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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 here.

(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 × Ã

C2 + 1 2

Ã C3

1.02895 + s + C3 1.04343 + s

!!

+

1

2 × 1

1.05289 + s × Ã

C2 + 1 2

Ã 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%+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

(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 above reveals that evaluating p0(s) directly is infeasible.

• Fortunately, the tree can be used to evaluate both p(s) and p0(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 p0A(s) as well, node A calculates p0A(s) = p0B(s) + p0C(s)

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

2(1 + r + s)2 . (81)

• This is easy if p0B(s) and p0C(s) are also computed at nodes B and C.

• Apply the above procedure inductively to yield p(s) and p0(s) at the root (p. 783).

• This is called the differential tree method.a

aLyuu (1999).

(7)

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

(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

(8)

### Spread of Nonbenchmark Bonds (continued)

• Let C represent the number of times the tree is traversed, which takes O(n2) time.

• The total running time is O(Cn2).

• In practice C is a small constant.

• The memory requirement is O(n).

(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. 787).

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

• It also differs from the yield spread (p. 110) and static spread (p. 111) of the nonbenchmark bond over an

(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: 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.

(13)

### 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.

(14)

### Fixed-Income Options

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

• Assume the Treasury pays annual interest.

• From p. 791 the three-year Treasury’s price minus the \$5 interest 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

(15)

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)

(16)

### Fixed-Income Options (continued)

• The option value is calculated to be \$1.458 on p. 791(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 if the

Treasury is worth \$98.579 without the accrued interest.

• The option value is computed to be \$0.096 on p. 791(b).

(17)

### 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 PV of 105).

• 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).

(18)

### 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.

(19)

### 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. 791 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.

(20)

### 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 rate rises (declines, respectively).

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

(21)

### Volatility Term Structures (continued)

• For example, based on the tree on p. 773, 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

µ0.05289 0.03526

= 20.273%.

(22)

### Volatility Term Structures (continued)

• Consider the three-year zero-coupon bond.

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

1

2 × 1

1.05289 ×

µ 1

1.04343 + 1 1.06514

= 0.90096.

• Thus its yield is

q 1

0.90096 − 1 = 0.053531.

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

1 × 1

×

µ 1

+ 1 ¶

= 0.93225.

(23)

### Volatility Term Structures (continued)

• Thus its yield is

q 1

0.93225 − 1 = 0.0357.

• The yield volatility is hence 1

2 ln

µ0.053531 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.

(24)

0 100 200 300 400 500 Time period

0.1 0.101 0.102 0.103 0.104

Spot rate volatility

(25)

### Volatility Term Structures (continued)

• 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. (77) on p. 753—and the ri are then simultaneously determined.

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

aBlack, Derman, and Toy (1990).

(26)

### Volatility Term Structures (concluded)

• Suppose the user supplies the volatility term structure which results in (v1, v2, v3, . . . ) for the tree.

• The volatility term structure one period from now will be determined by (v2, v3, v4, . . . ) not (v1, v2, v3, . . . ).

• The volatility term structure supplied by the user is hence not maintained through time.

• This issue will be addressed by other types of (complex) models.

(27)

## Foundations of Term Structure Modeling

(28)

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

— Roger Lowenstein, When Genius Failed (2000)

(29)

[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)

(30)

### 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.

(31)

### 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 .

(32)

### 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 .

(33)

### 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).

(34)

### 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.

(35)

### Fundamental Relations (continued)

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

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

P (t, T ) , T ≤ M. (82) – The forward price equals the future value at time T

of the underlying asset (see text for proof).

• Equation (82) holds whether the model is discrete-time or continuous-time.

(36)

### Fundamental Relations (continued)

• Forward rates and forward prices are related definitionally by

f (t, T, L) =

µ 1

F (t, T, T + L)

1/L

− 1 =

µ P (t, T ) P (t, T + L)

1/L

− 1 (83)

in discrete time.

– The analog to Eq. (83) under simple compounding is f (t, T, L) = 1

L

µ P (t, T )

P (t, T + L) − 1

.

(37)

### 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 (84)

by Eq. (82) on p. 811.

• 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 ) .

(38)

### Fundamental Relations (continued)

• So

f (t, T ) ≡ lim

∆t→0f (t, T, ∆t) = −∂P (t, T )/∂T

P (t, T ) , t ≤ T.

(85)

• Because Eq. (85) is equivalent to

P (t, T ) = eRtT f (t,s) ds, (86) the spot rate curve is

r(t, T ) = 1 T − t

Z T

t

f (t, s) ds.

(39)

### Fundamental Relations (concluded)

• The discrete analog to Eq. (86) 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

.

(40)

### 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). (87) – Relation (87) in fact follows from the risk-neutral

valuation principle.a

aTheorem 14 on p. 438.

(41)

### Risk-Neutral Pricing (continued)

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

• Rewrite Eq. (87) as

Etπ[ P (t + 1, T ) ]

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

– It says the current spot rate curve equals the expected spot rate curve one period from now discounted by the short rate.

(42)

### 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))

¸

. (88)

(43)

### Risk-Neutral Pricing (concluded)

• Equation (87) on p. 816 can also be expressed as Et[ P (t + 1, T ) ] = F (t, t + 1, T ).

– Verify that with, e.g., Eq. (82) on p. 811.

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

(44)

### Continuous-Time Risk-Neutral Pricing

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

h

eRtT r(s) ds i

, t < T. (89)

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

• When the local expectations theory holds, riskless arbitrage opportunities are impossible.

(45)

### 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 = t0.

(46)

### 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.

(47)

### Interest Rate Swaps (continued)

• The value of the swap at time t is thus Xn

i=1

Etπ h

eRtti r(s) ds(fi−1 − c) ∆t i

=

Xn i=1

Etπ

·

eRtti r(s) ds

µ 1

P (ti−1, ti) − (1 + c∆t)

¶¸

=

Xn i=1

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

= P (t, t0) − P (t, tn) − c∆t Xn

i=1

P (t, ti).

(48)

### 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.

(49)

### Swap Rate

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

Pn

i=1 P (t, ti) ∆t . (90)

• 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.

(50)

### 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

(51)

### The Binomial Model (continued)

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

µ ≡ qP u + (1 − q) P d

P − 1 = qu + (1 − q) d − 1.

(91)

• The variance of that return rate is b

σ2 ≡ q(1 − q)(u − d)2. (92)

• The bond whose maturity is only 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.

(52)

### The Binomial Model (continued)

• The market price of risk is defined as λ ≡ (bµ − r)/bσ.

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

(53)

### The Binomial Model (concluded)

• Now change the probability from q to p ≡ q − λp

q(1 − q) = (1 + r) − d

u − d , (93)

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.

(54)

### 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%

(55)

### 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. 832.

(56)

### 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.

(57)

### 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 an up move in rates.

(58)

### Numerical Examples (concluded)

• Solving the equation leads to p = 0.319.

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

(59)

### 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.

(60)

### 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.

(61)

### 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.

(62)

### 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.

(63)

### 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), F = (1 − p) × 92 + p × 98 = 93.914.

• On the other hand, the forward price for a one-year

forward contract on a one-year zero-coupon bond equals 90.703/96.154 = 94.331%.

• The forward price exceeds the futures price.

(64)

### Numerical Examples: Mortgage-Backed Securities

• Consider a 5%-coupon, two-year mortgage-backed

security without amortization, prepayments, and default risk.

• Its cash flow and price process are illustrated on p. 841.

• Its fair price is

M = (1 − p) × 102.222 + p × 107.941

1.04 = 100.045.

• Identical results could have been obtained via arbitrage considerations.

(65)

105

% 5

% & 102.222 (= 5 + (105/1.08))

105 %

0 M

105 &

& % 107.941 (= 5 + (105/1.02))

5

&

105

The left diagram depicts the cash flow; the right diagram illustrates the price process.

(66)

### Numerical Examples: MBSs (continued)

• Suppose that the security can be prepaid at par.

• It will be prepaid only when its price is higher than par.

• Prepayment will hence occur only in the “down” state when the security is worth 102.941 (excluding coupon).

• The price therefore follows the process, M

* 102.222

j 105

• The security is worth

M = (1 − p) × 102.222 + p × 105

= 99.142.

(67)

### Numerical Examples: MBSs (continued)

• The cash flow of the principal-only (PO) strip comes from the mortgage’s principal cash flow.

• The cash flow of the interest-only (IO) strip comes from the interest cash flow (p. 844(a)).

• Their prices hence follow the processes on p. 844(b).

• The fair prices are

PO = (1 − p) × 92.593 + p × 100

1.04 = 91.304,

IO = (1 − p) × 9.630 + p × 5

1.04 = 7.839.

(68)

PO: 100 IO: 5

% %

0 5

% & % &

100 5

0 0

0 0

& % & %

100 5

& &

0 0

(a)

92.593 9.630

% %

po io

& &

100 5

(b)

(69)

### Numerical Examples: MBSs (continued)

• Suppose the mortgage is split into half floater and half inverse floater.

• Let the floater (FLT) receive the one-year rate.

• Then the inverse floater (INV) must have a coupon rate of

(10% − one-year rate) to make the overall coupon rate 5%.

• Their cash flows as percentages of par and values are shown on p. 846.

(70)

FLT: 108 INV: 102

% %

4 6

% & % &

108 102

0 0

0 0

& % & %

104 106

& &

0 0

(a)

104 100.444

% %

flt inv

& &

104 106

(b)

(71)

### Numerical Examples: MBSs (concluded)

• On p. 846, the floater’s price in the up node, 104, is derived from 4 + (108/1.08).

• The inverse floater’s price 100.444 is derived from 6 + (102/1.08).

• The current prices are

FLT = 1

2 × 104

1.04 = 50,

INV = 1

2 × (1 − p) × 100.444 + p × 106

1.04 = 49.142.

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