2. A Basic Model of Bond Markets 19
2.4 Pricing by Replication and Discount Factors
In a market where there are no arbitrage opportunities, the following ques- tion can be posed: Given the profile of a certain cash flow in the future, what should its price be? Only in markets with no arbitrage opportunities can such a question make sense. As noted above, if arbitrage opportunities exist there may be more than one price assigned to a portfolio, as the law of one price does not hold. Moreover, one would not bother buying any security if one can make infinite amounts of money with no risk and no initial investment.
The question posed, however, may not have a solution. The absence of a
solution may occur in markets termedincomplete markets. An incomplete market is a market where, given a cash flow profile, there might not be a portfolio that generates this cash flow. A complete market is a market where, given a cash flow profile, there exists a portfolio that generates this cash flow.
Our aim in this section is to further introduce the pricing by replication approach. We initialized this approach when we spoke about forward con- tracts and the cost of carry model. In the current setting of a bond market the pricing by replication approach means that the price of a given cash flow should be the price of the portfolio generating such a cash flow. We shall return to the case ofincomplete marketsin the sequel but for now we assume and exemplify our approach in a complete market. This will also help with our understanding of the concept of incomplete markets.
Assuming a complete market, the price of a cash flow is obtained by minimizing the price of a portfolio generating the given cash flow. Note, that this time the cost of the portfolio is minimized instead of maximizing the proceeds. The same results will be obtained either way but it might be more intuitive to minimize the cost. Hence if we use our example of Chapter 1 (Table 1.1) to price the cash flow of 0 at time 1, $1 at time 2 and 0 at time 3 we need to solve the following optimization.
> simplex[minimize](94.5*B1+97*B2+89*B3,\
> {105*B1+10*B2+8*B3>=0, 110*B2+8*B3>=1,108*B3>=0});
B1=− 1
1155, B2= 1
110,B3=0
Let us see the cash flow produced by this portfolio. To this end we use our structure as before, namely:
> subs(B2=1/110,B1=(-1)/1155,B3 = 0,\
> [CashTime0=-94.5*B1-97*B2-89*B3,\
> CashTime1=105*B1+10*B2+8*B3,\
> CashTime2=0*B1+110*B2+8*B3,\
> CashTime3=0*B1+0*B2+108*B3]);
[CashTime0=−0.8000000000,CashTime1=0, CashTime2=1,CashTime3=0]
The implication is that a dollar at time 2 can be generated by buying a portfolio, the cost of which is 0.8. Thus if there are no arbitrage opportu- nities in the market, the cost (the present value) of a dollar at time 2 must be 0.8.
If this were the case and the price of a dollar at time 2 had been more than $0.8, say $0.9, one would act as follows: Borrow a dollar to be re- turned at time 2, using the price of 0.9. Receive $0.9, invest $0.8 in the above portfolio to generate a dollar at time 2. At time 2 take this dollar and pay off the loan. Arbitrage profit of $0.1 was generated at time 0.
If the price of a dollar at time 2 had been less than $0.8, say $0.7, one would act as follows: Borrow a dollar to be returned at time 2 using the price of 0.8. This can be accomplished by reversing the short position to a long position, and vice versa, in the portfolio that generates a dollar at time 2 as shown below.
> subs(B2=-1/110,B1=1/1155,B3 = 0,\
> [CashTime0=-94.5*B1-97*B2-89*B3,\
> CashTime1=105*B1+10*B2+8*B3,\
> CashTime2=0*B1+110*B2+8*B3,\
> CashTime3=0*B1+0*B2+108*B3]);
[CashTime0=0.8000000000, CashTime1=0, CashTime2=−1, CashTime3=0]
Thus one receives the $0.8 invests $0.7 to get a dollar at time 2. Use this dollar to pay off the loan. Arbitrage profit of $0.1 was generated at time 0.
Now it is also understood why, from a technical point of view, if the NA is not satisfied the question of the price of a dollar at time 2 is meaningless.
The minimization problem above would not be bounded, as one can gen- erate infinite arbitrage profit and the cost of the cash flow in question will be negative infinity.
If the NA is satisfied, finding a portfolio that generates the above cash flow does not require solving a minimization problem; it is sufficient to solve a system of equations as below:
> solve({105*B1+10*B2+8*B3=0, 110*B2+8*B3=1,108*B3=0});
B1=− 1
1155, B2= 1
110,B3=0
There are two advantages to solving the optimization problem. If we are given a market and we are not told whether or not the NA is satisfied, the above system of equations could have a solution but it will be meaningless.
However the optimization problem in this case is unbounded. Moreover, in an incomplete market the system of equations might have no solution while the optimization problem still produces useful information in this case, as we shall see later.
The market in our example therefore does satisfy the NA as the opti- mization problem produced a solution. One of the bonds in the market pays only in one period (bond one). Consequently, it is trivial to find the cost of the cash flow of a dollar at time 1 and zero at other times, while for the cost of a dollar at time 3 and zero at other times, an optimization problem should be solved. We present the optimization problems accomplishing these tasks next. Each optimization is followed by the substitutions of the solution to confirm the resultant cash flow of the solved portfolio.
> simplex[minimize](94.5*B1+97*B2+89*B3,\
> {105*B1+10*B2+8*B3>=1, 110*B2+8*B3>=0,108*B3>=0});
B1= 1
105, B2=0, B3=0
> subs(B3=0,B1=1/105,B2=0,\
> [CashTime0=-94.5*B1-97*B2-89*B3,\
> CashTime1=105*B1+10*B2+8*B3,\
> CashTime2=0*B1+110*B2+8*B3,\
> CashTime3=0*B1+0*B2+108*B3]);
[CashTime0=−0.9000000000,CashTime1=1, CashTime2=0,CashTime3=0]
> simplex[minimize](94.5*B1+97*B2+89*B3,\
> {105*B1+10*B2+8*B3>=0, 110*B2+8*B3>=0,108*B3>=1});
B1=− 4
6237, B2=− 1
1485, B3= 1 108
> subs(B2=-1/1485,B1=-4/6237,B3=1/108,\
> [CashTime0=-94.5*B1-97*B2-89*B3,\
> CashTime1=105*B1+10*B2+8*B3,\
> CashTime2=0*B1+110*B2+8*B3,\
> CashTime3=0*B1+0*B2+108*B3]);
[CashTime0=−0.6981481482,CashTime1=0, CashTime2=0,CashTime3=1]
The three cash flows just priced are “special” cash flows in this market, sometimes also referred to as elementary cash flows. The first cash flow is (1,0,0) i.e., receive $1 at timet1, $0 at timet2and $0 at timet3. Similarly the other two cash flows (interpreted as above) are (0,1,0) and (0,0,1). The table below summarizes the price of each cash flow.
Table 2.2 Three Elementary Cash Flows.
cost t1 t2 t3
0.9000 $1 $0 $0
0.8000 $0 $1 $0
0.6981 $0 $0 $1
In fact, now that we have the prices of these elementary cash flows, it is possible to value any cash flow in this market in a very simple way.
Consider the cash flow(c1,c2,c3)and suppose we can find three portfolios:
portfolio jdenoted byPorj with a price ofPPorj that pays $cj at timetj
and zero otherwise, for j=1,2,3. In this case the price of the cash flow
(c1,c2,c3)must equal (by the replication arguments)
3
∑
j=1
PPorj. However, the cash flow from Porj is actually cj times the cash flow from the jth elementary cash flow. Consequently, portfolio Porj that produces $cj at timetj is equivalent to buyingcj times the portfolio producing $1 at time tj. Portfolios producing the elementary cash flows will be referred to as elementary portfolios and will be denoted by EPorj, j=1,2,3. In our example the compositions of the elementary portfolios are summarized in Table 2.3.
Table 2.3 The Portfolios Generating the Elementary Cash Flows.
"(1,0,0)" "(0,1,0)" "(0,0,1)" Bond B1
0 B2
0 0 B3
1 105
1 110
4 6237
1 1155
1
1485
1 108
Indeed, we already know the prices of the elementary portfolios. In our example the portfolio producing $1 at timet1costs $0.9. Thus the portfolio producing $c1contingent at timet1will cost $0.9c1and the cost of the cash flow(c1,c2,c3)is simply
0.9c1+0.8c2+0.6981c3
(2.3)
The above argument also shows that the cash flow (c1,c2,c3) is gener- ated by buyingc1times the portfolioEPor1,c2 times the portfolioEPor2, andc3times the portfolioEPor3.
The reader may realize that prices of the portfolios producing the ele- mentary cash flows can be interpreted as discount factors. Consequently the cost (value) of a future cash flow is this present value, but the justifica- tion is provided using the NA and the replication approach.
We essentially described in this section two methods of finding the value, or price, of a certain cash flow. A direct replication approach in which a portfolio producing the cash flows at hand is solved. The cost of the portfolio must be the cost of the further cash flow or else arbitrage op- portunities will exist. Alternatively, each cash flow (c1,c2,c3) in this mar- ket can be valued by its present value, i.e., as 0.9c1+0.8c2+0.6981c3.
The next section elaborates on the meaning of discount factors and their intimate connection to the NA. This section ends with applying the two valuation techniques to an example showing that the results are the same.
Consider the cash flow (100,−20,45), its price (present value) should be
> 0.9*100-0.8*20+0.6981*45;
105.4145
This result can be confirmed by solving the optimization problem that finds the least-cost portfolio producing this cash flow.
> simplex[minimize](94.5*B1+97*B2+89*B3,\
> {105*B1+10*B2+8*B3>=100,110*B2+8*B3>=-20,108*B3>=45});
B1=652
693,B2=− 7
33,B3= 5 12
The cost of this portfolio is found by using our structure:
> subs(B1=652/693,B2=(-7)/33,B3 = 5/12,\
> [CashTime0=94.5*B1+97*B2+89*B3,\
> CashTime1=105*B1+10*B2+8*B3,\
> CashTime2=0*B1+110*B2+8*B3,\
> CashTime3=0*B1+0*B2+108*B3]);
[CashTime0=105.4166667, CashTime1=100, CashTime2=−20, CashTime3=45]
We see that indeed the two methods produce the same cost (but for round off errors).
Alternatively, confirmation can be carried out without the need to solve an optimization problem. Since we already know the units of bonds 1 and 2 and 3 in each of the elementary portfolios, we can find the portfolio producing this cash flow utilizing the information in the tables. The reader is asked to show this as well as the arbitrage strategies that exist if the price of the cash flow in question is not $105.416. We move next to the interpretation of the discount factors and their relation to the NA.