 93,989 and the value of the IO strip is 5,345. A. The sum of the IO and PO strips equals the value of the passtrough security. Exercise 5. Figure 5: Price of zeros (discounts) for different maturities a. The following table reports all zero coupon bonds using the bootstrap method. See Figure?? 35

37 T P z (0,T) b. The following shows the values of the BDT tree for the first 6 periods, using σ = i= % 5.87% 6.51% 7.40% 8.52% 9.74% 10.92% 4.41% 4.89% 5.56% 6.40% 7.32% 8.21% 3.68% 4.18% 4.81% 5.50% 6.17% 3.14% 3.62% 4.14% 4.64% 2.72% 3.11% 3.49% 2.34% 2.62% 1.97% c. On the Option Adjusted Spread (OAS): i. The price of the callable note is ii. The OAS is: 26 bps. iii. In order to make the OAS equal to zero, σ =0.15. d. The price of the callable note is The main reason is that volatilty 36

38 has a bigger impact in the Ho-Lee model, which in turn raises the value of the option and reduces the value of the callable security. e. The duration of the security is: f. Convexity: i. The dollar spot rate duration at the required nodes is: D $ i+1,j = and D $ i+1,j+1 = ii. Spot rate convexity is then: j. If instead of June, 2009 the lockout period finishes June, 2008 (i.e. instead of lasting two years it lasts one year), the convexity goes to

39 Solutions to Chapter 13 Exercise 2. a. The first 6 steps of the monthly Ho-Lee tree look like the following: i = % 5.29% 5.58% 5.87% 6.16% 6.44% 6.73% 4.71% 5.00% 5.29% 5.58% 5.87% 6.16% 4.42% 4.71% 5.00% 5.29% 5.58% 4.13% 4.42% 4.71% 5.00% 3.85% 4.13% 4.42% 3.56% 3.85% 3.27% b. The price of the interest rate barrier option would be The problem with the backward methodology is that we cannot distinguish the states that go down and out, since this is path dependant. c. Through the BDT model we get a price of , which is significantly lower than the previous value. This is because the BDT tends to be more concentrated around the center than the Ho-Lee model. This means that positive results are less extreme than in the Ho-Lee model. Thus the lower value. 38

40 Exercise 3. a. The trigger rates are: i r i % % % % % % % % % b. The value of the mortgage is: 100,477, which is still far from the value of the security obtained in the previous section: 100,071 (still outside of the confidence intervals). This may be because only 1,000 simulations were used. c. The value of the 3.5% passthrough security is: 99,154. d. Adding some probability to the model we get: i. Adding a 50 PSA we get that the price of the mortgage is now: 99,301. ii. If 80% of homeowners forget to exercise at the optimal time we get that the price of the passthrough is now: 99,294. e. This is good news for the bank issuing the mortgage since now it gets more money. As seen in the previous question, when homeowners don t exercise optimally it means that mortgages increase in value. f. The price of the IO strips is 9,259 with spot rate duration The price of the PO strips is 89,895 with spot rate duration Note that here only optimal exercise is considered. 39

41 Exercise 4. a. The results are the same as in chapter 11, exercise 5. b. See chapter 11, exercise 5. c. After around 600 simulations the average error over 40 periods goes below 10 cents on the value of a zero coupon bond with face value 100. d. The price from the tree model for a 1-year cap is Through the simulation approach (1000 simulations) we get , with a a standard error of The price from the tree is well within the confidence intervals. Sigma used to price a 1-year cap is e. The price of the Asian cap is with a standard error of Sigma for pricing a 2 year bond is f. Spot rate duration for the option is

42 Solutions to Chapter 14 Exercise 2. Results for both cases should be the same as figures in the book. Changes may appear from different values in the random number generator. Exercise 4. a. Consider the base case with the following parameters: γ =1, r = 5.40%, σ =1%,r(0) = 0.17% and dt = 1/252; we have: i. For γ: As this parameter increases, the speed by which the rate converges to the long-term rate, r, increases. Inversely as this parameter decreases, the rate converges slower to r (see Figure??). Figure 6: Vasicek simulations and expected value for different values of γ 41

43 ii. For σ: As this parameter increases the volatility of the process goes up. Inversely as it goes down so does the volatility of the process (see Figure??). Figure 7: Vasicek simulations and expected value for different values of σ b. The effects of having different values for r(0) can be seen in Figure??. We choose values that are higher, lower and the same as r Figure 8: Vasicek simulations and expected value for different values of r 0 42

44 Exercise 5. a. We have that: β = γdt and α = γ rdt. Forσ we can simply estimate the standard error of the regression and annualize it by dividing the standard error by dt. Using daily data on Treasury interest rates from 1/1/2008 to 3/5/2010 we obtain the following regression estimates: α = ; β = ; and σ SE = This leads to the following parameters: γ = ; r = ; and σ = In this particular case, this leads to a highly volatile process, which easily gives negative rates. The reason for this shortcoming is twofold: on the one hand we are using a period of particular interest rate volatility to estimate the model (the range of rates during this 2 year period was between 3.37% and 0.00%); on the other hand, r 0 is 0.11% which is already very close to zero. b. Figure?? plots the forecast with a simulation which shows the problems with the estimated model mentioned above. Figure 9: Vasicek model expected and simulated rates c. Figure?? shows the histograms for the forecasted values. 43

45 Figure 10: Histogram of rates at 1 year, 3 years and 5 years Exercise 7. Given the process: dr t = γ( r r t )dt + σdx t We have that m(r t,t)=γ( r r t )dt; ands(r t,t)=σ. Using Ito s lemma: [( ) ( ) F F dp t = + m(r t,t)+ 1 ( 2 ) ] ( ) F F t r 2 r 2 s(r t,t) 2 dt + s(r t,t)dx t r a. For F (r) =A + Br: So: F t =0; F r = B; 2 F r 2 =0 dp t = Bγ( r r t )dt + BσdX t = Bdr t b. For F (r) =e A Br F t =0; F r = BeA Br ; 2 F r 2 = B2 e A Br So: dp t = [ BPγ( r r t )+ 12 B2 Pσ 2 ] dt + BPσdX t dp t = P [ ] 1 2 B2 σ 2 dt Bdr t 44

46 Solutions to Chapter 15 Exercise 1. a. In order to value the zero coupons we use the risk neutral parameters. τ Z(t,T) τ Z(t,T)

47 b. Figure?? presents the yield curve for the bonds. Figure 11: Spot rate of bonds up to 10 years c. Spot rate duration is: τ B(t;T) τ B(t;T)

48 Exercise 2. a. Figure?? presents the effect on the term structure of interest rates due to changes in γ. Figure 12: Term structure of interest rates for three choices of γ (Vasicek) b. Figure?? presents the effect on the term structure of interest rates due to changes in r. Figure 13: Term structure of interest rates for three choices of r (Vasicek) 47

49 c. Figure?? presents the effect on the term structure of interest rates due to changes in σ. Figure 14: Term structure of interest rates for three choices of σ (Vasicek) Exercise 3. a. Figure?? presents the effect on the term structure of interest rates due to changes in γ. Figure 15: Term structure of interest rates for three choices of γ (CIR) 48

50 b. Figure?? presents the effect on the term structure of interest rates due to changes in r. Figure 16: Term structure of interest rates for three choices of r (CIR) c. Figure?? presents the effect on the term structure of interest rates due to changes in σ. Figure 17: Term structure of interest rates for three choices of σ (CIR) Exercise 6. Short-term bonds have higher yield volatility than long-term bonds (see Figure??). Return volatility and yield volatility are not the same since yields are a sort of average measure in order to make comparable differ- 49

51 ent maturities. Returns will be affected by the yield volatility times the time to maturity. Return volatility is always increasing in relation to time to maturity. Figure 18: Yield volatility with respect to τ Exercise 7. The Fundamental Pricing Equation states: Z t + Z r m (r, t)+ 1 2 Z 2 r 2 σ2 = rz where: m (r, t) =γ ( r r) The Vasicek formula for pricing a zero coupon bond: A(t;T ) B(t;T )r Z(r, t; T )=e where: B(t; T )= 1 ( ) 1 e γ (T t) γ ) A(t; T )=(B(t; T ) (T t)) ( r σ2 2(γ ) 2 σ2 B(t; T ) 2 4γ For the Left Hand Side of the Fundamental Pricing Equation we have: Z t =[A (t; T ) B (t; T )r] Z(r, t; T ) Z = B(t; T )Z(r, t; T ) r 2 Z r = B(t; T 2 )2 Z(r, t; T ) 50

52 Where: B (t; T )= e γ (T t) ) A (t; T )=(1+B (t; T )) ( r σ2 σ2 2(γ ) 2 2γ B (t; T )B(t; T ) Note that: B(t; T )= 1 γ (1 + B (t; T )) then: and γ B(t; T )=1+B (t; T ) γ B(t; T ) B (t; T )=1 So we can present A (t; T )as: ) A (t; T )=γ B(t; T ) ( r σ2 σ2 2(γ ) 2 2γ B (t; T )B(t; T ) = B(t; T ) [γ r ] σ2 2γ σ2 2γ B (t; T ) = B(t; T ) [γ r ] σ2 2γ (1 + B (t; T )) = B(t; T ) [γ r ] σ2 2γ γ B(t; T ) = B(t; T ) [γ r ] σ2 2 B(t; T ) So: A (t; T )=γ r B(t; T ) σ2 B(t; T )2 2 Which means that: Z [B(t; t = T )γ r ] σ2 2 B(t; T )2 B (t; T )r Z(r, t; T ) So we can present the Left Hand Side of the Fundamental Pricing Equation as: ] Z(r, t; T ) [B(t; T )γ r σ2 2 B(t; T )2 B (t; T )r B(t; T )γ ( r r)+ σ2 B(t; T )2 2 = Z(r, t; T )[B(t; T )γ r B (t; T )r] =Z(r, t; T )r [B(t; T )γ B (t; T )] Recall that: B(t; T )γ B (t; T )=1,so: LHS = Z(r, t; T )r 51

53 Exercise 9. a. Parameter Estimation: i. The risk neutral parameters are: γ = and r =7.279%. Note that σ is obtained through the regression estimate. ii. The risk natural parameters are: γ =0.1766, r =5.97% and σ = iii. The risk neutral model has a quicker convergence speed and a higher long-term rate than the risk natural model. This makes sense because we need to compensate economic agents in order for them to become risk neutral (we consider them to be risk averse). b. Sensitivity to interest rates: i. Because of the assumption r<15%, we can say that: P IF = P zero + P fixed P float In other words we price the inverse floater as a portfolio of different bonds, so PIF r can be presented as: P IF r = P zero r + P fixed r P float r For the Vasicek model, the price of a zero coupon bond is given by: A(t;T ) B(t;T )r Z(r, t; T )=e which means that: Z(r, t; T ) r We know all parameters for this so: = B(t; T )Z(r, t; T ) P zero r = Z(r, t;3) r =

54 The fixed coupon bond is composed by several zeros: P fixed r = 3 T =1 Z(r, t; T ) Z(r, t;3) c + = r r From the Z(r, 0; 1) we can obtain the annually compounded 1-year rate r 1 (0, 1) = 3.99% which we can use to obtain the value of a floating rate bond: P Float =(1+r 1 (0, 1))Z(0,r;1) So: Thus: P Float r P IF r =(1+r 1 (0, 1)) Z(0,r;1) r = = = ii. For convexity we have the same idea as in the previous exercise but instead we use: 2 Z(r, t; T ) r 2 = B(t; T ) 2 Z(r, t; T ) So: Thus: 2 P fixed r 2 = 2 P Float r 2 2 P zero r 2 3 T =1 = 2 Z(r, t;3) r 2 = c 2 Z(r, t; T ) r Z(r, t;3) r 2 = =(1+r 1 (0, 1)) 2 Z(0,r;1) r 2 = P IF r 2 = = iii. For the whole portfolio we simply weigh the sensitivities of interest rate of both assets, so we get: Π r =

55 c. The Fundamental Pricing Equation: 2 Π r 2 = i. The equation that Π(r, t; T ) must satisfy is: ii. In order to obtain Π t rπ = Π t + Π r η ( r r)+ 1 2 Π 2 r 2 σ2 we can rearrange the previous equation: Π t Π = rπ r η ( r r) 1 2 Π 2 r 2 σ2 From previous exercises we know both Π r , so: Π t = = and 2 Π r 2 = iii. In one day, the capital gain that can be contributed to the passage of time is: = This is the partial derivative of the value of the portfolio against time. Thus, we can interpret this as the following: each day, the value of the portfolio will rise by.0057 billion when interest rate does not change. Since this portfolio is not Delta Hedged, the Theta- Gamma relation does not hold. However, holding Delta constant, we can still see the impact that a low (or even negative) Gamma will tend to lead to high Theta. 54

56 d. ii. The following historgram plots the possible values for the portfolio in one year. 60 Histogram of Orange County portfolio distribution Frequency Portfolio Value Figure 19: Historgram of possible Orange County portfolio values in one year iii. The following table summarizes the results: VaR Initial 1-year 5% % As seen in the table the results experienced by Oragne County s portfolio where completely forseeable. 55

57 Solutions to Chapter 16 Exercise 1. We follow these steps: 1. Parameter Estimation: We use the data given in Table 15.4 to fit the Vasicek model, for which we obtain γ =0.0290, r =2.09% and σ = 1.78%. The graph shows the yield curve and its fitted values according to the Vasicek model. For σ one possiblity is to estimate the value from historical data or simply include it as a free parameter in the optimization process (see Figure??). Figure 20: Yield curve and Vasicek model (September 25, 2008) 56

58 2. Relative Pricing Error Discovery: We look for the bond where there is the highest difference between the model and the data prices. τ Z Data Z Model Diff Two bonds appear to be particularly promising τ =3.139 and τ =8.139, since they have the biggest gap between prices. We will use τ = 3.139, which seems to be overpriced. 57

59 3. Set-Up Trading Strategy: i. Since Z(3.139) seems to be overpriced we short sell it, making a profit of $ per bond. ii. Using the bond maturing at τ =4.889 we buy a synthetic position: Ẑ(3.139) = Δ Z(4.889) + C 0 = $ where: Δ= B(0, 3.139) ZModel (r, 0; 3.139) B(0, 4.889) Z(4.889) = C 0 = Z Model (r, 0; 3.139) 100 Δ Z(4.889) 100 = In other words, for each Z(3.139) bond we shorted, we buy of a Z(4.889) bond and take a $ cash position. This gives a $ profit per bond. Exercise 4. a. The price of the option is: Call (x 100) = $ b. In order to hedge the call we pick the zero coupon bond maturing at the same time. Two methodologies were proposed to obtain the hedge ratio for the replicating portfolio: i. Using equation (16.31) we obtain: Δ = and C 0 = ii. Through the numerical approximation we obtain: Δ = and C 0 = Which is almost identical to the number obtained with the previous methodology. c. Figure?? shows that the strategy does a good job replicating the option. 58

60 Figure 21: Value of the asset and of the replicating portfolio Exercise 7. In order to perform the trade we proceed as follows: 1. Parameter Estimation: i. Risk Natural Parameters. Recall that the Cox-Ingersoll-Ross (CIR) process for the short rate is defined by: dr =(η γr t ) dt + αr t dx As in the Vasicek case we would estimate risk neutral parameters through calibrating the model to the data (using the Non-Linear Least Squares methodology), yet the volatility parameter α we would derive from the actual data. Additionally, if we want to test the model against simulated values of interest rates we will eventually need the risk natural parameters as well. For this reason, we start by obtaining the risk natural parameters and the volatility parameter, before calibrating the model to the risk neutral parameters. As a reminder, recall that in the case of the Vasicek model we had: dr t = γ ( r r t ) dt + σdx t 59

61 and we estimated the risk natural parameters through a regression of the form (using data on the overnight repo rate from May 21, February 17, 2004): r t+dt = α + β 1 r t + σ S.E. ɛ where: α = γ rdt ; β 1 =(1 γdt), σ = σs.e. dt and ɛ N(0, 1). In the case of the CIR model this regression cannot be used because there is a r t in the variance term. In order to eliminate it we must divide both sides of the equation by r t : dr t = γ( r r t)dt + αdtɛ rt rt Decomposing dr t and rearranging a few terms, we have: r t+dt rt = γ rdt 1 rt +(1 γdt) r t rt + αdtɛ Which can be summarized as the following regression: r t+dt 1 r t = β 1 + β 2 + σ S.E. ɛ rt rt rt where: β 1 = γ rdt, β 2 =(1 γdt) andσ S.E. = αdt. Running a simple regression gives the following results (p-values in parenthesis), as well as the following parameters for the CIR model. α dt 1/252 (0.9224) β γ (0.7290) r β α (0.0000) σ S.E Note that neither the intercept nor β 1 are significant. The reason is that the regression is including the intercept, which is not present in the model above. We must run the regression without intercept in 60

62 order to have a better fit with our model. The results of doing so are the following: α 0 dt 1/252 (N/A) β γ (0.0026) r β α (0.0000) σ S.E ii. Risk Neutral Parameters. In order to price the bonds we need to find the risk neutral parameters (γ and r ). To do so we use the Non- Linear Least Squares approach to calibrate the parameters to the data given in Table 16.4 (we follow the same steps as in section 16.8). The minimization procedure yields r =16.53% and γ = Figure?? presents the spot rate from the CIR model. Figure 22: Spot rate 61