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# Consider a swaption on the swap of the previous slide

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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 &amp; 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% ¨ ¨¨ ¨ 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 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 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 &lt; 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 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 &amp; 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...
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