LIBOR Bond Pricing3/1/2016
I was working through Hull's Options, Futures and other Derivatives when Hull states that the price of a Bond paying semi annual coupons in line with six month LIBOR, and discounted using a six month LIBOR discount rate is par. To me this statement seems like it's probably true but I wasn't 100%. I couldn't find the proof online, but it turns out that it is true, here is my derivation.
Proof by induction:
Base case: $n=1$ value(n year bond) $= $PAR $ = \sum\limits_{i=1}^{2n1}\frac{(PAR)R/2}{(1+R/2)^i} + \sum\limits_{i=2n}^{2(n+1)1}\frac{(PAR)R/2}{(1+R/2)^i}$ $+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} + \left(\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right)$ Rearranging:$ = \left(\sum\limits_{i=1}^{2n1}\frac{(PAR)R/2}{(1+R/2)^i} + \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right) + \sum\limits_{i=2n}^{2(n+1)1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$ Then using the fact that a n year bond is priced at PAR:$ = PAR + \sum\limits_{i=2n}^{2(n+1)1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$ No we just take the PAR outside and cancel out to 1.$ = PAR\left(1 + \sum\limits_{i=2n}^{2(n+1)1}\frac{R/2}{(1+R/2)^i}+ \frac{(1+R/2)}{(1+R/2)^{(2(n+1))}} \frac{(1+R/2)}{(1+R/2)^{2n}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}} \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \left(\frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}}\right) \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1+ R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)}+ \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \left(\frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)} \right)+ \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{1 + R/2}{(1+R/2)^{(2n)}} \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{1}{(1+R/2)^{(2n 1)}} \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR$ 
AuthorI work as an actuary and underwriter at a global reinsurer in London. Categories
All
Archives
April 2024

Leave a Reply.