A., “Single Factor Heath-Jarrow-Morton Term Structure Models Based on Markov SpotInterest Rate Dynamics,” Journal of Financial and Quantitative Analysis, 30 (1995), 619-42.
Longstaff, F. A., and E. S. Schwartz, “Valuing American Options by Simulation: A Simple LeastSquares Approach,” Review of Financial Studies, 14, no. 1 (2001), 113-47.
Miltersen, K., K. Sandmann, and D. Sondermann, “Closed Form Solutions for Term StructureDerivatives with Lognormal Interest Rates,” Journal of Finance, 52, no. 1 (March 1997), 409-30.
Rebonato, R., /nterest Rate Option Models, 2nd edn., Wiley, Chichester, 1998.
Ritchken, P., and L. Sankarasubramanian, “Volatility Structures of Forward Rates and the Dynamicsof the Term Structure,” Mathematical Finance, 5 (1995), 55-72.
QUESTIONS AND PROBLEMS(ANSWERS IN SOLUTIONS MANUAL)
 
 
24.1. For the model in Section 24.1 when f(r) = r,a. What is the process followed by the bond price P(t, Tn the traditional risk-neutral world?b. What is the process followed by the bond’s yield in this risk-neutral world?c. For the parameters in Figure 24.1, what is the instantaneous correlation between the three-month and the ten-year zero rates?
24.2. For the model in Section 24.1, what are the coefficients of dz, and dz» in the process followed bythe instantaneous forward rate when f(r) =r?
24.3. Explain the difference between a Markov and a non-Markov model of the short rate.
24.4. Prove the relationship between the drift and volatility of the forward rate for the multifactorversion of HJM in equation (24.10).
24.5. Show from equation (24.9) that if the instantaneous forward rate F(t, T) in the HJM model has a
constant standard deviation then the process for r is the same as in the Ho Lee model.
24.6. Show that equations (23.17) and (23.18) are consistent with equations (24.8) and (24.9) whens(t, T) = e749,
24.7. It can be shown that in a one-factor model where the bond price volatility u(t, T, Q,) is a functiononly of ¢ and T, the process for r is Markov if and only if u(t, T) has the form x()[W(T) — yO].Use equations (24.8) and (24.9) to show that when c(t, 7) has this form the process for r isMarkov.
24.8. Provide an intuitive explanation of why a ratchet cap increases in value as the number of factorsincrease.
24.9. Show that equation (24.14) reduces to (24.6) as the 4; tend to zero.Interest Rate Derivatives: More Advanced Models 591
24.10.24.11.24.12.24.13,24,14,24.15.
Explain why a sticky cap is more expensive than a similar ratchet cap.
Explain why 10s and POs have opposite sensitivities to the rate of prepayments.
“An option-adjusted spread is analogous to the yield on a bond.” Explain this statement.Prove equation (24.19).
Prove the formula for V(t) on page 583.
Prove equation (24.22).
ASSIGNMENT QUESTIONS
 
 
24.16.
24.17.
In an annual-pay cap, the Black volatilities for caplets with maturities 1, 2, 3, and 5 years are18%, 20%, 22%, and 20%, respectively. Estimate the volatility of a one-year forward rate in theLIBOR market modei when the time to maturity is (a) 0 to i year, (b) 1 10 2 years, (c) 2 to 3 years,and (d) 3 to 5 years. Assume that the zero curve is flat at 5% per annum (annually compounded).Use DerivaGem to estimate flat volatilities for 2-, 3-, 4-, 5-, and 6-year caps.In the flexi cap considered in Section 24.3, the holder is obligated to exercise the first N in-the-money caplets. After that, no further caplets can be exercised (in the example, N = 5). Two otherways that flexi caps are\sometimes defined are:a. The holder can choose whether any caplet is exercised, but there is a limit of N on the totalnumber of caplets that can be exercised.b. Once the holder chooses to exercise a caplet, all subsequent in-the-money caplets must beexercised up to a maximum of N.Discuss the problems in valuing these types of flexi caps. Of the three types of flexi caps, whichwould you expect to be most expensive? Which would you expect to be least expensive?592 CHAPTER 24
APPENDIX 24A
The A(t, T), op, and 6(t) Functions in the Two-FactorHull-White Model
In this appendix, we provide some of the analytic results for the two-factor Hull-White modeldiscussed in Section 24.1 when f(r) =r.
The A(t, T) function isP(0, T)
In A(t, 7) = In PO.
 
+ Bt, T)F(O, t) —where
2od ~n= ral — ey BL, TY — po o9[B(0, t)C(0, 1) B(t, T) + v4 — 2]
— 4631C(0, 1° B(t, T) + ve — vs]e MHOIT (gla tbyt -1) eT (gett -1)
1 G+bla—-b) Qala —b)
1 twea (» + C(t, 7) — C(O, T) + 1 B(t, TY — 1BO, TY + .-
ab a
eat) _ ~)e athyt —]1 et —1
=~ a= bya tb) dala)
 
| t @ “1= C(0, t) — 4B, 1 +-414 = (15 ~ C101) 4800.0) ++)5 =p [4CU, TY — $C0, 77 + a]!w= pl ~ 3C(0, ty]where the Bt, 7) and C(t, T) functions are as in Section 24.1 and F(t, 7) is the instantaneous
forward rate at time ¢t for maturity T.The volatility function, op, is
fob = | {of1B(e, 7)— Bee, nF +o¥IC(e.7) — Cle. OF0+20 03[B(t, T) — B(t, HIIC(t, T) — CC, t))} dtThis shows that + has three components. Define
1!
_ —aT atU= Tan ey)
 
and
1 oT _ btbah” ~* )
 interest Rate Derivatives: More Advanced Models 593
The first component of o3 is
2an 201 ,-2at7b D (l-e™)The second is
 
uv v? Ua(See +5, 1) are al »)
The third is
 
200102, at -at,(U » at v (a+b)qe eae —D-ee 0)
Finally, the A(t) function isO(t) = F,(0, t) + aF(0, t) + ¢,(0, t) + aG(0, 4)where the subscript denotes a partial derivative and
H(t, T) = dof Bt, TY + 403CKt, TY + po 102 B(t, TYC(t, T)CHAPTER 25SWAPS REVISITED
Swaps have been central to the success of over-the-counter derivatives markets during the 1980sand 1990s. They have proved to be very flexible instruments for managing risk. Based on the rangeof different contracts that now trade and the total volume of business transacted each year, swapsare arguably one the most successful innovations in financial markets ever.
In Chapter 6 we discussed how plain vanilla interest rate swaps can be valued. The standardapproach can be summarized as: “‘Assume forward rates will be realized.’ The steps are:
1. Calculate the swap’s net cash flows on the assumption that LIBOR rates in the future equaltoday’s forward LIBOR rates.
2. Set the value of the swap equal to the present value of the net cash flows using today’sLIBOR zero curve for discounting.
We used this approach in Example 6.3. It involves regarding the swap as a portfolio of forwardrate agreements. As discussed in Chapter 6, it is important to refiect day count conventions andbusiness day conventions in the calculation of the cash flows and their timing. The LIBOR zerocurve is calculated as described in Section 6.3 from LIBOR deposit rates, Eurodollar futures, andquoted swap rates.
In this chapter we describe a number of nonstandard swaps. Some can be valued using the“assume forward rates will be realized’ approach; some require the application of the convexity,timing, and quanto adjustments we encountered in Chapters 21 and 22; and some containembedded options.
25.1 VARIATIONS ON THE VANILLA DEAL
 
Many interest rate swaps involve relatively minor variations to the plain vanilla structure inTable 6.3. In some swaps the notional principal changes with time in a predetermined way. Swapswhere the notional principal is an increasing function of time are known as step-up swaps. Swapswhere the notional principal is a decreasing function of time are known as amortizing swaps. Step-up swaps could be useful for a construction company that intends to borrow increasing amounts ofmoney at floating rates to finance a particular project and wants to swap it to fixed-rate funding.An amortizing swap could be used by a company that has fixed-rate borrowings with a certainprepayment schedule and wants to swap them to borrowings at a floating rate.
The principal, as well as the frequency of payments, can be different on the two sides of a swap.This is illustrated in Table 25.1, which shows a hypothetical swap between Microsoft and Citibank
594Swaps Revisited 595
 
Table 25.1 Extract from confirmation for a hypothetical swap where theprincipal and payment frequency are different on the two sides
 
 
Trade date 4-January-2001Effective date 11-January-2001Business day convention (all dates) Following business dayHoliday calendar US.
Termination date . 11-January-2006
Fixed amounts
Fixed-rate payer Microsoft
Fixed-rate notional principal USD 100 million
Fixed rate 6% per annumFixed-rate day count convention Actual/365
Fixed-rate payment dates Each |1-July and [1-January
commencing 11-July-2001 up toand including 11-January-2006
 
Floating amounts
Floating-rale payer Citibank
Floating-rate notional principal USD 120 million
Floating rate USD I-month LIBORFloating-rate day count convention Actual/360
Floating-rate payment dates 11-July-2001 and the I ith of
each month thereafter up toand including 11-January-2006
where the notional principal is $120 million on the floating side and $100 million on fixed side.Payments are made every month on the floating side and every six months on the fixed side.
These variations to the basic plain vanilla structure do not affect the valuation methodology. Wecap-still use the “assume forward rates are realized” approach.
/ The floating reference rate for a swap is not always LIBOR. For example, in some swaps it is thethree-month Treasury bill rate. A basis swap consists of exchanging cash flows calculated using onefloating reference rate for cash flows calculated using another floating reference rate. An examplewould be a swap where the three-month Treasury bill rate plus 60 basis points is exchanged forthree-month LIBOR with both being applied to a principal of $100 million. A basis swap could beused for risk management by a financial institution whose assets and liabilities are dependent ondifferent floating reference rates.
Swaps where the floating reference rate is not LIBOR can be valued using the “assume forwardrates are realized” approach. A zero curve other than LIBOR is necessary to calculate the net cashflows. However, we always discount the calculated cash flows at LIBOR.
25.2 COMPOUNDING SWAPS
 
Another variation on the plain vanilla swap is a compounding swap. Table 25.2 gives an example.There is only one payment date for both the floating-rate payments and the fixed-rate payments.This is at the end of the life of the swap. The floating rate of interest is LIBOR plus 20 basis points.596
CHAPTER 25
Table 25.2. Extract from confirmation for a hypothetical compounding
swap
Trade date
Effective date
Holiday calendar
Business day convention (all dates)Termination date
Fixed amounts
Fixed-rate payer
Fixed-rate notional principalFixed rate
Fixed-rate day count conventionFixed-raie paymeni daicFixed-rate compoundingFixed-rate compounding dates
4-January-200111-January-2001US.
Following business day
1 1-January-2006
Microsoft
USD 100 million
6% per annum
Actual/365
1i-January-2006
Applicable at 6.3%
Each 11-July and 1|1-Januarycommencing | 1-July-2001 up toand including 11-July-2005
 
Floating amounts
Floating-rate payer
Floating-rate notional principalFloating rate
Floating-rate day count conventionFloating-rate payment date
Floating-rate compounding
Floating-rate compounding dates
Citibank
USD 100 million
USD 6-month LIBORplus 20 basis points
Actual/360
1]-January-2006
Applicable at LIBORplus 10 basis points
Each 11-July and 11-Januarycommencing 1I1-July-2001 up toand including | 1-July-2005
Instead of being paid, the interest is compounded forward until the end of the life of the swap at arate of LIBOR plus 10 basis points. The fixed rate of interest is 6%. Instead of being paid, thisinterest is compounded forward at a fixed rate of interest of 6.3% until the end of the swap.We can use the “assume forward rates are realized’’ approach for valuing a compounding swap.It is straightforward to deal with the fixed side of the swap because the payment that we will make:at maturity is known with certainty. The “assume forward rates are realized’’ approach for thefloating part is justifiable because we can devise a series of forward rate agreements (KF RAs) wherethe floating-rate cash flows are exchanged for their values when each floating rate equals thecorresponding forward rate. To explain how this is done, suppose that fy is the time of the paymentdate immediately preceding the valuation date and that the payment dates following the valuationdate are at times t), t%,..., t,- Define t; = t;4; — t; (0 <i <n — 1) and other variables as follows:'
' All rates are here expressed with a compounding frequency reflecting their maturity. Three-month rates areexpressed with quarterly compounding, six-month rates are expressed with annual compounding, and so on.Swaps Revisited 597
R;: LIBOR rate from ¢; to t;,, (Ro is known)F,: Forward rate applicable to period between time ¢; and t,,; for é > 1 (all known)
5,: Spread above LIBOR at which interest is incurred on the floating side of the swap (20 basispoints in Table 25.2)
5): Spread above LIBOR at which floating interest compounds (10 basis points in Table 25.2)The value of the floating side of the swap at time t, is known. It isQ) = Qoll + (Ro + 52)t0] + L(Ro + 51)
The first term on the right-hand side is the result of compounding the floating payments from time tyto t). The second term is the floating payment at time ty.The value of the floating side at time ft is not known and depends on R,. It is
Q2 = Qi +(R, + 52)t] + LCR) + 5)t (25.1)
However, we can costlessly enter into two FRAs:
1. An FRA to exchange R, + 5) for F; +5 on a principal of Q,
 
2. An FRA to exchange R, +5, for F; +s, on a principal of L
The first FRA shows that the first term on the right-hand side of equation (25.1) is worthQ,[1 + (F, + 52)t,]. The second FRA shows that the second term on the right-hand side ofequation (25.1) is worth L(F, + 5,)t,. The value of the floating side of the swap at time ty istherefore
Q2 = Q,[1 + (F) + 52) ] + LF) + 5))t)A similar argument gives
Oj4) = Oi] + (F) + 52)t;) + LF + 5);for alli(1 <i<gn—-1).
Example 25.1 A compounding swap with annual resets has a life of three years. A fixed rate ispaid and a floating rate is received. The fixed interest rate is 4% and the floating interest rate is12-month LIBOR. The fixed side compounds at 3.9% and the floating side compounds at 12-monthLIBOR minus 20 basis points. The LIBOR zero curve is flat at 5% with annual compounding andthe notional principal is $100 million.
On the fixed side, interest of $4 million is earned at the end of the first year. This compounds to4x 1.039 = $4.156 million at the end of the second year. A second interest amount of $4 million isadded at the end of the second year, bringing the total compounded forward amount to $8.156million. This compounds to 8.156 x 1.039 = $8.474 million by the end of the third year, when thereis the third interest amount of $4 million. The cash flow at the end of the third year on the fixed sideof the swap is therefore $12.474 million.
On the floating side we assume all future interest rates equal the corresponding forward LIBORrates. Given the LIBOR zero curve, this means that we assume all future interest rates are 5% withannual compounding. The interest calculated at the end of the first year is $5 million. Compoundingthis forward at 4.8% (forward LIBOR minus 20 basis points) gives 5 x 1.048 = $5.24 million at theend of the second year. Adding in the interest, the compounded forward amount is $10.24 million.Compounding forward to the end of the third year, we get 10.24 x 1.048 = $10,731 million. Addingin the final interest gives $15.731 million.598 CHAPTER 25
We can value the swap by assuming that it leads to an inflow of $15.731 million and an outflow of$12.474 million at the end of year 3. The value of the swap is therefore
15.731 — 12.474
pst = 2814
or $2.814 million. (This analysis ignores day count issues.)
25.3 CURRENCY SWAPS
 
 
 
We introduced currency swaps in Chapter 6. These enable an interest rate exposure in one currencyto be swapped for an interest rate exposure in another currency. Usually two principals arespecified, one in each currency. The principals are exchanged at both the beginning and the endof the life of the swap as described in Section 6.5.
Suppose that the currencies involved in a currency swap are U.S. dollars (USD}-and Britishpounds (GBP). Ina fixed-for-fixed currency swap. a fixed rate of interest is specified in each currency.The payments on one side are determined by applying the fixed rate of interest in USD to the USDprincipal; the payments on the other side are determined by applying the fixed rate of interest in GBPto the GBP principal. We discussed the valuation of this type of swap in Section 6.6.
Another popular type of currency swap is floating-for-floating. In this, the payments on one sideare determined by applying USD LIBOR (possibly with a spread added) to the USD principal;similarly the payments on the other side are determined by applying GBP LIBOR (possibly with aspread added) to the GBP principal. A third type of swap is a cross-currency interest rate swap,where a floating rate in one currency is exchanged for a fixed rate in another currency.
Floating-for-floating and cross-currency interest rate swaps can be valued by assuming forwardrates are realized. Future LIBOR rates in each currency are assumed to equal today’s forward rates.This enables the cash flows in the currencies to be determined. The USD cash flows are disc6unted atthe USD LIBOR zero rate. The GBP cash flows are discounted at the GBP LIBOR zero rate. Thecurrent exchange rate is then used to translate the two present values to a common currency.
An adjustment to this procedure is sometimes made to reflect the realities of the market. Intheory, a new floating-for-floating swap should involve exchanging LIBOR in one currency forLIBOR in another currency (with no spreads added). In practice, macroeconomic effects give riseto spreads. Financial institutions often adjust the discount rates they use to allow for this. As anexample, suppose that market conditions are such that USD LIBOR is exchanged for Japaneseyen (JPY) LIBOR minus 20 basis points in new floating-for-floating swaps of all maturitics. In itsvaluations a U.S. financial institution would discount USD cash flows at USD LIBOR and itwould discount JPY cash flows at JPY LIBOR minus 20 basis points.? It would do this in allswaps that involved both JPY and USD cash flows.
25.4 MORE COMPLEX SWAPS
 
 
 
We now move on to consider some examples of swaps where the simple rule ‘‘assume forward rateswill be realized’’ does not work. In each case we must adjustments to forward rates.
? This adjustment is ad hoc, but, if it is not made, traders make an immediate profit or loss every time they trade anew JPY-USD floating-for-floating swap.Swaps Revisited 599
LIBOR-in-Arrears Swap
A plain vanilla interest rate swap is designed so that the floating rate of interest observed on onepayment date is paid on the next payment date. An alternative instrument that is sometimes tradedis a LIBOR-in-arrears swap. In this, the floating rate paid on a payment date equals the rateobserved on the payment date itself.
Suppose that the reset dates in the swap are f; fori =0,1,...,n, with t; =t,4, —t,. Define R; asthe LIBOR rate for the period between t; and t;,,, F; as the forward value of R;, and o; as thevolatility of this forward rate. (The value of o; is typically implied from caplet prices.) In a LIBOR-in-arrears swap, the payment on the floating side at time ¢; is based on R; rather than R,_). Asexplained in Section 22.6, it is necessary to make a convexity adjustment to the forward rate whenthe payment is valued. From equation (22.16), the valuation should be based on the assumptionthat the forward rate is
Fror tit;
F;i+ TLR,
 
(25.2)
rather than F;.
Example 25.2 In a LIBOR-in-arrears swap, the principal is $100 million. A fixed rate of 5% isreceived annually and LIBOR is paid. Payments are exchanged at the ends of years I, 2, 3, 4, and 5.The yield curve is flat at 5% per annum (measured with annual compounding). All caplet volatilitiesare 22% per annum.
The forward rate for each floating payment is 5%. If this were a regular swap rather than anin-arrears swap, its value would (ignoring day count conventions, etc.) be exactly zero. Because it is
an in-arrears swap, we must make convexity adjustments. In equation (25.2), F,; = 0.05, o, = 0.22,and t; = | for all i. The convexity adjustment changes the rate assumed at time t from 0.05 to0.05" x 0.22? x 1 x t;
0.05 + —Th005n1~— = 0.05 + 0.000105¢;
The floating rates for the payments at the ends of years 1, 2, 3, 4, and 5 should therefore be assumedto be 5.0105%, 5.021%, 5.0315%, 5.0420%, and 5.0525%, respectively. The net exchange on the firstpayment date is equivalent to a cash outflow of 0.0105% of $100 million or $10,500. Equivalent netcash flows for other exchanges are calculated similarly. The value of the swap is
10,500 21,000 31,500 42,000 52,5001.05 1.05 1.053 1.054 1.055
 
or —$131,947.
CMS and CRI Swaps
A constant maturity swap (CMS) is an interest rate swap where the floating rate equals the swaprate for a swap with a certain life. For example, the floating payments on a CMS swap might bemade every six months at a rate equal to the five-year swap rate. Usually there is a lag so that thepayment on a particular payment date is equal to the swap rate observed on the previous paymentdate. Suppose that rates are set at times fp, f), t2,... and payments are made at times f), f, f3,....The floating payment at time ¢;,, is
TLS;
where 7; = t;,, — 1; and S; is the swap rate at time t;.Suppose that y; is the forward value of the swap rate 5;. To value the payment at time ¢,,,, it600 CHAPTER 25
 
turns out to be correct to make a convexity adjustment to the forward swap rate, so that the swaprate is assumed to be
Gi (yy) Vit Fi Pi y iO Kiti1,22 i PEt iPS y Eo Bitot eee 25.3ME 2% wi * Gi(y) 1+ Ft, ( )
rather than y;. In this equation, o,; is the volatility of the forward swap rate, F; is the currentforward interest rate between times t; and t;,), a; is the volatility of this forward rate, and p; is thecorrelation between the forward swap rate and the forward interest rate. G;(x) is the price at time t;of a bond as a function of its yield x. The bond pays coupons at rate y, and has the same life andpayment frequency as the swap from which the CMS rate is calculated. Gj(x) and G/(x) are the firstand second partial derivatives of G; with respect to x. The volatilities o, ; can be implied from swapoptions; the volatilities ¢;; can be implied from caplet prices; the correlation p; can be estimatedfrom historical data.Equation (25.3) involves a convexity and a timing adjustment. The term
70)12,2, Gib,=y-o.;t:
2 nil Gi(y;)
 
is an adjustment similar the one we calculated in Example 22.6 of Section 22.6. It is based on theassumption that the swap rate, S;, leads to only one payment at time t;. The term
_ ViT FOF iC Kiti+ Fr;
is similar the one we calculated in equation (22.18) and allows for the fact that the paymentcalculated from S, is made at time f;,, rather than 4;.
Cyamsnln 902 Ing six OMS x fy apy rate is reccived and a Sxed rate of $% jsExample 25.3 Ina six-ycar CMS swap, the five-ycar swap ratc is reccived and a fixed rate of 5% is
paid on a notional principal of $100 million. The exchange of payments is semiannual (both on theunderlying five-year swap and on the CMS swap itself). The exchange on a payment date isdetermined from the swap rate on the previous payment date. The term structure is flat at 5% perannum with semiannual compounding. All options on five-year swaps have a 15% implied volatilityand all caplets with a six-month tenor have a 20% implied volatility. The correlation between eachcap rate and each swap rate is 0.7.
In this case, y; = 0.05, 0, ; = 0.15, t, = 0.5, F, = 0.05, of; = 0.20, and p; = 0.7 for all i. Also,
2.5 100G = ——— + —=) apap tay”
 
so that G;(y,;) = —437.603 and G/(y;) = 2261.23. Equation (25.3) gives the total convexity/timingadjustment as 0.0001197:;, or 1.197 basis points per year until the swap rate is observed. Forexample, for the purposes of valuing the CMS swap, the five-year swap rate in four years’ timeshould be assumed to be 5.0479% rather than 5% and the net cash flow received W the 4.5-year pointshould be assumed to be 0.5 x 0.000479 x 100,000,000 = $23,940. Other net cash flows are calcu-lated similarly. Taking their present value, we find the value of the swap to be $130,545.
  
A constant maturity Treasury swap (CMT swap) works similarly to a CMS swap except that thefloating rate is the yield on a Treasury bond with a specified life. The analysis of a CMT swap isSwaps Revisited 601
essentially the same as that for a CMS swap with S$, defined as the par yield on a