Mathematics of Financial Derivatives

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Mathematics of Financial Derivatives Lecture 11 Solesne Bourguin bourguin@math.bu.edu Boston University Department of Mathematics and Statistics Table of contents 1. Mechanics of interest rate swaps (continued) 2. The nature of swap rates 3. 1 Mechanics of interest rate swaps (continued)

Mechanics of interest rate swaps Role of financial intermediary Usually, two non-financial companies such as Intel and Microsoft do not get in touch directly to arrange a swap in the way indicated previously. They each deal with a financial institution, which makes money by taking a fixed fee of around 0.03% on the offsetting transactions taking place. This is illustrated in the following diagrams. 5.2% 4.985% 5.015% Intel Fin. Inst. Microsoft LIBOR LIBOR LIBOR + 0.1% LIBOR 0.2% 4.985% 5.015% Intel Fin. Inst. Microsoft LIBOR LIBOR 4.7% 2 Mechanics of interest rate swaps Market makers In practice, it is very unlikely that two companies will contact a financial institution at the same time and want to take opposite positions in exactly the same swap. For this reason, many large financial institutions act as market makers for swaps. This means that they are prepared to enter into a swap without having an offsetting swap with another counterparty. Hence market makers must carefully quantify and hedge the risks they are taking. Bonds, FRA or interest rate futures are examples of the instruments that can be used for hedging by swap market makers. 3 Mechanics of interest rate swaps Market makers (continued) The following table shows quotes for plain vanilla swaps that might be posted by a market maker. Maturity (years) Bid Offer Swap rate 2 6.03 6.06 6.045 3 6.21 6.24 6.225 4 6.35 6.39 6.370 5 6.47 6.51 6.490 7 6.65 6.68 6.665 10 6.83 6.87 6.850 As mentioned earlier, the bid-offer spread is 3 to 4 basis points. The average of the bid and the offer rates is known as the swap rate. 4

Mechanics of interest rate swaps Pricing of swaps using swap rates Consider a new swap where the fixed rate equals the current swap rate. The value of this swap is zero: this is an alternate definition of swap rates (if it was not zero, why would bid-offer spreads be centered around it?). We already saw that swaps can be characterized as the difference between a fixed-rate bond and a floating-rate bond. Define: B fix : Value of fixed-rate bond underlying the swap. B fl : Value of floating-rate bond underlying the swap. Since the swap is worth 0, it implies that B fix = B fl. 5 The nature of swap rates The nature of swap rates Some properties of swap rates It is natural to ask about the relation ship between LIBOR and swap rates. Recall that LIBOR is the rate at which AA-rated banks borrow for periods up to 12 months from other banks. A swap rate is the average of the fixed rate that a swap market maker is prepared to pay in exchange for receiving LIBOR and the fixed rate that it is prepared to receive in return for paying LIBOR. Like LIBOR rates, swap rates are not risk-free lending rates. However, they are reasonably close to risk-free in normal market conditions. 6

The nature of swap rates Some properties of swap rates (continued) A financial institution can earn the 5-year swap rate by doing the following: Lend the principal for the first 6 months to a AA borrower and then relend it for successive 6-months periods to other AA borrowers. Enter into a swap to exchange the LIBOR income for the 5-year swap rate. This shows that the 5-year swap rate is a rate with a credit risk corresponding to the situation where 10 consecutive 6-month LIBOR loans to AA companies are made. Remark For the 7-year swap rate, it would be 14 consecutive 6-month LIBOR loans to AA borrowers. 7 The nature of swap rates Important remark Note that 5-year swap rates are less than 5-year AA borrowing rates. This is because it is much more attractive to lend money for successive 6-month periods to borrowers who are always AA at the beginning of the periods than to lend it to one borrower for the whole 5 years when all we can be sure of is that the borrower is AA at the beginning of the 5 years. 8 The nature of swap rates Determining LIBOR/swap zero rates One problem with LIBOR rates is that direct observations are possible only for maturities out to 12 months. Traders can then use swap rates to extend the LIBOR curve further. Note that the value of a newly issued floating-rate bond that pays 6-month LIBOR is always equal to its principal value when the LIBOR/swap zero curve is used for discounting (because the bond provides a rate of interest of LIBOR and LIBOR is the discount rate). We saw that, for a newly issued swap where the fixed rate equals the swap rate, B fix = B fl. But as B fl equals the principal, it follows that B fix also equals the principal. 9

The nature of swap rates Determining LIBOR/swap zero rates (continued) Hence, swap rates define a set of par yield bonds. We can then apply the bootstrap method to deduce zero rates from these par yields. Example Suppose that the 6-month, 12-month, and 18-months LIBOR/swap zero rates have been determined as 4%, 4.5%, and 4.8% with continuous compounding and that the 2-year swap rate (for a swap where payment are made semiannually) is 5%. By what we just saw, this 5% swap rate means that a bond with a principal of $100 and a semiannual coupon of 5% per annum sells for par. It follows that, if R is the 2-year zero rate, 2.5e 0.04 0.5 + 2.5e 0.045 1.0 + 2.5e 0.048 1.5 + 102.5e R 2.0 = 100. Solving this equation yields R=4.953%. 10 Remark A swap is worth (close to) zero when it is first initiated. After it has been in existence for some time, its value may be positive or negative. Valuation in terms of bonds From the point of view of the floating-rate payer, a swap can be regarded as a long position in a fixed-rate bond and a short position in a floating-rate bond, so that V swap = B fix B fl. Similarly, from the point of view of the fixed-rate payer, a swap is a long position in a floating-rate bond and a short position in a fixed-rate bond, so that the value of the swap is V swap = B fl B fix. 11

Valuation in terms of bonds (continued) The value of the fixed rate bond B fix is easy to calculate. To value the floating-rate bond, we are going to use a crucial observation: the bond is worth the notional principal immediately after a payment. This is because at this time the bond is a fair deal where the borrower pays LIBOR for each subsequent accrual period. Suppose that the principal is L, the next exchange of payments is at time t, and the floating payment that will be made at time t (which was determined at the last payment date) is k. Remark We insist once more on the fact that both L and k are known amounts at the valuation date! 12 Valuation in terms of bonds (continued) Immediately after the payment, B fl = L as just explained. So immediately before the payment, B fl = L + k. This is a certain amount, so it can be discounted using the risk-free rate (here the LIBOR/swap zero curve) to get the present value of the bond. So B fl = (L + k )e r t, where r is the LIBOR/swap zero rate for maturity t. 13 Example Suppose that some time ago, a financial institution agreed to receive 6-month LIBOR and pay 3% per annum (with semiannual compounding) on a notional principal of $100 million. The swap has a remaining life of 1.25 years. The LIBOR rates (with continuous compounding) for 3-month, 9-month, and 15-month maturities are 2.8%, 3.2%, and 3.4%, respectively. The 6-month LIBOR rate at the last payment date was 2.9% (with semiannual compounding). 14

Example (continued) The calculations for valuing the swap in terms of bonds are summarized in the following table (amounts are expressed in millions). Time B fix B fl Discount factor Present value B fix Present value B fl 0.25 1.5 101.4500 0.9930 1.4895 100.7423 0.75 1.5 Irrelevant 0.9763 1.4644 Irrelevant 1.25 1.5 Irrelevant 0.9584 97.2766 Irrelevant Total: 100.2306 100.7423 Hence the value of the swap is V swap = 100.7423 100.2306 = $0.5117 million. Remark If we had considered the opposite position of paying fixed and receiving floating, the value of the swap would be -$0.5117 million. 15 Valuation in terms of FRAs A swap is worth (close to) zero when it is first initiated. After it has been in existence for some time, its value may be positive or negative. Observe that a swap can be characterized as a portfolio of FRA. Consider the swap between Microsoft and Intel from our example. The swap is a 3-year deal entered into on March 5, 2014, with semiannual payments. The first exchange of payments is known at the time the swap starts. The other five exchanges can be regarded as FRAs. 16 Valuation in terms of FRAs (continued) For instance, the exchange on March 5, 2015 is a FRA where interest at 5% is exchanged for interest at the 6-month rate observed in the market on September 5, 2014. The exchange on September 5, 2015 is a FRA where interest at 5% is exchanged for interest at the 6-months rate observed in the market on March 5, 2015. And so on... As we know how to value FRAs, we can hence value the swap. 17

Valuation in terms of FRAs (continued) The general procedure is described by the following. 1. Use the LIBOR/swap zero curve to calculate forward rates for each of the LIBOR rates that will determine swap s. 2. Calculate swap s on the assumption that the LIBOR rates will equal the forward rates. 3. Discount these swap s (using the LIBOR/swap zero curve) to obtain the swap value. 18 Example We take the same example as when we where valuing the swap in terms of bonds. Suppose that some time ago, a financial institution agreed to receive 6-month LIBOR and pay 3% per annum (with semiannual compounding) on a notional principal of $100 million. The swap has a remaining life of 1.25 years. The LIBOR rates (with continuous compounding) for 3-month, 9-month, and 15-month maturities are 2.8%, 3.2%, and 3.4%, respectively. The 6-month LIBOR rate at the last payment date was 2.9% (with semiannual compounding). 19 Example (continued) The calculations for valuing the swap in terms of FRAs are summarized in the following table (amounts are expressed in millions). Time Fixed Floating Net Discount factor Present value of net 0.25-1.5000 +1.4500-0.0050 0.9930-0.0497 0.75-1.5000 +1.7145 +0.2145 0.9763 +0.2094 1.25-1.5000 +1.8672 +0.3672 0.9584 +0.3519 Total: +0.5117 Hence the value of the swap is +$0.5117 million, which is consistent with what we found when valuing in terms of bond prices. 20