Binomial Interest Rate Tree

Introduction to Term Structure Modeling

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

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

— Warren Buffet, May 3, 2008

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.

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.

History

• The methodology was 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.

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.

aDerman (2004), “complexity without calibration is pointless.”

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. 988).

• In the figure on p. 988, node A coincides with the start of period j during which the short rate r is in effect.

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

r

* ^r

0.5

j _rh

0.5 A

B

C

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

Binomial Interest Rate Tree (continued)

• This may take one of two possible values:

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

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

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

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

• This model can be traced to Salomon Brothers.^a

aTuckman (2002).

Binomial Interest Rate Tree (continued)

• The short rate r can go to r_h and r_ with equal

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

• Hence the volatility of ln r after Δt time is^a σ = 1

2

√1

Δt ln

r_h r_



. (130)

• Above, σ is annualized, whereas r_ and r_h are period based.

aSee Exercise 23.2.3 in text.

Binomial Interest Rate Tree (continued)

• Note that

r_h

r_ = e^2σ

√Δt.

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

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

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

Binomial Interest Rate Tree (continued)

• In general there are j possible rates^a for period j, r_j, r_jv_j, r_jv_j², . . . , r_jv_j^j−1,

where

v_j = e^{Δ 2σj}

√Δt (131)

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

• We shall call r_j the baseline rates.

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

aNot j + 1.

Baseline rates

A C

B B

C

C

D

D D H₂ D

H L_{2 2}

H L_{3 3}

H L_{3 3}² H₁

H₃

Binomial Interest Rate Tree (concluded)

• In the limit, the short rate follows

r(t) = μ(t) e^{σ(t) W (t)}. (132) – The (percent) short rate volatility σ(t) is a

deterministic function of time.

• The expected value of r(t) equals μ(t) e^σ(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 achieves mean reversion to some long-term mean.

Memory Issues

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

• So only the baseline rates r_i and the multiplicative ratios v_i 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.

Set Things in Motion

• The abstract process is now in place.

• We need the yields to maturities 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.

Set Things in Motion (concluded)

• The term structure of (yield) volatilities^a can be estimated from:

– Historical data (historical volatility).

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

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

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

aOr simply the volatility (term) structure.

Model Term Structures

• The model price is computed by backward induction.

• Refer back to the figure on p. 988.

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

P_B + P_C

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

• We compute the values column by column (see next page).

• This takes O(n²) time and O(n) space.

HL

A C

B

Cash flows:

B

C

C

D

D

D D

+

+ 2 2 H

1 2

2 1

= B

+ 2 2

HL

2 3

2 1

= B

H

HL²

+ 2 2

HL

3 4

2 1

?

D

2₁

2₂

2₃

2₄

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 term structure.

Sample Term Structure

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

– This is calibration (the reverse of pricing).

• Assume the short rate volatility is such that v =^Δ r_h

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

An Approximate Calibration Scheme

• Start with the implied one-period forward rates.

• Equate the expected short rate with the forward rate.^a

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

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

• This forward rate can be derived from the market discount function via^b

f_j = d(j)

d(j + 1) − 1.

aSee Exercise 5.6.6 in text.

bSee Exercise 5.6.3 in text.

An Approximate Calibration Scheme (continued)

• Since the ith short rate r_jv_jⁱ⁻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



r_jv_jⁱ⁻¹ = f_j.

• Thus

r_j =

 2

1 + v_j

_j−1

f_j. (133)

• This binomial interest rate tree is trivial to set up (implicitly), in O(n) time.

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.

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:

An Approximate Calibration Scheme (concluded)

• Indeed, this bias is inherent: The tree overprices the bonds.^a

• Suppose we replace the baseline rates r_j by r_jv_j.

• Then the resulting tree underprices the bonds.^b

• The true baseline rates are thus bounded between r_j and r_jv_j.

aSee Exercise 23.2.4 in text.

bLyuu & C. Wang (F95922018) (2009, 2011).

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 (r_m) of the unknown baseline rate r_m for period m.

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

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

• It runs in O(n³) time.

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 (p. 206), the Arrow-Debreu price, or Green’s function.

– 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.

aJamshidian (1991).

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 = r^Δ _j. – The multiplicative ratio is v = v^Δ _j.

– P₁, P₂, . . . , P_j are the known state prices at earlier time j − 1.

– They have rates r, rv, . . . , rv^j−1 for period j.^a

• By definition, _j

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

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

aRecall p. 993.

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) =^Δ P₁

(1 + r) + P₂

(1 + rv) + P₃

(1 + rv²) +· · · + P_j

(1 + rv^j−1) (see next plot).

• So we solve

g(r) = 1

[ 1 + S(j) ]^j (134) for r.

1

1

P

rv

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 panel (a) next page).

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

• The calibrated tree is depicted on p. 1014.

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

H

HL

2 HL

2

2 1

= B

2 H

2 HL

1 2

2 1

= B = B

H

1

2 1

= B

21

2₂

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:

Binomial Interest Rate Tree Calibration (concluded)

• The Newton-Raphson method can be used to solve for the r in Eq. (134) on p. 1010 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. 1003.

bLyuu (1999).

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 panel (b) on p. 1013 as 0.232197, 0.460505, and 0.228308.

• Their sum gives the correct market discount factor 0.92101.

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. 1004 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. 1014 prices without bias the benchmark securities.

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.

Spread of Nonbenchmark Bonds (continued)

• We illustrate the idea with an example.

• Start with the tree on p. 1020.

• Consider a security with cash flow C_i 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).

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

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.

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 p_A(s) is computed at A.

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

p_A(s) = c + p_B(s) + p_C(s) 2(1 + r + s) , where c denotes the cash flow at A.

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) − p_B(s) + p_C(s) 2(1 + r + s)² .

(135)

• 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 p_A(s).^a

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

1 1 ? I

= B

1 1 ?L I

= B

1 1? ?L² ID

1 1 = I

= B

1 1 >L I

= B

1 1 > I

= B

1 1= ? IB²

1 1= ?L IB²

1 1

?

D

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 (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

Spread of Nonbenchmark Bonds (continued)

• Apply the above procedure inductively to yield p(s) and p^(s) at the root (p. 1024).

• This is called the differential tree method.^a – Similar ideas can be found in automatic

differentiation (AD)^b and backpropagation^c in artificial neural networks.

• The total running time is O(n²).

• The memory requirement is O(n).

aLyuu (1999).

bRall (1981).

cWerbos (1974); Rumelhart, Hinton, & Williams (1986).

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.

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. 1028).

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

• It also differs from the yield spread (p. 130) and static spread (p. 131) of the nonbenchmark bond over an otherwise identical benchmark bond.

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

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.

aLyuu (1999).

Fixed-Income Options

• Consider a 2-year 99 European call on the 3-year, 5%

Treasury.

• Assume the Treasury pays annual interest.

• From p. 1031 the 3-year Treasury’s price minus the $5 interest at year 2 could be $102.046, $100.630, or

$98.579 two years from now.

– The accrued interest is not included as it belongs to the original bondholder.

• Now compare the strike price against the bond prices.

• The call is in the money in the first two scenarios out of the money in the third.

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)

Fixed-Income Options (continued)

• The option value is calculated to be $1.458 on p. 1031(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. 1031(b).

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^a

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).

a

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 O_h − O_

P_h − 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.

Delta or Hedge Ratio (concluded)

• Delta measures the sensitivity of the option value to changes in the underlying bond price.

• So it shows how to hedge one with the other.

• Take the call and put on p. 1031 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.

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 y_h (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

y_h y_



. (136)

Volatility Term Structures (continued)

• For example, based on the tree on p. 1014, 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%.

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

2 × 1

1.03526 ×

 1

1.02895 + 1 1.04343



= 0.93225.

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.^a

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

aThe relation is reversed for price volatilities (duration).

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 a flat %10 volatility structure.

Volatility Term Structures (concluded)

• We started with v_i 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. (131) on p. 992—and the r_i are then simultaneously determined.

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

aBlack, Derman, & Toy (1990).

Foundations of Term Structure Modeling

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

— Roger Lowenstein, When Genius Failed (2000)

[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)

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.

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.^a

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

aAlternatively, the instantaneous spot rate, or short rate, at time t.

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.^a

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 .

aIn other words, the (T − t)-period spot rate at time t.

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).

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. (137)

• 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.

Fundamental Relations (continued)

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

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

P (t, T ) , T ≤ M. (138) – The forward price equals the future value at time T

of the underlying asset.^a

• Equation (138) holds whether the model is discrete-time or continuous-time.

aSee Exercise 24.2.1 of the textbook for proof.

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 (139)

in discrete time.

• The analog to Eq. (139) under simple compounding is f (t, T, L) = 1

L

 P (t, T )

P (t, T + L) − 1

 .

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 (140)

by Eq. (138) on p. 1050.

• 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 ) .

Fundamental Relations (continued)

• So

f (t, T ) =^Δ −∂ ln P (t, T )

∂T = −∂P (t, T )/∂T

P (t, T ) , t ≤ T.

(141)

• Because Eq. (141) is equivalent to P (t, T ) = e⁻

_T

t f (t,s) ds, (142) the spot rate curve is

r(t, T ) =  _T

t f (t, s) ds T − t .

Fundamental Relations (concluded)

• The discrete analog to Eq. (142) 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

.