Thursday, May 22, 2014

Fixed Income Derivatives: Options on Bonds

In this lesson, we moved to pricing options on the zero coupon bonds that we explored in the last lesson.

We will need the short rate lattice and the zcb lattice for our examples and these are shown below:














We then priced a European call option on this zero coupon bond.
The maturity of the option is t=2 periods and the strike price is $84.
Therefore, the payoff of the option is: max(0, Z(2,.,4)-84) where the . in the j position indicated whichever state the bond ultimately ends up.

We can simply find the payoff of the bond from the lattice above: if the bond is worth $83.08, the option will be worthless since the strike is $84. If the bond is worth $87.35, the option is worth $3.35, and if the bond is worth $90.64, then the option is worth $6.64.  These are the final payoffs of the option and we can input them into our lattice and work backwards using the risk-neutral probabilities to calculate the price of the option, pictured below:














We then priced an American put option on this same zero coupon bond.
The maturity of the option is t=3 periods and the strike price is $88.
We did the exact same thing, except we stopped at each node to see whether it was optimal to exercise the option at that node.  We can see from the bond price lattice that $88 is less than each of the values of the bond at t=3, so it would never be optimal to hold the put option until maturity.  Therefore the payoff of the put option would be 0 at each ending node.  Below is the lattice for the American put option:













We calculated 4.92, 0.65, 8.73, 3.57, and 10.78 by simply finding the true value of the option:
4.92 = max(88-83.08, 1/(1+.0938)* [0.5(0) + 0.5(0)])
0.65 = max(88-87.35, 1/(1+.0675)* [0.5(0) + 0.5(0)])
8.73 = max(88-79.27, 1/(1+.075)* [0.5(4.92) + 0.5(0.65)])
3.57 = max(88-84.43, 1/(1+.054)* [0.5(0) + 0.5(0.65)])
10.78 = max(88-77.22, 1/(1+.06)* [0.5(8.73) + 0.5(3.57)])

We can see that it would be optimal here to exercise early at every single node, which isn't a very realistic scenario but interesting all the same.




























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