MTH6154 Financial Mathematics I Interest Rates and Present Value Analysis

Transcription

1 16 MTH6154 Financial Mathematics I Interest Rates and Present Value Analysis Contents 2 Interest Rates Definitions Rate of Return Annualised Rate of Return Interest rates Variable Interest Rates The Instantaneous Interest Rate The Yield Curve Present value analysis An Example on transferring cash flows on the time line Inflation Discount Factors Internal Rate of Return Definition Existence and uniqueness of the internal rate of return An Example Interest Rates 2.1 Definitions Rate of Return Suppose we invest an amount P, called the principal, which at a later date results in a value of P final. The rate of return, is the percentage change of this amount: R = P final P P The rate of return is sometimes simply called the return of the investment. Note that the expression above can also be written as P final = P 1 + R, which is useful if R is known and we wish to calculate P final. Examples 2.1. We consider 2 parts in this example:.

2 17 1. We invested P = 15, in a property which is now valued at P final = 2,. The return on our investment is: R = 2, 15, 15, =.3333 = 33.33%. 2. How much cash will result from a placement of $12, in a investment that guarantees a 2% return? We simply substitute in the formula above Annualised Rate of Return P final = P 1 + R = $12, = $14,4. The rate of return described above is often a too crude number to use in evaluating the performance of an investment. This is because it does not take into account how much time it took for the investment to accrue its value. For example, the rate of a return of 33.33% described in Example 2.1, part 2.11, would be spectacular if gained in two weeks. It would be less spectacular if the increase took 4 years. The annualised rate of return is defined to be the rate of return divided by the time in years, T, it took to accomplish: r = R T. Note that the expression in the previous section can be written as P final = P 1 + rt. Examples We invested 1,5 in some shares one month ago. The investment is now worth 1,35. Therefore the rate of return of our investment is: R = = and the annualised rate of return, based on T = 1/12, is: r =.1 1/12 =.1 = 1%, = 1.2 = 12%. 2. How much money will we obtain if we invest $2,2 for six months in an investment that guarantees an annualised rate of return of 3%? We substitute in the equation above: P final = P 1 + rt = $ = $2, Remark 2.3. The main reason to use annualised rates of return is to standardise or rescale rates of return so that they can be compared. For example, it is difficult to understand whether a return of R =.1% is a good or bad return for a one day investment. The annualised rate is r = R/T =.1/1/365 = 3.65% which is a very generous annualised rate in the current low interest rate world.

3 Interest rates Suppose we borrow an amount P, called the principal, with interest at annualised rate r. This rate could be rate r per time period Usually this is equal to one year We shall adopt the convention from now on, that time period = 1 year. This means that, if after 1 year we have to repay the loan, we need to pay the principal plus the interest, that is P + rp = P 1 + r. By definition, if interest is compounded every 1 -th of a year, then after one year we owe n P 1 + n r n. 1 Note that, if the interest rate is compounded annually, we put n = 1 and then the amount owed equals P 1 + r, exactly as stated above. Example 2.4. Suppose you borrow an amount P, to be repaid after one year with interest at rate r per annum, compounded semi-annually. This corresponds to choosing n = 2 in 1. Then, the following facts happen sequentially. After half a year you are charged a simple interest at rate r/2 per half-year, which is then added on to the principal. Thus, after 6 months you owe P 1 + r, 2 This is again charged interest at rate r/2 for the second half-year period, therefore after 12 months you owe P 1 + r 1 + r = P 1 + r If n = 4 in 1 we say that interest is compounded quarterly, and if n = 12 we say that interest is compounded monthly. Roughly speaking, if interest is compounded, you pay interest on the interest. Example 2.5. Suppose that you borrow 1 at interest rate 18% compounded monthly. How much do you owe after one year? Solution. Let P = 1 be the amount borrowed. After one year you owe = = The compounded interest rate r is also called the nominal rate. In this case, the effective interest rate r eff is defined to be r eff = amount paid after one year P P For the interest rate compounded every 1 -th of a year, we have n r eff = P 1 + r/nn P P = 1 + r n n 1. Example 2.6. Suppose that you borrow 1 at interest rate 18% compounded monthly. What is the effective interest rate?.

4 19 Solution. We know from previous Example that after one year you owe This implies that the effective interest rate is r eff = =.195 = 19.5%. 1 Example 2.7. Credit card company A charges a 24.5% nominal daily rate whereas credit card company B charges a 24.7% nominal monthly rate. Which company offers the best deal? Solution. We cannot directly compare the nominal interest rates offered as they refer to different periods. To compare them we calculate the effective rate for both deals. For company A we have: r eff = For company B we have: r eff = % 1 = % 12 1 = We conclude that company B offers a better deal even though the nominal rates advertised might suggest the opposite. Imagine now that interest is compounded at every millisecond interval. In this limiting case for n, we shall say that interest is compounded continuously. To be precise, when interest is compounded continuously at a nominal rate r per annum, the amount to be repaid after one year is equal to lim 1 P + r n = P e r. n n This implies that for a continuously compounded interest the effective interest rate is r eff = P er P P = e r 1. Example 2.8. Suppose that a bank offers a continuously compounded interest at nominal rate 5%. What is the effective interest rate? Solution. The effective interest rate in this case is r eff = e.5 1 =.5127 = 5.127%. More generally, if the compounded interest is charged at nominal rate r and is compounded every 1 -th of a year, then after t years an investor owes n P 1 + r n 1 + r n 1 + r n = P 1 + r nt ; } n {{ n n } n t times and if it is compounded continuously, then after t years an investor owes P } e r e r {{ e} r = P e rt. t times

5 2 Example 2.9 The doubling rule. If a bank offers a continuously compounded interest at nominal rate r, how long does it take for the amount of money in the bank to double? Solution. Let t denote the time in years after which the amount P in the bank will double. Then P e rt = 2P e rt = 2 rt = log 2. therefore t = log r r For example, if this nominal interest rate is r = 1%, then it will take 69.3 years for the amount in the bank to double. 2.2 Variable Interest Rates We briefly considered interest rates in section 3.1. In this section we resume this analysis using the notion of instantaneous interest rate The Instantaneous Interest Rate Let us write r = rt and call rt the instantaneous interest rate. We are interested in the amount P t that is accumulated in a bank account at time t, given that the amount P is deposited at time. An equivalent problem can be stated as follows. Suppose that you borrow P from a bank at time and the continuously compounded variable interest rate is rt. How much do you owe to the bank at any time t >? It turns out that the as yet unknown function P t satisfies a differential equation which we shall derive. Theorem 2.1. Suppose that rt is a piece-wise continuous function. Then P t = P trt. 2 Proof. Let P t be the amount that has been accumulated in the bank account at time t and let h be a small time period. We assume that the interest rate rt does not change too drastically between times t and t + h due to continuity. Then at time t + h we have so P t + h P t + P trth, P t + h P t P trth, thus P t + h P t P trt. h If we let h become smaller and smaller, then the difference between the right and left hand side of the above equation becomes smaller and smaller, so Thus P t + h P t lim h h which is the desired differential equation. P t = P trt, = P trt.

6 21 Let us now solve this equation. This is a separable first order differential equation which can be solved as follows. Writing P t P t = rt and integrating both sides yields t Thus hence and so P u P u du = t ru du. t [log P u] u=t u= = ru du, log P t = log P + t ru du, t P t = P exp ru du, 3 which is the desired relation between the amount P t in your bank at time t and the timevarying interest rate function rt. Note that if rt = r, that is, the interest rate is constant, then P t = P e rt. Thus, the general formula 3 reduces to the familiar formula for continuously compounded interest if the interest rate is constant, exactly as expected The Yield Curve In the context of a time-varying interest rate rt, the concept of the yield curve rt, defined by rt = 1 t t ru du, turns out to be useful. Note that rt is just the average interest rate on the interval, t. We have t P t = P exp ru du = P exprt t. Example Let the time-varying instantaneous interest rate be given by rt = t r 1 + t 1 + t r 2, where < r 1 < r 2 < 1. In order to get a rough idea what the graph of this function looks like note that r = r 1 and lim t rt = r 2. Moreover, rewriting we see that, for t >, rt = r 2 r 2 r t, r t = r 2 r t 2 >.

7 22 and r t = 2 r 2 r t 3 <. Thus rt is a concave function for positive arguments, starting at r 1 for t =, growing monotonically thereafter, and reaching r 2 asymptotically as t tends to infinity. The yield curve is rt = 1 t r 2 r 2 r 1 du = 1 r2 t r 2 r 1 log1 + t = r 2 r 2 r 1 log1 + t. t 1 + u t t Note that lim rt = r 2. t We close this section with one more example. Example Suppose rt = r + A cos ωt, where < A < r and ω >, which can be thought of as a simple model of fluctuating interest rates. Now, if at time you have P pounds in the bank, then at time t you have t P t = P exp ru du t = P exp r + A cos ωu du [ ] u=t A sin ωu = P exp r u + ω u= A sin ωt = P exp r t +. ω The yield curve is Note that rt = 1 t t ru du = 1 t 2.3 Present value analysis In this section we answer the question: r t + lim rt = r. t A sin ωt A sin ωt = r +. ω ωt How much is it worth today, a payment received in the future? In order to get a better feeling, suppose that the current interest rate is 1%. If somebody gives you 1 pounds today, you can put it in the bank and after a year you will have 11 pounds. Thus 1 pounds received today have a value of 11 pounds in a year s time, or conversely, 11 pounds received in a year s time have a present value of 1 pounds. More generally, suppose you can borrow and lend money at nominal rate r per annum. Then, suppose that you borrow the amount 1 + r i V pounds now, then you can put it in the bank and at the end of year i you will have in the bank 1 + r i V 1 + r i = V pounds.

8 23 Thus, V pounds received in i years at interest rate r is worth 1 + r i V in today s money. Thus, we define the present value of V pounds at the end of year i to be P V V = 1 + r i V. Similarly, if the compounded interest is compounded every 1 -th of a year at nominal rate n r, the present value of V pounds in i years is P V V = 1 + r n inv. We can generalise the notion of present value to a cash flow stream a = a 1, a 2,..., a n, that is, a sequence of payments, where a i is the payment made at the end of year i. We claim that the present value of the cash flow stream a = a 1, a 2,..., a n is P V a, given by P V a = a i 1 + r i. In order to see this, observe that the cash flow stream a = a 1, a 2,..., a n can be replicated by putting the amount P V a in the bank at time and then by making successive withdrawals a 1, a 2,..., a n every year. At the end of year 1 the amount in the bank is P V a1 + r a 1 = a i 1 + r i 1 + r a 1 = a 1 + a r + a r a n 1 + r n 1 a 1 = a r + a r a n 1 + r n 1. At the end of year 2 the amount in the bank is a2 1 + r + a r + + a n 1 + r a r n 1 2 = a r + a r a n 1 + r n 2. Continuing in this way, we see that at the end of year n, all withdrawals have been made and no money is left in the bank. This way we managed to replicate exactly the payments of the cash flow stream a, by using the initial amount of P V a. Example You are offered three different jobs. The salary paid at the end of each year in thousands of pounds is Job A B C Which job pays best if the nominal rate is r =.1, r =.2, and r =.3?

9 24 Solution. We shall compare the present values of the cash flow streams. Now, the present value for job A is P V A = r r r r r 5, while the present value for job B is P V B = r r r r r 5. Finally, the present value for job C is P V C = r r r r r 5. Taking the different interest rates into account we get the following present values. Thus, r PV A PV B PV C if r = 1%, then A pays best, then B, then C; if r = 2%, then B pays best, then A, then C; if r = 3%, then C pays best, then B, then A. It is possible to consider cash flow streams which go on forever. Before doing so, recall some facts about geometric series. Theorem Let b 1 and let n be a positive integer. Then i= b i = 1 + b + b b n = 1 bn+1 1 b = bn+1 1 b 1. Proof. Let Then so Thus as claimed. x = 1 + b + b b n. bx = b + b b n + b n+1, x bx = 1 b n+1. x = 1 bn+1 1 b, A useful consequence of the theorem above, is the following corollary.

10 25 Corollary If b < 1, then Proof. Note that if b < 1, then Thus i= b i = lim n i= i= b i = 1 1 b. lim n bn =. b i = lim n 1 b n+1 1 b = 1 1 b = 1 1 b. Another corollary follows. Corollary If r < 1, then Proof. if b < 1, then b i=a r i = ra r b+1 1 r. b b a r i = r a r i = r a 1 rb a+1 1 r i=a i= = ra r b+1. 1 r We now return to the present value of cash flow streams that go forever. Example A perpetuity entitles its owner to be paid an amount c at the end of each year indefinitely. What is its present value? Solution. In this case the cash flow stream is a = c, c, c,... and its present value is P V a = c 1 + r i = c 1 + r = c 1 + r r i r i i= c 1 = 1 + r r 1 c = 1 + r 1 = c r.

11 26 Remark Note that the present value c of the perpetuity can be replicated in the r following way. Initially, put c in the bank. After one year you have c 1 + r in the bank and r r you withdraw c, after which you have c r 1 + r c = c r + c c = c r. After one more year end of second year you have c 1 + r in the bank and you withdraw c, r after which you have c r 1 + r c = c r + c c = c r. Observe that the amount of money in the bank does not change, remains c after each withdrawal and you can continue withdrawing money in this way at the end of every year, r forever An Example on transferring cash flows on the time line. We close this section with the following example. Example Suppose that John is self-employed and wants to save for his retirement in 2 years. From the time of his retirement onwards he wants to withdraw 1, every month at the beginning of each month for 3 years. What amount of money does he have to save every month at the beginning of each month for the next 2 years to fund his retirement? Suppose that the nominal interest rate is 6% compounded monthly. There exist various possible solutions to this example. Here is one of them. Solution. Let A be the unknown amount deposited monthly. The idea is that the forward value of the deposits at the end of year 2 should equal the discounted value of all the withdrawals at the beginning of year 21. First, we calculate the forward value of all the deposits at the end of year 2. Set the monthly compounding factor as α = 1 + r/12 = 1.5. If A is deposited at the beginning of each month i, then the forward value of each such deposit by the end of year 2 will be Aα 24 i, for i =, 1, 2,..., 239. Hence by the end of the 2 year period there will be 24 monthly deposits 2 years with the total value Aα 24 + Aα Aα 2 + Aα = Aα α24 1 α 1. Next, set the monthly discounting factor β = α 1, since it is the inverse of the monthly compounding. If 1, is withdrawn at the beginning of each month 24 + j, then the value of each such withdrawal at the beginning of month 241 beginning of year 21 is 1β j, for j =, 1,..., 359. Hence, the total value of the 36 withdrawals 3 years at the beginning of the 3 year period is 1 + 1β + 1β β 359 = 1 1 β36 1 β. So, we equate the forward value of the deposits with the discounted value of all the withdrawal, to get A α α24 1 α 1 1 β36 = 1 1 β and, taking into account that α/α 1 = 1/1 β, we obtain A = 1 1 β36 α 24 1 = 36.9.

12 27 Here is a second solution. Solution 2. Let A denote the as yet unknown amount of pounds deposited every month. We will first calculate the present value P V A of all his deposits. Write β = Then, since there are 24 months in 2 years, = P V A = A + Aβ + Aβ Aβ 239 = A1 + β + β β 239 = A 1 β24 1 β. Let W = 1 denote his monthly withdrawals in pounds. Since his first withdrawal will take place in 24 months time, and since there are 36 months in 3 years, the present value P V W of all his withdrawals is P V W = W β 24 + W β W β = W β β + + β β36 = W β 1 β. Clearly, the present value of his deposits must equal the present value of his withdrawals. Thus so hence A 1 β24 1 β A = W β P V A = P V W, = W β24 1 β36 1 β, 24 1 β36 = 36.9, 1 β24 that is, he has to deposit 361 every month to fund his retirement Inflation We now have a quick look at inflation, the phenomenon of prices increasing as a whole over time. Inflation can be quantified by reference to an index, for example the Retail Price Index RPI, which gives the price of a basket of goods over a period of time. Alternatively, inflation can be quantified by the rate at which prices increase, denoted by r inf. For example, if the yearly inflation rate is 3% then what costs 1 pounds now, will cost 13 pounds next year. From a different point of view, the purchasing power of 13 pounds in one year is 1 pounds today. How do we adjust the nominal interest rate to take into account inflation? In order to answer this question, let r inf denote the rate of inflation and let r denote the nominal interest rate.

13 28 - Suppose you put P in the bank now. - Then after one year you will have 1 + rp in the bank. - But the purchasing power of the amount 1 + rp in one year from now, taking into account the inflation, is the same as the amount 1 + rp 1 + r inf 1 today. So the real interest rate real in the sense that it reflects the decrease of purchasing power denoted by r a, also known as inflation adjusted interest rate, is given by r a = therefore in practice we use 1+r 1+r inf P P P = 1 + r 1 + r inf 1 = r r inf 1 + r inf r r inf, r a r r inf. Example 2.2. Suppose that the current nominal interest rate is 5% and the rate of inflation is 3%. Then the inflation adjusted interest rate is approximately 5% 3% = 2% Discount Factors We can write present value for all types of interest by the use of discount factors. Definition The discount factor to T years, DT, is the inverse of what an account holding P = 1 will grow to in time T. Remarks If r is the yearly compounded interest rate i.e. the one year rate, then the discount factor to T = n years is Dn = r n. 2. If r is the monthly compounded interest rate i.e. the one month rate, then n months corresponds to T = n/12 years, and the discount factor to n months is n 1 D = r 1 n If r is the continuously compounded rate, then at time T a deposit will grow by a factor of exprt, so that the discount factor to T years is DT = e rt. 4. If rt is the variable interest rate, then at time T a deposit will grow by a factor of e rt t, where rt is the yield curve, so that the discount factor to T years is In Example 2.11, DT = e rt T. DT = exp rt T = exp r 2 T +r 2 r 1 log1+t = exp r 2 T 1+T r 2 r 1. In Example 2.12, A sin ωt DT = exp r T. ω

14 29 5. Note that when interest rates are positive which is most often the case the discount factor is smaller than 1 hence the wording discount. This means that the value of having money in the future is less than the value of having it now; the reason being that we are missing on the opportunity of earning interest rate on that money. Definition The present value PV of N units of cash paid at time T is defined to be P V = DT N. I.e. the cash-flow discounted to the present by the discount factor. Definition If we expect to receive N 1, N 2,..., N n units of cash on times T 1, T 2,..., T n then the present value of these cash-flows is just the sum of each individual PV P V = DT i N i. 2.4 Internal Rate of Return Definition Suppose you invest amount a and after one year you get back amount b. Recall from Section that the rate of return, denoted by r, of this investment is defined by r = b a a or r = b a 1 or a = b 1 + r. More generally, if the initial investment a yields the yearly cash flow stream b 1, b 2,..., b n, then we define the internal rate of return r to be a number such that 1 < r < and b i a = 1 + r = b i i 1 + r i, that is, the internal rate of return is the hypothetical one-year interest rate such that the present value of the cash flow stream is equal to the initial investment. Or put differently again, the rate of return r is a solution of the equation fr =, for the function fr = a + b i 1 + r i Existence and uniqueness of the internal rate of return The definition above is meaningful, if the equation fr = has a unique solution 1 < r <. The following result proves this precise property.

15 3 Theorem If a > and b i > for i = 1,..., n, then the equation fr = a = b i 1 + r i. 4 has a unique solution r in the range 1,. Proof. This results from the following easy calculations: f is a strictly decreasing function. This is so because b i are positive as then: f r = b i i <. 1 + r i+1 lim r 1 fr = + as lim fr = lim r 1 r 1 a + b i 1 + r i = a + b i lim 1 + r 1 r i = +. lim r + fr = a < as lim fr = r + lim r + a + b i 1 + r i = a + b i lim 1 + r + r i = a. Therefore f is a function that is strictly monotonically decreasing on 1, and its values descend from + to a <. As f is a continuous function, by Bolzano, there must be a point r 1, where the function f attains a zero, that is where 4 is satisfied. The solution is unique as the function is monotonic An Example Solving equation 4 can be non-trivial as it is related to solving a polynomial equation of degree n for which no closed solutions are available in general. However the case n = 2 can be solved easily as illustrated by the following example. Example Suppose that at the beginning of 212 an investor bought shares of a certain company worth 1,, which managed to sell again at the beginning of 214 for 1,5. At the end of 212 and 213 the investor also received dividends of 1 and 2, respectively. a What is the internal rate of return of this investment? b What is the internal rate of return of this investment if the investor had to pay 1% income tax on the dividends and 18% capital gains tax? Solution. a The internal rate of return is a solution of the equation 1 = r r r 2 = r r 2.

16 31 By rewriting this, we get 11 + r r 17 =, which is a quadratic equation in 1 + r with solutions This implies that the solutions r are and and reject, since r >. Thus the internal rate of return of this investment is r = 3.94%. b Since the two dividend payments were taxed at 1%, the investor received only 1 1 = 9 at the end of 212 and 2 2 = 18 at the end of 213. As the investor made a profit of 1,5 1, = 5 after selling the shares, he had to pay.18 5 = 9 capital gains tax. This reduces the investor s final pay-off to 15 9 = 1,41. Thus the internal rate of return in this case is a solution of the equation 1 = r 1 + r 2 Again, this is a quadratic equation in r with solutions.335 and reject, since r >. Thus the rate of return of this investment is r = 3.35%. Note that this amount is smaller than the one in a to take account of the reduced profitability arising from tax payments.