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Introduction to Term Structure Modeling

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Introduction to Term Structure Modeling

(2)

The fox often ran to the hole by which they had come in, to ﬁnd out if his body was still thin enough to slip through it.

— Grimm’s Fairy Tales

(3)

And the worst thing you can have is models and spreadsheets.

— Warren Buﬀet, May 3, 2008

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Outline

• Use the binomial interest rate tree to model stochastic term structure.

– Illustrates the basic ideas underlying future models.

– Applications are generic in that pricing and hedging methodologies can be easily adapted to other models.

• Although the idea is similar to the earlier one used in option pricing, the current task is more complicated.

– The evolution of an entire term structure, not just a single stock price, is to be modeled.

– Interest rates of various maturities cannot evolve arbitrarily, or arbitrage proﬁts may occur.

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Issues

• A stochastic interest rate model performs two tasks.

– Provides a stochastic process that deﬁnes future term structures without arbitrage proﬁts.

– “Consistent” with the observed term structures.

(6)

History

• Methodology founded by Merton (1970).

• Modern interest rate modeling is often traced to 1977 when Vasicek and Cox, Ingersoll, and Ross developed simultaneously their inﬂuential models.

• Early models have ﬁtting problems because they may not price today’s benchmark bonds correctly.

• An alternative approach pioneered by Ho and Lee (1986) makes ﬁtting the market yield curve mandatory.

• Models based on such a paradigm are called (somewhat misleadingly) arbitrage-free or no-arbitrage models.

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Binomial Interest Rate Tree

• Goal is to construct a no-arbitrage interest rate tree consistent with the yields and/or yield volatilities of zero-coupon bonds of all maturities.

– This procedure is called calibration.a

• Pick a binomial tree model in which the logarithm of the future short rate obeys the binomial distribution.

– Exactly like the CRR tree.

• The limiting distribution of the short rate at any future time is hence lognormal.

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Binomial Interest Rate Tree (continued)

• A binomial tree of future short rates is constructed.

• Every short rate is followed by two short rates in the following period (p. 836).

• In the ﬁgure on p. 836, node A coincides with the start of period j during which the short rate r is in eﬀect.

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r

* r

0.5

j rh 0.5

A

B

C

period j − 1 period j period j + 1 time j − 1 time j

(10)

Binomial Interest Rate Tree (continued)

• At the conclusion of period j, a new short rate goes into eﬀect for period j + 1.

• This may take one of two possible values:

– r: the “low” short-rate outcome at node B.

– rh: the “high” short-rate outcome at node C.

• Each branch has a 50% chance of occurring in a risk-neutral economy.

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Binomial Interest Rate Tree (continued)

• We shall require that the paths combine as the binomial process unfolds.

• The short rate r can go to rh and r with equal

risk-neutral probability 1/2 in a period of length ∆t.

• Hence the volatility of ln r after ∆t time is σ = 1

2

1

∆t ln

(rh r

)

(see Exercise 23.2.3 in text).

• Above, σ is annualized, whereas r and rh are period

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Binomial Interest Rate Tree (continued)

• Note that

rh

r = e

∆t

.

• Thus greater volatility, hence uncertainty, leads to larger rh/r and wider ranges of possible short rates.

• The ratio rh/r may depend on time if the volatility is a function of time.

• Note that rh/r has nothing to do with the current short rate r if σ is independent of r.

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Binomial Interest Rate Tree (continued)

• In general there are j possible ratesa in period j, rj, rjvj, rjvj2, . . . , rjvjj−1,

where

vj ≡ ej∆t (95)

is the multiplicative ratio for the rates in period j (see ﬁgure on next page).

• We shall call rj the baseline rates.

• The subscript j in σj is meant to emphasize that the short rate volatility may be time dependent.

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Baseline rates

A C

B B

C

C

D

D D D r2

r v2 2

r v3 3

r v3 32 r1

r3

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Binomial Interest Rate Tree (concluded)

• In the limit, the short rate follows the following process, r(t) = µ(t) eσ(t) W (t), (96) in which the (percent) short rate volatility σ(t) is a

deterministic function of time.

• The expected value of r(t) equals µ(t) eσ(t)2(t/2).

• Hence a declining short rate volatility is usually imposed to preclude the short rate from assuming implausibly high values.

• Incidentally, this is how the binomial interest rate tree

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Memory Issues

• Path independency: The term structure at any node is independent of the path taken to reach it.

• So only the baseline rates ri and the multiplicative ratios vi need to be stored in computer memory.

• This takes up only O(n) space.a

• Storing the whole tree would take up O(n2) space.

– Daily interest rate movements for 30 years require roughly (30 × 365)2/2 ≈ 6 × 107 double-precision ﬂoating-point numbers (half a gigabyte!).

aThroughout, n denotes the depth of the tree.

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Set Things in Motion

• The abstract process is now in place.

• We need the annualized rates of return of the riskless bonds that make up the benchmark yield curve and their volatilities.

• In the U.S., for example, the on-the-run yield curve

obtained by the most recently issued Treasury securities may be used as the benchmark curve.

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Set Things in Motion (concluded)

• The term structure of (yield) volatilitiesa can be estimated from:

– Historical data (historical volatility).

– Or interest rate option prices such as cap prices (implied volatility).

• The binomial tree should be consistent with both term structures.

• Here we focus on the term structure of interest rates.

aOr simply the volatility (term) structure.

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Model Term Structures

• The model price is computed by backward induction.

• Refer back to the ﬁgure on p. 836.

• Given that the values at nodes B and C are PB and PC, respectively, the value at node A is then

PB + PC

2(1 + r) + cash flow at node A.

• We compute the values column by column without

explicitly expanding the binomial interest rate tree (see next page).

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rv

A C

B

Cash flows:

B

C

C

D

D

D D

C

C P P

r

1 2

2 1

C P P

rv

2 3

2 1

r

rv2

C P P

rv

3 4

2 1

2

h

P1

P2

P3

P4

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Term Structure Dynamics

• An n-period zero-coupon bond’s price can be computed by assigning \$1 to every node at period n and then

applying backward induction.

• Repeating this step for n = 1, 2, . . . , one obtains the market discount function implied by the tree.

• The tree therefore determines a term structure.

• It also contains a term structure dynamics.

– Taking any node in the tree as the current state induces a binomial interest rate tree and, again, a

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Sample Term Structure

• We shall construct interest rate trees consistent with the sample term structure in the following table.

• Assume the short rate volatility is such that v ≡ rh/r = 1.5, independent of time.

Period 1 2 3

Spot rate (%) 4 4.2 4.3

One-period forward rate (%) 4 4.4 4.5

Discount factor 0.96154 0.92101 0.88135

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An Approximate Calibration Scheme

• Start with the implied one-period forward rates and then equate the expected short rate with the forward rate (see Exercise 5.6.6 in text).

• For the ﬁrst period, the forward rate is today’s one-period spot rate.

• In general, let fj denote the forward rate in period j.

• This forward rate can be derived from the market discount function via fj = (d(j)/d(j + 1)) − 1 (see Exercise 5.6.3 in text).

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An Approximate Calibration Scheme (continued)

• Since the ith short rate rjvji−1, 1 ≤ i ≤ j, occurs with probability 2−(j−1) (j−1

i−1

), this means

j i=1

2−(j−1)

(j − 1 i − 1

)

rjvji−1 = fj.

• Thus

rj =

( 2

1 + vj

)j−1

fj. (97)

• The binomial interest rate tree is trivial to set up.

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An Approximate Calibration Scheme (continued)

• The ensuing tree for the sample term structure appears in ﬁgure next page.

• For example, the price of the zero-coupon bond paying

\$1 at the end of the third period is

1

4 × 1 1.04 ×

( 1

1.0352 ×

( 1

1.0288

+ 1

1.0432 )

+ 1

1.0528 ×

( 1

1.0432

+ 1

1.0648 ))

or 0.88155, which exceeds discount factor 0.88135.

• The tree is thus not calibrated.

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4.0%

3.52%

2.88%

5.28%

4.32%

6.48%

Baseline rates

A C

B B

C

C

D

D

D

D

period 2 period 3 period 1

4.0% 4.4% 4.5%

Implied forward rates:

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An Approximate Calibration Scheme (concluded)

• Indeed, this bias is inherent: The tree overprices the bonds (see Exercise 23.2.4 in text).

• Suppose we replace the baseline rates rj by rjvj.

• Then the resulting tree underprices the bonds.a

• The true baseline rates are thus bounded between rj and rjvj.

aLyuu and Wang (F95922018) (2009, 2011).

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Issues in Calibration

• The model prices generated by the binomial interest rate tree should match the observed market prices.

• Perhaps the most crucial aspect of model building.

• Treat the backward induction for the model price of the m-period zero-coupon bond as computing some function f (rm) of the unknown baseline rate rm for period m.

• A root-ﬁnding method is applied to solve f(rm) = P for rm given the zero’s price P and r1, r2, . . . , rm−1.

• This procedure is carried out for m = 1, 2, . . . , n.

• It runs in O(n3) time.

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Binomial Interest Rate Tree Calibration

• Calibration can be accomplished in O(n2) time by the use of forward induction.a

• The scheme records how much \$1 at a node contributes to the model price.

• This number is called the state price or the Arrow-Debreu price.

– It is the price of a state contingent claim that pays

\$1 at that particular node (state) and 0 elsewhere.

• The column of state prices will be established by moving forward from time 0 to time n.

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Binomial Interest Rate Tree Calibration (continued)

• Suppose we are at time j and there are j + 1 nodes.

– The unknown baseline rate for period j is r ≡ rj. – The multiplicative ratio is v ≡ vj.

– P1, P2, . . . , Pj are the known state prices at earlier time j − 1, corresponding to rates r, rv, . . . , rvj−1 for period j.

• By deﬁnition,j

i=1 Pi is the price of the (j − 1)-period zero-coupon bond.

• We want to ﬁnd r based on P1, P2, . . . , Pj and the price of the j-period zero-coupon bond.

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Binomial Interest Rate Tree Calibration (continued)

• One dollar at time j has a known market value of 1/[ 1 + S(j) ]j, where S(j) is the j-period spot rate.

• Alternatively, this dollar has a present value of g(r) P1

(1 + r) + P2

(1 + rv) + P3

(1 + rv2) +· · ·+ Pj

(1 + rvj−1).

• So we solve

g(r) = 1

[ 1 + S(j) ]j (98)

for r.

(32)

Binomial Interest Rate Tree Calibration (continued)

• Given a decreasing market discount function, a unique positive solution for r is guaranteed.

• The state prices at time j can now be calculated (see ﬁgure (a) next page).

• We call a tree with these state prices a binomial state price tree (see ﬁgure (b) next page).

• The calibrated tree is depicted on p. 861.

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A C B

B

C

C

D

D D

D 4.00%

3.526%

2.895%

0.480769

0.460505

0.228308 A

C B

C

C

D

D D D

B

period 2 period 3 period 1

4.0% 4.4% 4.5%

Implied forward rates:

0.480769

1

0.112832

(b)

0.333501

0.327842

0.107173 0.232197

(a) 1

r

rv

P rv

2

2 1

P r

P rv

1 2

2 1

2 1

P r

1

2 1

a f

P1

P2

(34)

4.00%

3.526%

2.895%

5.289%

4.343%

6.514%

A

C

B

C

C

D

D D D

B

period 2 period 3 period 1

4.0% 4.4% 4.5%

Implied forward rates:

(35)

Binomial Interest Rate Tree Calibration (concluded)

• The Newton-Raphson method can be used to solve for the r in Eq. (98) on p. 858 as g(r) is easy to evaluate.

• The monotonicity and the convexity of g(r) also facilitate root ﬁnding.

• The total running time is O(n2), as each root-ﬁnding routine consumes O(j) time.

• With a good initial guess,a the Newton-Raphson method converges in only a few steps.b

aSuch as the rj = (1+v2

j )j−1 fj on p. 851.

(36)

A Numerical Example

• One dollar at the end of the second period should have a present value of 0.92101 by the sample term structure.

• The baseline rate for the second period, r2, satisﬁes 0.480769

1 + r2 + 0.480769

1 + 1.5 × r2 = 0.92101.

• The result is r2 = 3.526%.

• This is used to derive the next column of state prices shown in ﬁgure (b) on p. 860 as 0.232197, 0.460505, and 0.228308.

• Their sum gives the correct market discount factor

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A Numerical Example (concluded)

• The baseline rate for the third period, r3, satisﬁes 0.232197

1 + r3 + 0.460505 1 + 1.5 × r3

+ 0.228308 1 + (1.5)2 × r3

= 0.88135.

• The result is r3 = 2.895%.

• Now, redo the calculation on p. 852 using the new rates:

1

4 × 1

1.04 × [ 1

1.03526 × ( 1 1.02895

+ 1 1.04343

) + 1

1.05289 × ( 1 1.04343

+ 1 1.06514

)],

which equals 0.88135, an exact match.

• The tree on p. 861 prices without bias the benchmark securities.

(38)

Spread of Nonbenchmark Bonds

• Model prices calculated by the calibrated tree as a rule do not match market prices of nonbenchmark bonds.

• The incremental return over the benchmark bonds is called spread.

• If we add the spread uniformly over the short rates in the tree, the model price will equal the market price.

• We will apply the spread concept to option-free bonds next.

(39)

Spread of Nonbenchmark Bonds (continued)

• We illustrate the idea with an example.

• Start with the tree on p. 867.

• Consider a security with cash ﬂow Ci at time i for i = 1, 2, 3.

• Its model price is p(s), which is equal to

1

1.04 + s × [

C1 + 1

2 × 1

1.03526 + s × (

C2 + 1 2

( C3

1.02895 + s

+ C3

1.04343 + s ))

+

1

2 × 1

1.05289 + s × (

C2 + 1 2

( C3

1.04343 + s

+ C3

1.06514 + s ))]

.

• Given a market price of P , the spread is the s that

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4.00%+s

3.526%+s

2.895%+s

5.289%+s

4.343%+s

6.514%+s

A C

B

period 2 period 3 period 1

4.0% 4.4% 4.5%

Implied forward rates:

B

C

C

D

D D D

(41)

Spread of Nonbenchmark Bonds (continued)

• The model price p(s) is a monotonically decreasing, convex function of s.

• We will employ the Newton-Raphson root-ﬁnding method to solve p(s) − P = 0 for s.

• But a quick look at the equation for p(s) reveals that evaluating p(s) directly is infeasible.

• Fortunately, the tree can be used to evaluate both p(s) and p(s) during backward induction.

(42)

Spread of Nonbenchmark Bonds (continued)

• Consider an arbitrary node A in the tree associated with the short rate r.

• In the process of computing the model price p(s), a price pA(s) is computed at A.

• Prices computed at A’s two successor nodes B and C are discounted by r + s to obtain pA(s) as follows,

pA(s) = c + pB(s) + pC(s) 2(1 + r + s) , where c denotes the cash ﬂow at A.

(43)

Spread of Nonbenchmark Bonds (continued)

• To compute pA(s) as well, node A calculates pA(s) = pB(s) + pC(s)

2(1 + r + s) pB(s) + pC(s)

2(1 + r + s)2 . (99)

• This is easy if pB(s) and pC(s) are also computed at nodes B and C.

• Apply the above procedure inductively to yield p(s) and p(s) at the root (p. 871).

• This is called the diﬀerential tree method.a

aLyuu (1999).

(44)

1 1 c

s

1 1 cv

s

1 1c cv2 sh

1 1 a

s

1 1 bv

s

1 1 b

s

1 1a b sf2

1 1a c sf2

1 1a cv sf2

1 1

cv2 s

h

2

1 1a bv sf2

1 1a a sf2

A C

B B

C

C

D

D D D

A C

B B

C

C

D

D D D

(a) (b)

A

C

p sBa f

B

p s c p s p s r s

A

B C

a f a f

( ) ( ) 2 1

p s p s p s r s

p s p s r s

A

B C B C

a f a f a f

( ) ( ) ( ) ( )

2 1 2 1 2

pCa fs pCa fs p sBa f

a

b

c

a

b

c

r

(45)

Spread of Nonbenchmark Bonds (continued)

• The total running time is O(n2).

• The memory requirement is O(n).

(46)

Spread of Nonbenchmark Bonds (continued)

Number of Running Number of Number of Running Number of partitions n time (s) iterations partitions time (s) iterations

500 7.850 5 10500 3503.410 5

1500 71.650 5 11500 4169.570 5

2500 198.770 5 12500 4912.680 5

3500 387.460 5 13500 5714.440 5

4500 641.400 5 14500 6589.360 5

5500 951.800 5 15500 7548.760 5

6500 1327.900 5 16500 8502.950 5

7500 1761.110 5 17500 9523.900 5

8500 2269.750 5 18500 10617.370 5

9500 2834.170 5 . . . . . . . . . . . .

75MHz Sun SPARCstation 20.

(47)

Spread of Nonbenchmark Bonds (concluded)

• Consider a three-year, 5% bond with a market price of 100.569.

• Assume the bond pays annual interest.

• The spread can be shown to be 50 basis points over the tree (p. 875).

• Note that the idea of spread does not assume parallel shifts in the term structure.

• It also diﬀers from the yield spread (p. 116) and static spread (p. 117) of the nonbenchmark bond over an

(48)

4.50%

100.569 A

C B

5 5 105

Cash flows:

B

C

C

D

D D 4.026% D

3.395%

5.789%

4.843%

7.014%

105

105

105

105 106.552

105.150

103.118 106.754

103.436

(49)

More Applications of the Diﬀerential Tree: Calibrating Black-Derman-Toy (in seconds)

Number Running Number Running Number Running

of years time of years time of years time

3000 398.880 39000 8562.640 75000 26182.080 6000 1697.680 42000 9579.780 78000 28138.140 9000 2539.040 45000 10785.850 81000 30230.260 12000 2803.890 48000 11905.290 84000 32317.050 15000 3149.330 51000 13199.470 87000 34487.320 18000 3549.100 54000 14411.790 90000 36795.430 21000 3990.050 57000 15932.370 120000 63767.690 24000 4470.320 60000 17360.670 150000 98339.710 27000 5211.830 63000 19037.910 180000 140484.180 30000 5944.330 66000 20751.100 210000 190557.420 33000 6639.480 69000 22435.050 240000 249138.210 36000 7611.630 72000 24292.740 270000 313480.390

75MHz Sun SPARCstation 20, one period per year.

(50)

More Applications of the Diﬀerential Tree: Calculating Implied Volatility (in seconds)

American call American put

Number of Running Number of Number of Running Number of partitions time iterations partitions time iterations

100 0.008210 2 100 0.013845 3

200 0.033310 2 200 0.036335 3

300 0.072940 2 300 0.120455 3

400 0.129180 2 400 0.214100 3

500 0.201850 2 500 0.333950 3

600 0.290480 2 600 0.323260 2

700 0.394090 2 700 0.435720 2

800 0.522040 2 800 0.569605 2

Intel 166MHz Pentium, running on Microsoft Windows 95.

(51)

Fixed-Income Options

• Consider a two-year 99 European call on the three-year, 5% Treasury.

• Assume the Treasury pays annual interest.

• From p. 879 the three-year Treasury’s price minus the \$5 interest at year 2 could be \$102.046, \$100.630, or

\$98.579 two years from now.

• Since these prices do not include the accrued interest, we should compare the strike price against them.

• The call is therefore in the money in the ﬁrst two

scenarios, with values of \$3.046 and \$1.630, and out of

(52)

A

C B

B

C

C

D

D D D

105

105

105

105 4.00%

101.955 1.458

3.526%

102.716 2.258

2.895%

102.046 3.046

5.289%

99.350 0.774

4.343%

100.630 1.630

6.514%

98.579 0.000

(a)

A

C B

B

C

C

D

D D D

105

105

105

105 4.00%

101.955 0.096

3.526%

102.716 0.000

2.895%

102.046 0.000

5.289%

99.350 0.200

4.343%

100.630 0.000

6.514%

98.579 0.421

(b)

(53)

Fixed-Income Options (continued)

• The option value is calculated to be \$1.458 on p. 879(a).

• European interest rate puts can be valued similarly.

• Consider a two-year 99 European put on the same security.

• At expiration, the put is in the money only when the Treasury is worth \$98.579 without the accrued interest.

• The option value is computed to be \$0.096 on p. 879(b).

(54)

Fixed-Income Options (concluded)

• The present value of the strike price is PV(X) = 99 × 0.92101 = 91.18.

• The Treasury is worth B = 101.955.

• The present value of the interest payments during the life of the options is

PV(I) = 5 × 0.96154 + 5 × 0.92101 = 9.41275.

• The call and the put are worth C = 1.458 and P = 0.096, respectively.

• Hence the put-call parity is preserved:

C = P + B − PV(I) − PV(X).

(55)

Delta or Hedge Ratio

• How much does the option price change in response to changes in the price of the underlying bond?

• This relation is called delta (or hedge ratio) deﬁned as Oh − O

Ph − P .

• In the above Ph and P denote the bond prices if the short rate moves up and down, respectively.

• Similarly, Oh and O denote the option values if the short rate moves up and down, respectively.

(56)

Delta or Hedge Ratio (concluded)

• Since delta measures the sensitivity of the option value to changes in the underlying bond price, it shows how to hedge one with the other.

• Take the call and put on p. 879 as examples.

• Their deltas are

0.774 − 2.258

99.350 − 102.716 = 0.441, 0.200 − 0.000

99.350 − 102.716 = −0.059,

respectively.

(57)

Volatility Term Structures

• The binomial interest rate tree can be used to calculate the yield volatility of zero-coupon bonds.

• Consider an n-period zero-coupon bond.

• First ﬁnd its yield to maturity yh (y, respectively) at the end of the initial period if the short rate rises

(declines, respectively).

• The yield volatility for our model is deﬁned as (1/2) ln(yh/y).

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Volatility Term Structures (continued)

• For example, based on the tree on p. 861, the two-year zero’s yield at the end of the ﬁrst period is 5.289% if the rate rises and 3.526% if the rate declines.

• Its yield volatility is therefore 1

2 ln

(0.05289 0.03526

)

= 20.273%.

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Volatility Term Structures (continued)

• Consider the three-year zero-coupon bond.

• If the short rate rises, the price of the zero one year from now will be

1

2 × 1

1.05289 ×

( 1

1.04343 + 1 1.06514

)

= 0.90096.

• Thus its yield is

1

0.90096 − 1 = 0.053531.

• If the short rate declines, the price of the zero one year from now will be

1 × 1

×

( 1

+ 1 )

= 0.93225.

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Volatility Term Structures (continued)

• Thus its yield is

1

0.93225 − 1 = 0.0357.

• The yield volatility is hence 1

2 ln

(0.053531 0.0357

)

= 20.256%, slightly less than the one-year yield volatility.

• This is consistent with the reality that longer-term bonds typically have lower yield volatilities than shorter-term bonds.

• The procedure can be repeated for longer-term zeros to obtain their yield volatilities.

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0 100 200 300 400 500 Time period

0.1 0.101 0.102 0.103 0.104

Spot rate volatility

(62)

Volatility Term Structures (concluded)

• We started with vi and then derived the volatility term structure.

• In practice, the steps are reversed.

• The volatility term structure is supplied by the user along with the term structure.

• The vi—hence the short rate volatilities via Eq. (95) on p. 840—and the ri are then simultaneously determined.

• The result is the Black-Derman-Toy model of Goldman Sachs.a

aBlack, Derman, and Toy (1990).

(63)

Foundations of Term Structure Modeling

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[Meriwether] scoring especially high marks in mathematics — an indispensable subject for a bond trader.

— Roger Lowenstein, When Genius Failed (2000)

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[The] ﬁxed-income traders I knew seemed smarter than the equity trader [· · · ] there’s no competitive edge to being smart in the equities business[.]

— Emanuel Derman, My Life as a Quant (2004) Bond market terminology was designed less to convey meaning than to bewilder outsiders.

— Michael Lewis, The Big Short (2011)

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Terminology

• A period denotes a unit of elapsed time.

– Viewed at time t, the next time instant refers to time t + dt in the continuous-time model and time t + 1 in the discrete-time case.

• Bonds will be assumed to have a par value of one — unless stated otherwise.

• The time unit for continuous-time models will usually be measured by the year.

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Standard Notations

The following notation will be used throughout.

t: a point in time.

r(t): the one-period riskless rate prevailing at time t for repayment one period later

(the instantaneous spot rate, or short rate, at time t).

P (t, T ): the present value at time t of one dollar at time T .

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Standard Notations (continued)

r(t, T ): the (T − t)-period interest rate prevailing at time t stated on a per-period basis and compounded once per period—in other words, the (T − t)-period spot rate at time t.

F (t, T, M ): the forward price at time t of a forward

contract that delivers at time T a zero-coupon bond maturing at time M ≥ T .

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Standard Notations (concluded)

f (t, T, L): the L-period forward rate at time T implied at time t stated on a per-period basis and compounded once per period.

f (t, T ): the one-period or instantaneous forward rate at time T as seen at time t stated on a per period basis and compounded once per period.

• It is f(t, T, 1) in the discrete-time model and f (t, T, dt) in the continuous-time model.

• Note that f(t, t) equals the short rate r(t).

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Fundamental Relations

• The price of a zero-coupon bond equals

P (t, T ) =



(1 + r(t, T ))−(T −t), in discrete time, e−r(t,T )(T −t), in continuous time.

• r(t, T ) as a function of T deﬁnes the spot rate curve at time t.

• By deﬁnition,

f (t, t) =



r(t, t + 1), in discrete time, r(t, t), in continuous time.

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Fundamental Relations (continued)

• Forward prices and zero-coupon bond prices are related:

F (t, T, M ) = P (t, M )

P (t, T ) , T ≤ M. (100) – The forward price equals the future value at time T

of the underlying asset (see text for proof).

• Equation (100) holds whether the model is discrete-time or continuous-time.

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Fundamental Relations (continued)

• Forward rates and forward prices are related deﬁnitionally by

f (t, T , L) =

( 1

F (t, T, T + L)

)1/L

− 1 =

( P (t, T ) P (t, T + L)

)1/L

− 1 (101)

in discrete time.

– The analog to Eq. (101) under simple compounding is

f (t, T, L) = 1 L

( P (t, T )

P (t, T + L) − 1 )

.

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Fundamental Relations (continued)

• In continuous time,

f (t, T, L) = −ln F (t, T, T + L)

L = ln(P (t, T )/P (t, T + L))

L (102)

by Eq. (100) on p. 898.

• Furthermore,

f (t, T, ∆t) = ln(P (t, T )/P (t, T + ∆t))

∆t → −∂ ln P (t, T )

∂T

= −∂P (t, T )/∂T P (t, T ) .

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Fundamental Relations (continued)

• So

f (t, T ) ≡ lim

∆t→0f (t, T, ∆t) = −∂P (t, T )/∂T

P (t, T ) , t ≤ T.

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• Because Eq. (103) is equivalent to P (t, T ) = e

T

t f (t,s) ds

, (104)

the spot rate curve is

r(t, T ) = 1 T − t

T t

f (t, s) ds.

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Fundamental Relations (concluded)

• The discrete analog to Eq. (104) is

P (t, T ) = 1

(1 + r(t))(1 + f (t, t + 1))· · · (1 + f(t, T − 1)).

• The short rate and the market discount function are related by

r(t) = ∂P (t, T )

∂T

T =t

.

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Risk-Neutral Pricing

• Assume the local expectations theory.

• The expected rate of return of any riskless bond over a single period equals the prevailing one-period spot rate.

– For all t + 1 < T ,

Et[ P (t + 1, T ) ]

P (t, T ) = 1 + r(t). (105) – Relation (105) in fact follows from the risk-neutral

valuation principle.a

aTheorem 16 on p. 463.

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Risk-Neutral Pricing (continued)

• The local expectations theory is thus a consequence of the existence of a risk-neutral probability π.

• Rewrite Eq. (105) as

Etπ[ P (t + 1, T ) ]

1 + r(t) = P (t, T ).

– It says the current market discount function equals the expected market discount function one period from now discounted by the short rate.

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Risk-Neutral Pricing (continued)

• Apply the above equality iteratively to obtain

P (t, T )

= Etπ

[ P (t + 1, T ) 1 + r(t)

]

= Etπ

[ Et+1π [ P (t + 2, T ) ] (1 + r(t))(1 + r(t + 1))

]

= · · ·

= Etπ

[ 1

(1 + r(t))(1 + r(t + 1))· · · (1 + r(T − 1)) ]

. (106)

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Risk-Neutral Pricing (concluded)

• Equation (105) on p. 903 can also be expressed as Et[ P (t + 1, T ) ] = F (t, t + 1, T ).

– Verify that with, e.g., Eq. (100) on p. 898.

• Hence the forward price for the next period is an unbiased estimator of the expected bond price.

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Continuous-Time Risk-Neutral Pricing

• In continuous time, the local expectations theory implies P (t, T ) = Et

[ e

T

t r(s) ds ]

, t < T. (107)

• Note that etT r(s) ds is the bank account process, which denotes the rolled-over money market account.

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Interest Rate Swaps

• Consider an interest rate swap made at time t with payments to be exchanged at times t1, t2, . . . , tn.

• The ﬁxed rate is c per annum.

• The ﬂoating-rate payments are based on the future annual rates f0, f1, . . . , fn−1 at times t0, t1, . . . , tn−1.

• For simplicity, assume ti+1 − ti is a ﬁxed constant ∆t for all i, and the notional principal is one dollar.

• If t < t0, we have a forward interest rate swap.

• The ordinary swap corresponds to t = t .

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Interest Rate Swaps (continued)

• The amount to be paid out at time ti+1 is (fi − c) ∆t for the ﬂoating-rate payer.

• Simple rates are adopted here.

• Hence fi satisﬁes

P (ti, ti+1) = 1

1 + fi∆t.

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Interest Rate Swaps (continued)

• The value of the swap at time t is thus

n i=1

Etπ [

e

ti

t r(s) ds

(fi−1 − c) ∆t]

=

n i=1

Etπ [

e

ti

t r(s) ds

( 1

P (ti−1, ti) − (1 + c∆t) )]

=

n i=1

[ P (t, ti−1) − (1 + c∆t) × P (t, ti) ]

= P (t, t0) − P (t, tn) − c∆t

n i=1

P (t, ti).

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Interest Rate Swaps (concluded)

• So a swap can be replicated as a portfolio of bonds.

• In fact, it can be priced by simple present value calculations.

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Swap Rate

• The swap rate, which gives the swap zero value, equals Sn(t) P (t, t0) − P (t, tn)

n

i=1 P (t, ti) ∆t . (108)

• The swap rate is the ﬁxed rate that equates the present values of the ﬁxed payments and the ﬂoating payments.

• For an ordinary swap, P (t, t0) = 1.

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The Term Structure Equation

• Let us start with the zero-coupon bonds and the money market account.

• Let the zero-coupon bond price P (r, t, T ) follow dP

P = µp dt + σp dW.

• At time t, short one unit of a bond maturing at time s1 and buy α units of a bond maturing at time s2.

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The Term Structure Equation (continued)

• The net wealth change follows

−dP (r, t, s1) + α dP (r, t, s2)

= (−P (r, t, s1) µp(r, t, s1) + αP (r, t, s2) µp(r, t, s2)) dt + (−P (r, t, s1) σp(r, t, s1) + αP (r, t, s2) σp(r, t, s2)) dW.

• Pick

α P (r, t, s1) σp(r, t, s1) P (r, t, s2) σp(r, t, s2).

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The Term Structure Equation (continued)

• Then the net wealth has no volatility and must earn the riskless return:

−P (r, t, s1) µp(r, t, s1) + αP (r, t, s2) µp(r, t, s2)

−P (r, t, s1) + αP (r, t, s2) = r.

• Simplify the above to obtain

σp(r, t, s1) µp(r, t, s2) − σp(r, t, s2) µp(r, t, s1)

σp(r, t, s1) − σp(r, t, s2) = r.

• This becomes

µp(r, t, s2) − r

σp(r, t, s2) = µp(r, t, s1) − r σp(r, t, s1)

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The Term Structure Equation (continued)

• Since the above equality holds for any s1 and s2, µp(r, t, s) − r

σp(r, t, s) ≡ λ(r, t) (109) for some λ independent of the bond maturity s.

• As µp = r + λσp, all assets are expected to appreciate at a rate equal to the sum of the short rate and a constant times the asset’s volatility.

• The term λ(r, t) is called the market price of risk.

• The market price of risk must be the same for all bonds

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The Term Structure Equation (continued)

• Assume a Markovian short rate model, dr = µ(r, t) dt + σ(r, t) dW.

• Then the bond price process is also Markovian.

• By Eq. (14.15) on p. 202 in the text,

µp = (

∂P

∂T + µ(r, t) ∂P

∂r + σ(r, t)2 2

2P

∂r2 )

/P,

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σp = (

σ(r, t) ∂P

∂r )

/P, (110)

subject to P (· , T, T ) = 1.

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The Term Structure Equation (concluded)

• Substitute µp and σp into Eq. (109) on p. 916 to obtain

∂P

∂T + [ µ(r, t)− λ(r, t) σ(r, t) ] ∂P

∂r + 1

2 σ(r, t)2 2P

∂r2 = rP.

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• This is called the term structure equation.

• Once P is available, the spot rate curve emerges via r(t, T ) = −ln P (t, T )

T − t .

• Equation (111) applies to all interest rate derivatives, the diﬀerence being the terminal and the boundary

zero-coupon bond prices, forward rates, or the short rate. • Bond price and forward rate models are usually non-Markovian

– Taking any node in the tree as the current state induces a binomial interest rate tree and, again, a term structure.... Binomial Interest Rate

• The binomial interest rate tree can be used to calculate the yield volatility of zero-coupon bonds.. • Consider an n-period

– at a premium (above its par value) when its coupon rate c is above the market interest rate r;. – at par (at its par value) when its coupon rate is equal to the market

– at a premium (above its par value) when its coupon rate c is above the market interest rate r;. – at par (at its par value) when its coupon rate is equal to the market

• Extension risk is due to the slowdown of prepayments when interest rates climb, making the investor earn the security’s lower coupon rate rather than the market’s higher rate.

• Extension risk is due to the slowdown of prepayments when interest rates climb, making the investor earn the security’s lower coupon rate rather than the market’s higher rate..

• Extension risk is due to the slowdown of prepayments when interest rates climb, making the investor earn the security’s lower coupon rate rather than the market’s... Prepayment