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(1)

• 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.

(2)

### Spread of Nonbenchmark Bonds (continued)

• We illustrate the idea with an example.

• 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 solves P = p(s).

(3)

4.00%+I

3.526%+I

2.895%+I

5.289%+I

4.343%+I

6.514%+I

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

(4)

### 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.

(5)

### 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.

(6)

### Spread of Nonbenchmark Bonds (continued)

• To compute pA(s) as well, node A calculates pA(s) = pB(s) + pC(s)

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

(111)

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

• When A is a terminal node, simply use the payoﬀ function for pA(s).a

aContributed by Mr. Chou, Ming-Hsin (R02723073) on May 28, 2014.

(7)

1 1 ? I

1 1 ?L I

1 1? ?L2 ID

1 1 = I

1 1 >L I

1 1 > I

1 1= > IB2

1 1= ? IB2

1 1= ?L IB2

1 1

?L2 I

### D

2

1 1= >L IB2

1 1= = IB2

A C

B B

C

C

D

D D D

A C

B B

C

C

D

D D D

(a) (b)

A

C

F I*= B

B F I ? F I F I

)= B *2 1=( ) H I+( )B

F I F I F I H I

F I F I

)= B = B = B*2 1( ) +( ) 2 1*( )H I+( )2

F I+= B

F I+= B

F I*= B

=

>

?

=

>

?

H

(8)

### Spread of Nonbenchmark Bonds (continued)

• Apply the above procedure inductively to yield p(s) and p(s) at the root (p. 950).

• This is called the diﬀerential tree method.a

• The total running time is O(n2).

• The memory requirement is O(n).

aLyuu (1999).

(9)

### 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.

(10)

### 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. 954).

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

• It also diﬀers from the yield spread (p. 119) and static spread (p. 120) of the nonbenchmark bond over an otherwise identical benchmark bond.

(11)

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

(12)

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

a

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.

aLyuu (1999).

(13)

### Fixed-Income Options

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

• Assume the Treasury pays annual interest.

• From p. 957 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

(14)

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)

(15)

### Fixed-Income Options (continued)

• The option value is calculated to be \$1.458 on p. 957(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. 957(b).

(16)

### 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).

(17)

### 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.

(18)

### 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. 957 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.

(19)

### 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).

(20)

### Volatility Term Structures (continued)

• For example, based on the tree on p. 940, 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%.

(21)

### 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 

0.900961 − 1 = 0.053531.

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

1

2 × 1

1.03526 ×

 1

1.02895 + 1 1.04343



= 0.93225.

(22)

### Volatility Term Structures (continued)

• Thus its yield is 

0.932251 − 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.a

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

aThe relation is reversed for price volatilities (duration).

(23)

0 100 200 300 400 500 Time period

0.1 0.101 0.102 0.103 0.104

Spot rate volatility

Short rate volatility given ﬂat %10 volatility term structure.

(24)

### 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. (107) on p. 918—and the ri are then simultaneously determined.

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

aBlack, Derman, and Toy (1990).

(25)

## Foundations of Term Structure Modeling

(26)

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

— Roger Lowenstein, When Genius Failed (2000)

(27)

[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)

(28)

### 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.

(29)

### 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 .

(30)

### 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 .

(31)

### 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).

(32)

### 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.

(33)

### Fundamental Relations (continued)

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

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

P (t, T ) , T ≤ M. (112) – The forward price equals the future value at time T

of the underlying asset (see text for proof).

• Equation (112) holds whether the model is discrete-time or continuous-time.

(34)

### 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 (113)

in discrete time.

– The analog to Eq. (113) under simple compounding is f (t, T, L) = 1

L

 P (t, T )

P (t, T + L) − 1

 .

(35)

### 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 (114)

by Eq. (112) on p. 976.

• Furthermore,

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

Δt → −∂ ln P (t, T )

∂T

= −∂P (t, T )/∂T P (t, T ) .

(36)

### Fundamental Relations (continued)

• So

f (t, T ) ≡ lim

Δt→0f (t, T, Δt) = −∂P (t, T )/∂T

P (t, T ) , t ≤ T.

(115)

• Because Eq. (115) is equivalent to P (t, T ) = e

T

t f (t,s) ds, (116) the spot rate curve is

r(t, T ) =

 T

t f (t, s) ds T − t .

(37)

### Fundamental Relations (concluded)

• The discrete analog to Eq. (116) 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

.

(38)

### 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). (117) – Relation (117) in fact follows from the risk-neutral

valuation principle.a

aTheorem 17 on p. 509.

(39)

### Risk-Neutral Pricing (continued)

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

• Rewrite Eq. (117) 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.

(40)

### 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))



. (118)

(41)

### Risk-Neutral Pricing (concluded)

• Equation (117) on p. 981 can also be expressed as Et[ P (t + 1, T ) ] = F (t, t + 1, T ).

– Verify that with, e.g., Eq. (112) on p. 976.

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

aBut the forward rate is not an unbiased estimator of the expected future short rate (p. 932).

(42)

### 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. (119)

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

(43)

### Interest Rate Swaps

• Consider an interest rate swap made at time t (now) 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 Δt 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 = t0.

(44)

### Interest Rate Swaps (continued)

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

• Simple rates are adopted here.

• Hence fi satisﬁes

P (ti, ti+1) = 1

1 + fiΔt.

(45)

### 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) Δt

=

n i=1

Etπ

e

ti

t r(s) ds

 1

P (ti−1, ti) − (1 + cΔt)



=

n i=1

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

= P (t, t0) − P (t, tn) − cΔt

n i=1

P (t, ti).

(46)

### 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.

(47)

### 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) Δt . (120)

• 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.

(48)

### 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.

(49)

### 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).

(50)

### 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) after rearrangement.

(51)

### 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) (121) 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 to preclude arbitrage opportunities.

(52)

### 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,

(122)

σp =



σ(r, t) ∂P

∂r



/P, (122)

subject to P ( · , T, T ) = 1.

(53)

### The Term Structure Equation (concluded)

• Substitute μp and σp into Eq. (121) on p. 994 to obtain

∂P

∂T + [ μ(r, t) − λ(r, t) σ(r, t) ] ∂P

∂r + 1

2 σ(r, t)2 2P

∂r2 = rP.

(123)

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

(54)

### The Binomial Model

• The analytical framework can be nicely illustrated with the binomial model.

• Suppose the bond price P can move with probability q to P u and probability 1 − q to P d, where u > d:

P

* P d 1 − q

q j Pu

(55)

### The Binomial Model (continued)

• Over the period, the bond’s expected rate of return is

μ ≡ qP u + (1 − q) P d

P − 1 = qu + (1 − q) d − 1.

(124)

• The variance of that return rate is

2 ≡ q(1 − q)(u − d)2. (125)

(56)

### The Binomial Model (continued)

• In particular, the bond whose maturity is one period

away will move from a price of 1/(1 + r) to its par value

\$1.

• This is the money market account modeled by the short rate r.

• The market price of risk is deﬁned as λ ≡ (μ − r)/σ.

• As in the continuous-time case, it can be shown that λ is independent of the maturity of the bond (see text).

(57)

### The Binomial Model (concluded)

• Now change the probability from q to p ≡ q − λ

q(1 − q) = (1 + r) − d

u − d , (126) which is independent of bond maturity and q.

– Recall the BOPM.

• The bond’s expected rate of return becomes pP u + (1 − p) P d

P − 1 = pu + (1 − p) d − 1 = r.

• The local expectations theory hence holds under the new probability measure p.

(58)

### Numerical Examples

• Assume this spot rate curve:

Year 1 2

Spot rate 4% 5%

• Assume the one-year rate (short rate) can move up to 8% or down to 2% after a year:

4%

* 8%

j 2%

(59)

### Numerical Examples (continued)

• No real-world probabilities are speciﬁed.

• The prices of one- and two-year zero-coupon bonds are, respectively,

100/1.04 = 96.154, 100/(1.05)2 = 90.703.

• They follow the binomial processes on p. 1003.

(60)

### Numerical Examples (continued)

90.703

* 92.593 (= 100/1.08)

j 98.039 (= 100/1.02) 96.154

* 100 j 100

The price process of the two-year zero-coupon bond is on the left; that of the one-year zero-coupon bond is on the right.

(61)

### Numerical Examples (continued)

• The pricing of derivatives can be simpliﬁed by assuming investors are risk-neutral.

• Suppose all securities have the same expected one-period rate of return, the riskless rate.

• Then

(1 − p) × 92.593

90.703 + p × 98.039

90.703 − 1 = 4%,

where p denotes the risk-neutral probability of a down move in rates.

(62)

### Numerical Examples (concluded)

• Solving the equation leads to p = 0.319.

• Interest rate contingent claims can be priced under this probability.

(63)

### Numerical Examples: Fixed-Income Options

• A one-year European call on the two-year zero with a

\$95 strike price has the payoﬀs, C

* 0.000 j 3.039

• To solve for the option value C, we replicate the call by a portfolio of x one-year and y two-year zeros.

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### Numerical Examples: Fixed-Income Options (continued)

• This leads to the simultaneous equations, x × 100 + y × 92.593 = 0.000, x × 100 + y × 98.039 = 3.039.

• They give x = −0.5167 and y = 0.5580.

• Consequently,

C = x × 96.154 + y × 90.703 ≈ 0.93 to prevent arbitrage.

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### Numerical Examples: Fixed-Income Options (continued)

• This price is derived without assuming any version of an expectations theory.

• Instead, the arbitrage-free price is derived by replication.

• The price of an interest rate contingent claim does not depend directly on the real-world probabilities.

• The dependence holds only indirectly via the current bond prices.

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### Numerical Examples: Fixed-Income Options (concluded)

• An equivalent method is to utilize risk-neutral pricing.

• The above call option is worth

C = (1 − p) × 0 + p × 3.039

1.04 ≈ 0.93,

the same as before.

• This is not surprising, as arbitrage freedom and the existence of a risk-neutral economy are equivalent.

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### Numerical Examples: Futures and Forward Prices

• A one-year futures contract on the one-year rate has a payoﬀ of 100 − r, where r is the one-year rate at

maturity:

F

* 92 (= 100 − 8) j 98 (= 100 − 2)

• As the futures price F is the expected future payoﬀ (see text or p. 510),

F = (1 − p) × 92 + p × 98 = 93.914.

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### Numerical Examples: Futures and Forward Prices (concluded)

• The forward price for a one-year forward contract on a one-year zero-coupon bond isa

90.703/96.154 = 94.331%.

• The forward price exceeds the futures price.b

aSee Eq. (112) on p. 976.

bRecall p. 454.

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## Equilibrium Term Structure Models

(70)

8. What’s your problem? Any moron can understand bond pricing models.

— Top Ten Lies Finance Professors Tell Their Students

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### Introduction

• This chapter surveys equilibrium models.

• Since the spot rates satisfy

r(t, T ) = −ln P (t, T ) T − t ,

the discount function P (t, T ) suﬃces to establish the spot rate curve.

• All models to follow are short rate models.

• Unless stated otherwise, the processes are risk-neutral.

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### The Vasicek Model

a

• The short rate follows

dr = β(μ − r) dt + σ dW.

• The short rate is pulled to the long-term mean level μ at rate β.

• Superimposed on this “pull” is a normally distributed stochastic term σ dW .

• Since the process is an Ornstein-Uhlenbeck process, E[ r(T ) | r(t) = r ] = μ + (r − μ) e−β(T −t) from Eq. (59) on p. 567.

aVasicek (1977).

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### The Vasicek Model (continued)

• The price of a zero-coupon bond paying one dollar at maturity can be shown to be

P (t, T ) = A(t, T ) e−B(t,T ) r(t), (127) where

A(t, T ) =

exp

(B(t,T )−T +t)(β2μ−σ2/2)

β2 σ2B(t,T )2

if β = 0, exp

σ2(T −t)3 6

if β = 0.

and

B(t, T ) =

⎧⎨

1−e−β(T −t)

β if β = 0, T − t if β = 0.

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### The Vasicek Model (concluded)

• If β = 0, then P goes to inﬁnity as T → ∞.

• Sensibly, P goes to zero as T → ∞ if β = 0.

• Even if β = 0, P may exceed one for a ﬁnite T .

• The spot rate volatility structure is the curve (∂r(t, T )/∂r) σ = σB(t, T )/(T − t).

• When β > 0, the curve tends to decline with maturity.

• The speed of mean reversion, β, controls the shape of the curve.

• Indeed, higher β leads to greater attenuation of volatility with maturity.

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2 4 6 8 10 Term 0.05

0.1 0.15 0.2

Yield

humped

inverted

normal

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### The Vasicek Model: Options on Zeros

a

• Consider a European call with strike price X expiring at time T on a zero-coupon bond with par value \$1 and maturing at time s > T .

• Its price is given by

P (t, s) N (x) − XP (t, T ) N (x − σv).

aJamshidian (1989).

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### The Vasicek Model: Options on Zeros (concluded)

• Above

x ≡ 1

σv ln

 P (t, s) P (t, T ) X



+ σv 2 , σv ≡ v(t, T ) B(T, s),

v(t, T )2

⎧⎨

σ2[1−e−2β(T −t)]

, if β = 0 σ2(T − t), if β = 0 .

• By the put-call parity, the price of a European put is XP (t, T ) N (−x + σv) − P (t, s) N(−x).

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### Binomial Vasicek

• Consider a binomial model for the short rate in the time interval [ 0, T ] divided into n identical pieces.

• Let Δt ≡ T/n and

p(r) ≡ 1

2 + β(μ − r)√ Δt

.

• The following binomial model converges to the Vasicek model,a

r(k + 1) = r(k) + σ√

Δt ξ(k), 0 ≤ k < n.

aNelson and Ramaswamy (1990).

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### Binomial Vasicek (continued)

• Above, ξ(k) = ±1 with

Prob[ ξ(k) = 1 ] =

⎧⎪

⎪⎨

⎪⎪

p(r(k)) if 0 ≤ p(r(k)) ≤ 1 0 if p(r(k)) < 0

1 if 1 < p(r(k))

.

• Observe that the probability of an up move, p, is a decreasing function of the interest rate r.

• This is consistent with mean reversion.

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### Binomial Vasicek (concluded)

• The rate is the same whether it is the result of an up move followed by a down move or a down move followed by an up move.

• The binomial tree combines.

• The key feature of the model that makes it happen is its constant volatility, σ.

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### The Cox-Ingersoll-Ross Model

a

• It is the following square-root short rate model:

dr = β(μ − r) dt + σ√

r dW. (128)

• The diﬀusion diﬀers from the Vasicek model by a multiplicative factor

r .

• The parameter β determines the speed of adjustment.

• The short rate can reach zero only if 2βμ < σ2.

• See text for the bond pricing formula.

aCox, Ingersoll, and Ross (1985).

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### Binomial CIR

• We want to approximate the short rate process in the time interval [ 0, T ].

• Divide it into n periods of duration Δt ≡ T/n.

• Assume μ, β ≥ 0.

• A direct discretization of the process is problematic because the resulting binomial tree will not combine.

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### Binomial CIR (continued)

• Instead, consider the transformed process x(r) ≡ 2√

r/σ.

• It follows

dx = m(x) dt + dW, where

m(x) ≡ 2βμ/(σ2x) − (βx/2) − 1/(2x).

• Since this new process has a constant volatility, its associated binomial tree combines.

(84)

### Binomial CIR (continued)

• Construct the combining tree for r as follows.

• First, construct a tree for x.

• Then transform each node of the tree into one for r via the inverse transformation

r = f (x) ≡ x2σ2 4 (p. 1028).

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x + 2

Δt f(x + 2

Δt)

 

x +

Δt f(x +

Δt)

   

x x f(x) f(x)

   

x −

Δt f(x −

Δt)

 

x − 2

Δt f(x − 2

Δt)

(86)

### Binomial CIR (concluded)

• The probability of an up move at each node r is p(r) ≡ β(μ − r) Δt + r − r

r+ − r . (129)

– r+ ≡ f(x +

Δt) denotes the result of an up move from r.

– r ≡ f(x −

Δt) the result of a down move.

• Finally, set the probability p(r) to one as r goes to zero to make the probability stay between zero and one.

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### Numerical Examples

• Consider the process,

0.2 (0.04 − r) dt + 0.1√

r dW,

for the time interval [ 0, 1 ] given the initial rate r(0) = 0.04.

• We shall use Δt = 0.2 (year) for the binomial approximation.

• See p. 1031(a) for the resulting binomial short rate tree with the up-move probabilities in parentheses.

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(0.472049150276)0.04

0.05988854382 (0.44081188025)

0.03155572809 (0.489789553691)

0.02411145618 (0.50975924867)

0.0713328157297 (0.426604457655)

0.08377708764

0.01222291236 0.01766718427

(0.533083330907) (0.472049150276)0.04

0.0494442719102

(0.455865503068) 0.0494442719102 (0.455865503068)

0.03155572809 (0.489789553691)

0.05988854382

0.04

0.02411145618

(a)

(b) 0.992031914837

0.984128889634 0.976293244408 0.968526861261 0.960831229521

0.992031914837 0.984128889634

0.976293244408 0.992031914837 0.990159879565

0.980492588317 0.970995502019 0.961665706744

0.993708727831 0.987391576942 0.981054487259 0.974702907786

0.988093738447 0.976486896485 0.965170249273

0.990159879565 0.980492588317

0.995189317343 0.990276851751 0.985271123591

0.993708727831 0.987391576942 0.98583472203 0.972116454453

0.996472798388 0.992781347933

0.983384173756

0.988093738447

0.995189317343

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### Numerical Examples (continued)

• Consider the node which is the result of an up move from the root.

• Since the root has x = 2

r(0)/σ = 4, this particular node’s x value equals 4 +

Δt = 4.4472135955.

• Use the inverse transformation to obtain the short rate x2 × (0.1)2

4 ≈ 0.0494442719102.

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### Numerical Examples (concluded)

• Once the short rates are in place, computing the probabilities is easy.

• Note that the up-move probability decreases as interest rates increase and decreases as interest rates decline.

– I suspect that

p(r) = A

Δt

r + B − C√

rΔt for some A, B, C > 0.a

• This phenomenon agrees with mean reversion.

• Convergence is quite good (see text).

aThanks to a lively class discussion on May 28, 2014.

• The XYZ.com bonds are equivalent to a default-free zero-coupon bond with \$X par value plus n written European puts on Merck at a strike price of \$30.. – By the

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

• 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

• P u is the price of the i-period zero-coupon bond one period from now if the short rate makes an up move. • P d is the price of the i-period zero-coupon bond one period from now

• The XYZ.com bonds are equivalent to a default-free zero-coupon bond with \$X par value plus n written European puts on Merck at a strike price of \$30. – By the

6 《中論·觀因緣品》，《佛藏要籍選刊》第 9 冊，上海古籍出版社 1994 年版，第 1

volume suppressed mass: (TeV) 2 /M P ∼ 10 −4 eV → mm range can be experimentally tested for any number of extra dimensions - Light U(1) gauge bosons: no derivative couplings. =&gt;