This preview shows page 1. Sign up to view the full content.
Unformatted text preview: d then discount them appropriately:
– so (r5 − c )+ at t = 6 is worth (r5 − c )+ /(1 + r5 ) at t = 5.
A sample calculation:
0.015 = max(0, .0354 − .02)
1 + .0354
4 Pricing a Caplet
.138
Expiration at t = 6
Strike = 2% ¨
¨
¨¨ .099
.103
¨
¨¨ ¨¨¨
¨
.080 ¨
.076 ¨
.068
¨
¨
¨¨
¨
¨¨ ¨¨
¨
.064 ¨
.059
.053 ¨
.045
¨
¨
¨
¨¨
¨¨
¨¨ ¨¨¨
¨
¨
¨
.052 ¨
.047 ¨
.041 ¨
.035 ¨
.028
¨
¨
¨
¨
¨¨
¨¨
¨
¨
¨¨ .038¨¨¨ .032¨¨¨ .026¨¨¨ .021¨¨¨ .015
.042
¨
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:
0.021 = 1
1
1
× 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 ShortRate lattice
18.31% ¨
¨¨
¨ 13.18%
14.65%
¨¨
¨
¨
¨
¨¨
9.49%
11.72% ¨ 10.55% ¨
¨
¨
¨
¨¨
¨¨
¨¨
¨
¨
¨
6.83%
9.38% ¨
8.44% ¨
7.59% ¨
¨
¨
¨¨
¨¨
¨
¨
¨¨.75% ¨¨¨.08% ¨¨¨.47% ¨¨.92%
4
7.5% ¨
6
6
5
¨
¨
¨¨
¨¨
¨
¨¨
¨¨
¨
¨
¨
¨
¨
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 interestrate 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
Strike = ﬁxed rate = 5% ¨
¨
¨¨0723
.1648
.
¨
¨¨ ¨¨¨
¨ .1014
.1793 ¨
¨ .0410
¨
¨
¨
¨¨
¨
¨
¨
¨
.1686 ¨
.1021 ¨
.0512 ¨
.0172
¨
¨
¨
¨
¨¨ ¨¨¨ ¨¨¨ ¨¨¨
¨ .0829
−.0008
.1403
.0400
.0122
¨
¨
¨
¨
¨
¨¨
¨¨
¨¨
¨¨ ¨¨¨
¨ .0496¨¨ .0137¨ −.0085¨¨−.0174¨¨−.0141
.0990
¨
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
(.0938 − .05) +
× 0.1793 +
× 0.1021
0.1686 =
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 cashﬂows of swap are not included at those times. 4 Pricing Swaptions
.1125
Fixed rate in Swap = 5%
Underlying Swap Expiration at t = 6
Option Expiration at t = 3 ¨ .1648 ¨
¨ ¨
¨
.0723 ¨
¨
¨¨
¨¨
.0410
.1793 ¨
.1014 ¨
¨
¨¨
¨¨
¨¨
¨
¨
¨
¨
¨
.0172
.1286 ¨
.1021 ¨
.0512 ¨
¨
¨
¨¨
¨¨
¨
¨
¨¨
¨
¨
¨ .0665 ¨¨ .0400 ¨¨ .0122 ¨¨ −.0008
.0908 ¨
¨
¨
¨
¨¨
¨¨
¨
¨
¨
¨¨
¨
¨
¨ .0406¨¨ .0191¨¨
¨ −.0174 ¨¨ .0141
.0620
0¨
¨
¨−
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
.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
Pie,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.
Call such a security an elementary security and Pie,j is its state price.
Can see that elementary security prices sat...
View
Full
Document
 Fall '13
 MartinHaugh

Click to edit the document details