Liabilities of $1$ each are due at then ends of periods $1$ and $2$. There are three securities available to produce asset income to cover these liabilities, as follows:
(i) A bond due at the end of period 1 with coupon at rate .01 per period, valued at a periodic yield of 14%
(ii) A bond due at the end of period 2 with coupon at rate .02 per period, valued at a periodic yield of 15%
(iii) A bond due at the end of period 2 with coupon at rate .20 per period, valued at a periodic yield of 14.95%
Determine the cost of the portfolio that exactly-matches asset income to liabilities due using
(a) bonds (i) and (ii) only.
(b) bonds (i) and (iii) only
(c) Show that the combination of securities in (b) minimizes the cost of all exact-marching portfolios made up of a combination of the three securities. Note that the minimum cost exact-matching portfolio does not use the highest yielding security in this case.
My attempt was to match the assets with liabilities for t=1 and t=2 here is how I did it:
$L_1$ at $t=1$ is $1(1.14^{-1})=0.877$
$L_2$ at $t=2$ is $1(1.15^{-2})=0.75614$
$X(1.01)(1.14^{-1})+Y(0.02)(1.15^{-1})=0.877$
$Y(1.02)(1.15^{-2})=0.75614$
Solving these leads to $X=0.97$ and $Y=0.98$
$PV=0.97(1.01)(1.14^{-1})+0.98(0.02(1.15^{-1})+1.02(1.15^{-2})=1.633$
Granted that review this afterward I realize I am just turning around and not actually solving the problem. Could anyone help me out and explain what are the proper steps and why?
Myy solution is to solve for bond 2 first, because to reach the exact matching, we need to ensure that A=L at all time, and the 1st bond couldn't cover the liability at the end of period 2. Then substitute the result of bond 2 into the equation of period 1. Hope this will be helpful.