Financial Engineering & Risk Management

Introduction to Term Structure Lattice Models

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
 Fixed Income Markets
Fixed income markets are enormous and in fact bigger than equity markets.
According to SIFMA, in Q3 2012 the total outstanding amount of US bonds was
$35.3 trillion:
Government $10.7 30.4%
Municipal $3.7 10.5%
Mortgage $8.2 23.3%
Corporate $8.6 24.3%
Agency $2.4 6.7%
Asset-backed $1.7 4.8%
Total $35.3 tr 100%
in comparison, size of US equity markets is approx $26 trillion.
Fixed income derivatives markets are also enormous
includes interest-rate and bond derivatives, credit derivatives, MBS and ABS
will focus here on interest-rate and bond derivatives
using binomial lattice models.
(The slides and Excel spreadsheet should be sufficient but Chapter 14 of Luenberger is an excellent reference
for the material in this section.)
2
Binomial Models for the Short Rate

Will use binomial lattice models as our vehicle for introducing:

1. the mechanics of the most important fixed income derivative securities
- bond futures (and forwards)
- caplets and caps, floorlets and floors
- swaps and swaptions
2. the philosophy behind fixed income derivatives pricing
more on this soon.
Fixed-income models are inherently more complex than security models
- need to model evolution of entire term-structure of interest rates.

The short-rate, rt , is the variable of interest in many fixed income models

including binomial lattice models
rt is the risk-free rate that applies between periods t and t + 1
it is a random process but rt is known by time t.

3
The Philosophy of Fixed Income Derivatives Pricing

We will simply specify risk-neutral probabilities for the short-rate, rt

without any reference to the true probabilities of the short-rate
This is in contrast to the binomial model for stocks where we specified p and
1p
and then used replication arguments to obtain q and 1 q.
We will price securities in such a way that guarantees no-arbitrage
Will match market prices of liquid securities via a calibration procedure
often the most challenging part.
Will see that derivatives pricing in practice is really about extrapolating
from liquid security prices to illiquid security prices.

4
Binomial Models for the Short-Rate

 r3,3
 
 
r2,2  r3,2 
 
   

r1,1  r2,1  r3,1 
  
   
r0,0  r1,0 r2,0 r3,0 
   
t=0 t=1 t=2 t=3 t=4

We will take zero-coupon bond (zcb) prices to be our basic securities

k
will use Zi,j for time i, state j price of a zcb that matures at time k
k
Would like to specify binomial model by specifying all Zi,j s at all nodes
possible but awkward if we want to ensure no-arbitrage.
Instead will specify the short-rate, ri,j at each node Ni,j
the risk-free rate that applies to the next period.

5
Binomial Models for the Short-Rate

r3,3
  
 
r2,2  r3,2 
 
qu  
r1,1  qd r2,1 r3,1

   
r0,0  r1,0  r2,0  r3,0 
   
t=0 t=1 t=2 t=3 t=4

Let Zi,j be the date i, state j price of a non-coupon paying security.

Will use risk-neutral pricing to price every security so that:
1
Zi,j = [qu Zi+1,j+1 + qd Zi+1,j ] (1)
1 + ri,j

where qu and qd are the risk-neutral probabilities of an up- and down-move

so qd + qu = 1 and must have qd > 0 and qu > 0.
There can be no arbitrage when we price using (3). Why?
6
Binomial Models for the Short Rate

If the security pays a coupon, Ci+1,j , at date i + 1 and state j then

1
Zi,j = [qu (Zi+1,j+1 + Ci+1,j+1 ) + qd (Zi+1,j + Ci+1,j )] (2)
1 + ri,j

where Zi+1,. is now the ex-coupon price at date i + 1.

If we use (3) or (2) to price securities in the lattice model then arbitrage is
not possible
- Regardless of what probabilities we use! Why is this?

In fact it is very common to simply set qu = qd = 1/2

- and to calibrate other parameters to market prices.
We will assume qu = qd = 1/2 in our examples.
7
Financial Engineering & Risk Management
The Cash Account and Pricing Zero-Coupon Bonds

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
Binomial Models for the Short-Rate

r3,3

 
 
r2,2  r3,2 
 
qu  

r1,1  q r 2,1  r3,1 
 d  
   
r0,0  r1,0 r2,0 r3,0 
   
t=0 t=1 t=2 t=3 t=4

We use risk-neutral pricing to price every non-coupon paying security:

1
Zi,j = [qu Zi+1,j+1 + qd Zi+1,j ] (3)
1 + ri,j

qu > 0 and qd > 0 are the risk-neutral probabilities of an up- and

down-move, respectively, of the short-rate.
There can be no arbitrage when we price using (3). Why?

2
The Cash-Account
The cash-account is a particular security that in each period earns interest at
the short-rate
- we use Bt to denote its value at time t and assume that B0 = 1.
The cash-account is not risk-free since Bt+s is not known at time t for any
s>1
- it is locally risk-free since Bt+1 is known at time t.
Note that Bt satisfies Bt = (1 + r0,0 )(1 + r1 ) . . . (1 + rt1 )
- so that Bt /Bt+1 = 1/(1 + rt ).
Risk-neutral pricing for a non-coupon paying security then takes the form:
1
Zt,j = [qu Zt+1,j+1 + qd Zt+1,j ]
1 + rt,j
 
Zt+1
= EQ
t
1 + rt,j
 
Bt
= EQ
t Z t+1 (4)
Bt+1

3
Risk-Neutral Pricing with the Cash-Account

Therefore for a non-coupon paying security, (4) is equivalent to

 
Zt Zt+1
= EQt (5)
Bt Bt+1

We can iterate (5) to obtain

 
Zt Zt+s
= EQ
t (6)
Bt Bt+s

for any non-coupon paying security and any s > 0.

4
Risk-Neutral Pricing with the Cash-Account

Risk-neutral pricing for a coupon paying security takes the form:

1
Zt,j = [qu (Zt+1,j+1 + Ct+1,j+1 ) + qd (Zt+1,j + Ct+1,j )]
1 + rt,j
 
Q Zt+1 + Ct+1
= Et (7)
1 + rt,j
We can rewrite (7) as
 
Zt Q Ct+1 Zt+1
= Et + (8)
Bt Bt+1 Bt+1
More generally, we can iterate (8) we obtain
" t+s #
Zt Q
X Cj Zt+s
= Et + (9)
Bt B
j=t+1 j
Bt+s

Pricing using (9) ensures no-arbitrage

note that (6) is a special case of (9).
5
 A Sample Short-Rate lattice
18.31%


14.65%  13.18%

   
11.72%  10.55%  9.49%
  
   
  
9.38%  8.44%  7.59%  6.83%
   
    
7.5% 
 6.75% 
 6.08% 
 5.47% 
 4.92%
       
    
6% 5.4% 4.86% 4.37% 3.94% 3.54%

t=0 t=1 t=2 t=3 t=4 t=5

The short-rate, r, grows by a factor of u = 1.25 or d = .9 in each period

not very realistic but more than sufficient for our purposes.

6
 Pricing a ZCB that Matures at Time t=4
100


89.51  100
 

  

83.08 92.22 100
  

 
  
79.27  87.35  94.27  100
   
   
77.22  84.43 90.64  95.81 100

t=0 t=1 t=2 t=3 t=4

 
1 1 1
e.g. 83.08 = 89.51 + 92.22 .
1 + .0938 2 2

Can compute the term-structure by pricing ZCBs of every maturity and then
backing out the spot-rates for those maturities
- so s4 = 6.68% assuming per-period compounding, i.e., 77.22(1 + s4 )4 = 100.

7
 Pricing a ZCB that Matures at Time t=4
100


89.51  100
 

  

83.08 92.22 100
  

 
  
79.27  87.35  94.27  100
   
   
77.22  84.43 90.64  95.81 100

t=0 t=1 t=2 t=3 t=4

Therefore can compute compute Z01 , Z02 , Z03 and Z04

and then compute s1 , s2 , s3 and s4 to obtain the term-structure of
interest rates at time t = 0.
At t = 1 we will compute new ZCB prices and obtain a new term-structure
model for the short-rate, rt , therefore defines a model for the term-structure!

8
Financial Engineering & Risk Management
Fixed Income Derivatives: Options on Bonds

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
Our Sample Short-Rate lattice

18.31%


14.65%  13.18%
 
 
11.72%  10.55%  9.49%
  
    

  
9.38%  8.44%  7.59%  6.83%
      
 



 
7.5%  6.75%  6.08%  5.47%  4.92%
  



  
    
6%  5.4%  4.86%  4.37%  3.94%  3.54%

t=0 t=1 t=2 t=3 t=4 t=5

2
Pricing a ZCB that Matures at Time t=4

100


89.51  100
 

   
83.08 92.22 100
     
  
79.27  87.35  94.27  100
   
      
77.22 84.43 90.64 95.81 100

t=0 t=1 t=2 t=3 t=4

 
1 1 1
e.g. 83.08 = 89.51 + 92.22 .
1 + .0938 2 2

3
Pricing a European Call Option on the ZCB

0
Strike = $84
 Option Expiration at t = 2

1.56
 3.35
4
Option Payoff = max 0, Z2,. 84



 
 Underlying ZCB Matures at t = 4

  
2.97  4.74 6.64

t=0 t=1 t=2

 
1 1 1
e.g. 1.56 = 0 + 3.35 .
1 + .075 2 2

4
 Pricing an American Put Option on a ZCB
0 Strike = $88
 Expiration at t = 3
 4
Payoff at t = 3 is max(0, 88 Z3,. )
4.92 0
Underlying ZCB Matures at t = 4
 
 
8.73 0.65 0
  
  
10.78 3.57 0 0

t=0 t=1 t=2 t=3

  
1 1 1
e.g. 4.92 = max 88 83.08 , 0 + 0 .
1 + .0938 2 2
  
1 1 1
e.g. 8.73 = max 88 79.27 , 4.92 + 0.65 .
1 + .075 2 2

Turns out its optimal early-exercise everywhere

not a very realistic example.
5
Financial Engineering & Risk Management
Fixed Income Derivatives: Bond Forwards

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
Our Sample Short-Rate lattice

18.31%


14.65%  13.18%
 
 
11.72%  10.55%  9.49%
  
    

  
9.38%  8.44%  7.59%  6.83%
      
 



 
7.5%  6.75%  6.08%  5.47%  4.92%
  



  
    
6%  5.4%  4.86%  4.37%  3.94%  3.54%

t=0 t=1 t=2 t=3 t=4 t=5

2
Pricing a Forward on a Coupon-Bearing Bond
Delivery at t = 4 of a 2-year 10% coupon-bearing bond.
We assume delivery takes place just after a coupon has been paid.
In the pricing lattice we use backwards induction to compute the t = 4
ex-coupon price of the bond.
Let G0 be the forward price at t = 0 and let Z46 be the ex-coupon bond price
at t = 4. Then risk-neutral pricing implies
 6 
Q Z4 G0
0 = E0
B4
where B4 is the value of the cash-account at t = 4.
Rearranging terms and using the fact that G0 is known at date t = 0 we
obtain
EQ [Z 6 /B4 ]
G0 = 0Q 4 . (10)
E0 [1/B4 ]
Recall that EQ
0 [1/B4 ] is time t = 0 price of a ZCB maturing at t = 4 but
with a face value $1
have already calculated this to be .7722.
3
 Pricing a Forward on a Coupon-Bearing Bond
110

102.98 
110
 

91.66 107.19 110
  
 
 98.44110.46

 110
  
 
103.83  
 112.96  110
 
    
  108.00114.84 110
     
         
    111.16 116.24 110

t=0 t=1 t=2 t=3 t=4 t=5 t=6

First find ex-coupon price, Z46 , of the bond at time t = 4:

 
1 1 1
e.g. 98.44 = 107.19 + 110.46 .
1 + .1055 2 2
4
 Pricing a Forward on a Coupon-Bearing Bond

91.66


85.08  98.44
  
   
81.53 93.27 103.83
    
  
79.99  90.45  99.85  108.00
  
  
     
79.83 89.24 97.67 104.99 111.16

t=0 t=1 t=2 t=3 t=4

6
Now work backwards in lattice to compute EQ
0 [Z4 /B4 ] = 79.83.
Can now use (13) to obtain
79.83
G0 = = 103.38.
0.7722

5
Financial Engineering & Risk Management
Fixed Income Derivatives: Bond Futures

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
Pricing Futures Contracts

Let Fk be the date k price of a futures contract that expires after n periods.
Let Sk denote the time k price of the security underlying the futures
contract.
Then Fn = Sn , i.e., at expiration the futures price and the underlying
security price must coincide.
Can compute the futures price at t = n 1 by recalling that anytime we
enter a futures contract, the initial value of the contract is 0.
Therefore the futures price, Fn1 , at date t = n 1 must satisfy (why?)
 
0 Fn Fn1
= EQ n1 .
Bn1 Bn

2
Pricing Futures Contracts

Since Bn and Fn1 are both known at date t = n 1, we therefore have

Fn1 = EQ
n1 [Fn ].

By the same argument, we obtain

Fk = EQ
k [Fk+1 ] for 0 k < n.

Can then use the law of iterated expectations to obtain

F0 = EQ
0 [Fn ] .

Since Fn = Sn we have
F0 = EQ
0 [Sn ] (11)
holds regardless of whether or not underlying security pays coupons etc.
In contrast corresponding forward price, G0 , satisfies
EQ
0 [Sn /Bn ]
G0 = . (12)
EQ
0 [1/Bn ]

3
 A Futures Contract on a Coupon-Bearing Bond
Futures contract written on same coupon bond as earlier forward contract
Underlying coupon bond matures at time t = 6
91.66
Futures expiration at t = 4 

95.05 98.44
 
 
98.09  101.14 103.83
  



  
100.81 103.52 105.91 108.00
   
    
103.22 105.64 107.75 109.58 111.16

t=0 t=1 t=2 t=3 t=4

Note that the forward price, 103.38, and futures price, 103.22, are close
but not equal!

4
Financial Engineering & Risk Management
Fixed Income Derivatives: Caplets and Floorlets

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
 Pricing a Caplet

A caplet is similar to a European call option on the interest rate, rt .

Usually settled in arrears but they may also be settled in advance.

If maturity is and strike is c, then payoff of a caplet (settled in arrears) at
time is
(r 1 c)+
so the caplet is a call option on the short rate prevailing at time 1,
settled at time .

A floorlet is the same as a caplet except the payoff is (c r 1 )+ .

A cap consists of a sequence of caplets all of which have the same strike.

A floor consists of a sequence of floorlets all of which have the same strike.

2
Our Short-Rate lattice

18.31%


14.65%  13.18%
 
 
11.72%  10.55%  9.49%
  
    

  
9.38%  8.44%  7.59%  6.83%
      
 



 
7.5%  6.75%  6.08%  5.47%  4.92%
  



  
    
6%  5.4%  4.86%  4.37%  3.94%  3.54%

t=0 t=1 t=2 t=3 t=4 t=5

3
 Pricing a Caplet
.138
Expiration at t = 6 
Strike = 2%  
.103 .099
   
.080 
 .076 
 .068
  
  
.064  .059  .053  .045
   
      
.052

 .047

 .041
  .035

 .028

    
.042  .038 .032 .026  .021  .015

t=0 t=1 t=2 t=3 t=4 t=5

Note that it is easier to record the time t = 6 cash flows at their time 5 predecessor
nodes, and then discount them appropriately:
so (r5 c)+ at t = 6 is worth (r5 c)+ /(1 + r5 ) at t = 5.
A sample calculation:
max(0, .0354 .02)
0.015 =
1 + .0354
4
 Pricing a Caplet
.138
Expiration at t = 6 
Strike = 2%  
.103 .099
 

.080  .076  .068
  
   

.064 .059 .053 .045
 


 
 
.052  .047  .041  .035  .028
    
      

.042  .038 .032  .026 .021  .015

t=0 t=1 t=2 t=3 t=4 t=5

Now work backwards in the lattice to find the price at t = 0.

A sample calculation:
1 1 1
h i
0.021 = 0.028 + 0.015
1.0394 2 2

5
Financial Engineering & Risk Management
Fixed Income Derivatives: Swaps and Swaptions

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
 Our Short-Rate lattice
18.31%


14.65%  13.18%

   
11.72%  10.55%  9.49%
  
   
  
9.38%  8.44%  7.59%  6.83%
   
    
7.5% 
 6.75% 
 6.08% 
 5.47% 
 4.92%
       
    
6% 5.4% 4.86% 4.37% 3.94% 3.54%

t=0 t=1 t=2 t=3 t=4 t=5

Want to price an interest-rate swap with fixed rate of 5% that expires at t = 6

first payment at t = 1 and final payment at t = 6
payment of (ri,j K ) made at time t = i + 1 if in state j at time i.
2
 Pricing Swaps
.1125
Expiration, i.e. last payment, at t = 6 
First payment, at t = 1  
.1648 .0723
Strike = fixed rate = 5%  
  
.1793  .1014  .0410
  
  
.1686  .1021  .0512  .0172
   
      
.1403  .0829  .0400  .0122  .0008
    
    
.0990  .0496 .0137 .0085 .0174 .0141
 

t=0 t=1 t=2 t=3 t=4 t=5

Note that it is easier to record the time t cash flows at their time t 1 predecessor
nodes, and then discount them appropriately:
so (r5,5 K ) at t = 6 is worth (r5,5 K )/(1 + r5,5 ) = .0723 at t = 5.
A sample calculation:
1 1 1
h i
0.1686 = (.0938 .05) + 0.1793 + 0.1021
1.0938 2 2
3
Pricing Swaptions

A swaption is an option on a swap.

Consider a swaption on the swap of the previous slide
- will assume that the option strike is 0%
not to be confused with the strike, i.e. fixed rate, of underlying swap
- and the swaption expiration is at t = 3.
Swaption value at expiration is therefore max(0, S3 ) where S3 underlying
swap price at t = 3.
Value at dates 0 t < 3 computed in usual manner by working backwards
in the lattice
but underlying cash-flows of swap are not included at those times.

4
 Pricing Swaptions
.1125
Fixed rate in Swap = 5% 
Underlying Swap Expiration at t = 6 
.1648 .0723
Option Expiration at t = 3  
 
.1793  .1014 .0410
  

    
.1286  .1021  .0512  .0172
    
 
   
.0008
.0908 
 .0665 
 .0400 
 .0122 

    
      
.0620  .0406  .0191  0  .0174  .0141

t=0 t=1 t=2 t=3 t=4 t=5

Swaption price is computed by determining payoff at maturity, i.e t = 3 and then

working backwards in the lattice.
A sample calculation:
1 1 1
h i
.0908 = .1286 + .0665
1 + .075 2 2
5
Financial Engineering & Risk Management
The Forward Equations

M. Haugh G. Iyengar
Department of Industrial Engineering and Operations Research
Columbia University
The Forward Equations
e
Pi,j denotes the time 0 price of a security that pays $1 at time i, state j and
0 at every other time and state.
e
Call such a security an elementary security and Pi,j is its state price.
Can see that elementary security prices satisfy the forward equations

e e
e
Pk,s1 Pk,s
Pk+1,s = + , 0<s <k +1 (13)
2(1 + rk,s1 ) 2(1 + rk,s )
e
e 1 Pk,0
Pk+1,0 =
2 (1 + rk,0 )
e
e 1 Pk,k
Pk+1,k+1 = .
2 (1 + rk,k )
e
with P0,0 = 1.

2
 Deriving the Forward Equations

  
0  
?
 1


   
? 0
    

   0  
 0
t=0 t=1 t=2 t=3 t=4 t=5

Consider the security that pays $1 only at t = 3 and only in state 2

e
value of this security is P3,2 by definition.
But can also work backwards in lattice to price it. Its value at node N2,2 is
1 1 1 1
h i
0 + 1 =
1 + r2,2 2 2 2(1 + r2,2 )

its value at node N2,0 is 0, and its value at node N2,1 is

1 1 1 1
h i
1 + 0 = .
1 + r2,1 2 2 2(1 + r2,1 )
e 1 e 1 e e
Therefore P3,2 = 2(1+r2,2 )
P2,2 + 2(1+r2,1 )
P2,1 + 0 P2,0 .
3
 Our Short-Rate lattice

18.31%


14.65%  13.18%
 
 
11.72%  10.55%  9.49%
  
    

  
9.38%  8.44%  7.59%  6.83%
      
 



 
7.5%  6.75%  6.08%  5.47%  4.92%
  



  
    
6%  5.4%  4.86%  4.37%  3.94%  3.54%

t=0 t=1 t=2 t=3 t=4 t=5

Now compute the forward prices by iterating the equations forward starting with
e
P0,0 = 1.

4
 ... and the Corresponding Elementary Prices
e
Key: Value at node Ni,j is Pi,j 0.0196


0.0449 0.1041
 
 
0.1003  0.1868  0.2193
  
  
.2194  0.3079  0.2901  0.2293
   
  
.4717
 .4432 0.3143  0.1992 0.1190
   
    
1 .4717 .2238 0.1067 0.0511 0.0246

t=0 t=1 t=2 t=3 t=4 t=5

Sample calculations:
e e
Pk,s1 Pk,s
.3079 = +
2(1 + rk,s1 ) 2(1 + rk,s )
.4432 .2194
= +
2(1 + .0675) 2(1 + .0938)

5
 Derivative Prices Via Elementary Prices

Given the elementary prices the calculation of some security prices becomes very
straightforward:

e.g. Can calculate Z04 as

Z04 = 100 (.0449 + .1868 + .2901 + .1992 + .0511)

= 77.22

as calculated before.

6
 Derivative Prices Via Elementary Prices
Consider a forward-start swap that begins at t = 1 and ends at t = 3
notional principal is $1 million
fixed rate in the swap is 7%
payments at t = i for i = 2, 3 are based as usual on fixed rate minus floating
rate that prevailed at t = i 1
The forward feature of the swap is that it begins at t = 1
first payment is then at t = 2 since payments are made in arrears.

Question: What is the value, V0 , of the forward swap today at t = 0?

Solution: The value is given by

(.07 .0938) (.07 .0675) (.07 .0486)
V0 = .2194 + .4432 + .2238
1.0938 1.0675 1.0486
(.07 .075) (.07 .054)
+ .4717 + .4717
1.075 1.054
= $5, 800.