# 1 Interest rates, and risk-free investments

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1 Interest rates, and risk-free investments Copyright c 2005 by Karl Sigman. Interest and compounded interest Suppose that you place x 0 (\$) in an account that offers a fixed (never to change over time) annual interest rate of r > 0 (\$ per year). x 0 is called the principle, and one year later at time t = you will have the amount y = x 0 ( + r) (your payoff) having made a profit of rx 0. Everything here is deterministic, in that the amount y is not random; there is no uncertainty hence no risk associated with this investment. This is an example of a risk-free investment. Formally, the variance of the payoff is zero; V ar(y ) = 0. If you reinvest y for another year, then at time t = 2 you would have the amount y 2 = y (+r) = x 0 (+r) 2. In general, if you keep doing this year after year, then at time t = n years, your payoff would be y n = x 0 (+r) n, n 0. Reinvesting the payoff at the end of one time period for yet another is known as compounding the interest. In our example interest was compounded annually, but compounding could be done biannually, quarterly, monthly, daily and so on. For example, if done biannually, then after a half a year, at time t = 0.5, we would earn interest at rate 0.5r yielding x 0 ( + 0.5r). Then a half a year later at time t = interest would be earned again at rate 0.5r on the amount x 0 ( + 0.5r) finally yielding a payoff at time t = of the amount y = x 0 ( + 0.5r)( + 0.5r) = x 0 ( + 0.5r) 2 = x 0 ( r 2 + r) > x 0 ( + r). Thus as expected, biannual compounding is more profitable than annual compounding, and as is easily seen, the profit increases as the time period for compounding get smaller. For example, compounding quarterly yields payoff x 0 ( r) 4 > x 0 ( + 0.5r) 2. In general, if the time period is of size /k (k > 0 an integer), compounding takes place k times per year at a rate r/k per period and yields a payoff at time t = of the amount y (k) = x 0 ( + r k )k. () If this payoff is reinvested for yet another year, then the payoff at time t = 2 is x 0 ( + r k )k ( + r k )k = x 0 ( + r k )2k and in general, after n years yields x 0 ( + r k )nk. This is important for monthly compounding in which the monthly rate is r/2 and the payoff after n years is x 0 ( + r 2 )2n. Over an arbitrary interval of time, (0, t], compounding a total of k times would mean compounding every t/k units of time at rate rt/k and yielding a payoff at time t of y t (k) = x 0 ( + rt k )k. (2) Investing in stocks on the other hand is an example of a risky investment because the price of stocks evolves over time in a random way, apriori unknown to us and not possible to predict with certainty. The stock price at a later time thus has a non-zero variance. As with gambling, when investing in stock there is even a risk of losing some or all of your principle.

2 Letting the compounding interval get smaller and smaller is equivalent to letting k, in which case the compounding becomes continuous compounding yielding a payoff at time t of lim y t(k) = lim x 0( + rt k k k )k = x 0 e rt, (3) where l Hospital s rule is used to show that ( + rt k )k e rt as follows: First take the natural logarithm so that equivalently, we need to show that as k, ln ( + rt k ) = rt. k Taking the derivative wrt k of numerator and denominator yields (after some cancellation) which indeed converges to rt as k. Thus: rt( + rt k ), An initial investment of x 0 at time t = 0, under continuous compounded interest at rate r, is worth x 0 e rt at time t 0..2 Doubling your money If the annual interest rate is r, and you invest x 0 under continuous compounding, then how long will you have to wait until you have doubled your money? We wish to find the value of t (years) for which x 0 e rt = 2x 0, or, equivalently, for which e rt = 2. Solving we conclude t = ln (2)/r = (.693)/r. For example, if r = 0.04, then t 7 years. Notice how then, we could double yet again in another 7 years and so on, concluding that x 0 would increase by a factor of 4 in 34 years; a factor of 8 in 5 years and so on. This is simply an example of geometric growth, and corresponds to the steep rise in the graph of the function f(t) = e rt, t 0..3 Effective and nominal interest rates The annual rate r discussed previously is called the nominal interest rate. As we have seen, by compounding we can do better, and this motivates computing the effective interest rate, that is, the rate required without compounding to yield the same payment, at the end of one year, with compounding at rate r. For example, if compounding is done biannually, then the effective rate is the solution r to the equation + r = ( + 0.5r) 2, or r = r r 2. So if r = 0.07, then r = In general, when compounding k times per year, it is the solution r to + r = ( + r/k) k, or r = ( + r/k) k, and when compounding continuously it is the solution to + r = e r, or r = e r. For example if r = 0.07 and the compounding is monthly, then r = ( + (0.07)/2) 2 = If compounding is continuous, then r = e 0.07 =

4 If compounded continuously, then Plugging in r = 0.05 yields 8, 56. 0, 000e r + 0, 000e 2r. 2. Suppose today is January, and you move into a new apartment in which you must pay a monthly rent of \$000 on the first of every month, starting January. You plan to move out after exactly one year. Then at the end of your year you will have payed a total of \$2, 000. But suppose that you want to pay off all this 2 month future debt now on January. How much should you pay assuming that the nominal annual interest rate is r = 0.06 and interest will be compounded monthly? What if r = 0.0? Repeat when compounding is continuous. We need to compute the present value of the 2 month debt, where the monthly interest rate is r/2: derived as follows: ( i= ) ( + r/2) i ( ) = 000 ( + r/2) i, You must pay 000 on January, followed by 000 on the first of every month for the next months. On February, the 000 to be payed has present value 000/( + r/2) since this is one month in the future. On March, the 000 to be payed has present value 000/( + r/2) 2 since this is two months in the future. Continuing in this fashion we finally get to the last monthly payment on December ( months from now) with present value 000/( + r/2). Summing up all 2 present values yields P V. In effect, you could take the amount P V < 2000, pay your landlord 000 from it now (January ), then place the rest, P V 000, in the bank at interest rate r (compounded monthly): At the beginning of each month you would withdraw 000 and pay your landlord. Then on December there would be exactly 000 left in the account, you would pay this last month s rent to your landlord and close the account. Recalling the geometric series formula a i = an+ a, we can simplify our formula for PV by letting a = /( + r/2), n = ; ( ) (+r/2) (+r/2) ( ) = 000 (+r/2) 2 r/2 (+r/2) = 000( 2 r + ) ( 4 ) ( + r/2) 2.

5 Plugging in r = 0.06 yields 000(20)(.05809) =, 677. If r = 0.0 this amount drops to 000(2)(.07664) = 469. When compounding is continuous, 000 at time i months from now has present value 000e rt with t = i/2; 000e ri 2. Thus 000 ( ) e ri 2 ( e r = 000 e r 2 Plugging in r = 0.06 yields, 676. Plugging in r = 0.0 yields, You win a very special lottery which agrees to pay you \$00 each year forever starting today. Even after you die, the payments will continue to your relatives/friends. Assuming an interest rate of r = 0.08 compounded annually (and assumed constant forever), what is the present value of this lottery? What if compounded continuously? For any interest rate r compounded annually, the present value of this lottery is given by the infinite sum 00 ( + r) i = 00( + /r). via the sum of a geometric series (see The Appendix.) Thus 00/(0.08) + 00 = 350. The point here is that \$00 to be earned i years from now has a present value of only 00/( + r) i. (Note how the present value of the \$00 becomes worthless as i tends to infinity.) This means that if you are given the amount x 0 = 350 now all at once, you could arrange to receive the \$00 lottery amounts each year by simply placing x 0 in an account earning interest at rate r = 0.08 and withdrawing \$00 each year. The account would never go broke. Under continuous compounding 00e ri = 00/( e r ), via the sum of a geometric series. Thus 00/( e 0.08 ) = 300. Contracts like this with a forever into the future payment scheme are called perpetual annuities. For obvious reasons, such contracts are not common. ). 5

6 .5 Net present values Assessing the present value of an investment, for purposes of comparison with another investment, involves allowing for negative amounts (money spent). In this case we call the present value the net present value (NPV). The investment with the largest NPV (but of course positive) is the better investment using this criterion. Letting x i denote the money earned/spent at year i yields a vector (x 0, x, x 2,..., x n ) representing the investment up to time t = n, and its NPV (when interest is compounded yearly at rate r) is given by N x i ( + r) i. For example ( 5,, 3, 0, ) means you spent 5 initially (at time t = 0) then earned at t =, 3 at t = 2, nothing at t = 4 and spent at t = 5. The NPV for this is 5 + ( + r) + 3( + r) 2 ( + r) 4. Examples for NPV. (Opening a winery:) Suppose you plan to open your own winery. An initial investment of (millions of dollars), is required. There are two possibilities to choose from: wait 2 years later to sell your first batch of wine earning 3, or earn 5 by waiting until 4 years later (aging the wine makes it more valuable). Annual interest rate is r =.0 compounded yearly. What should you do? + r =., and thus (, 0, 3) versus (, 0, 0, 0, 5) in terms of NPV yields: NP V = + 3/(.) 2 =.479. NP V 2 = + 5/(.) 4 = 2.4. Thus it is better to wait 4 years, since this has a higher NPV. APPENDIX: Geometric series For any number a, and any integer n 0, a i = an+ a, (4) as is easily derived via algebra; ( + a + a a n )( a) = a n+. If a <, then lim n a n = 0, thus a i = a. (5) In our interest rate examples, a = /( + r) in which case a = + /r, yielding = ( + r) i = ( + ( r ) ) ( + r) n+, (6) and = ( + r) i = + r. (7) 6

7 In many applications, n starts at n = instead of n = 0, the difference being a i = + a i. i= This leads to the modified formulas and i= ( + r) i = ( ) r ( + r) n, (8) i= ( + r) i = r. (9).6 Loan Application; Annuity formulas As an application of the previous section, suppose you wish to take out a loan for \$P, at monthly interest rate r requiring you to make payments each month (starting month from now), for n months. How much must you pay per month? Denoting this by A. Since the sum of the PV s of your n monthly payments must equal \$P we conclude that so that we can solve for A P = A ( + r) i= i = A r ( A = r( + r)n P ( + r) n. ) ( + r) n, In a perpetual situation, where n = (e.g., payments must be made each month forever), then P = A/r, so A = rp. 7

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