MTH6154 Financial Mathematics I Interest Rates and Present Value Analysis

16 MTH6154 Financial Mathematics I Interest Rates and Present Value Analysis Contents 2 Interest Rates and Present Value Analysis 16 2.1 Definitions.................................... 16 2.1.1 Rate of Return.............................. 16 2.1.2 Annualised Rate of Return........................ 17 2.1.3 Interest rates............................... 18 2.2 Variable Interest Rates.............................. 20 2.2.1 The Instantaneous Interest Rate..................... 20 2.2.2 The Yield Curve............................. 21 2.3 Present value analysis.............................. 22 2.3.1 An Example on transferring cash flows on the time line......... 26 2.3.2 Inflation................................. 27 2.3.3 Discount Factors............................. 28 2.4 Internal Rate of Return............................. 29 2.4.1 Definition................................ 29 2.4.2 Existence and uniqueness of the internal rate of return......... 29 2.4.3 An Example............................... 30 2.5 Immunisation of cash-flows........................... 31 2.5.1 The effective duration.......................... 32 2.5.2 Reddington Immunisation........................ 33 2.5.3 Immunisation with compound interest.................. 36 2 Interest Rates and Present Value Analysis 2.1 Definitions 2.1.1 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),.

17 which is useful if R is known and we wish to calculate P final. Examples 2.1. We consider 2 parts in this example: 1. We invested P = 150,000 in a property which is now valued at P final = 200,000. The return on our investment is: R = 200,000 150,000 150,000 = 0.3333 = 33.33%. 2. How much cash will result from a placement of $12,000 in a investment that guarantees a 20% return? We simply substitute in the formula above 2.1.2 Annualised Rate of Return P final = P 1 + R) = $12,000 1 + 0.2) = $14,400. 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 40 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 2.2. 1. We invested 1,500 in some shares one month ago. The investment is now worth 1,350. Therefore the rate of return of our investment is: R = 1350 1500 1500 = 150 1500 and the annualised rate of return, based on T = 1/12, is: r = 0.1 1/12 = 0.1 = 10%, = 1.2 = 120%. 2. How much money will we obtain if we invest $2,200 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 ) = $2200 1 + 0.03 1 ) = $2,233. 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 = 0.01% is a good or bad return for a one day investment. The annualised rate is r = R/T = 0.0001/1/365) = 3.65% which is a very generous annualised) rate in the current low interest rate world.

18 2.1.3 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 2. 2 2 2) 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 100 at interest rate 18% compounded monthly. How much do you owe after one year? Solution. Let P = 100 be the amount borrowed. After one year you owe 100 1 + 0.18 ) 12 = 100 1.195 = 119.5. 12 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/n)n P P = 1 + r n) n 1. Example 2.6. Suppose that you borrow 100 at interest rate 18% compounded monthly. What is the effective interest rate?.

19 Solution. We know from previous Example that after one year you owe 119.5. This implies that the effective interest rate is 119.5 100 r eff = = 0.195 = 19.5%. 100 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 + 1 ) 365 365 24.5% 1 = 0.277516. 1 + 1 12 24.7% ) 12 1 = 0.276973. 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 0.05 1 = 0.05127 = 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

20 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 2 0.693. 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. 2.2.1 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 0) is deposited at time 0. An equivalent problem can be stated as follows. Suppose that you borrow P 0) from a bank at time 0 and the continuously compounded variable interest rate is rt). How much do you owe to the bank at any time t > 0?) It turns out that the as yet unknown function P t) satisfies a differential equation which we shall derive. Theorem 2.10. Suppose that rt) is a piece-wise continuous function. Then P t) = P t)rt). 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 t)rt)h, P t + h) P t) P t)rt)h, thus P t + h) P t) P t)rt). 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 0 h which is the desired differential equation. P t) = P t)rt), = P t)rt).

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 0 P u) P u) du = t 0 ru) du. t [log P u)] u=t u=0 = ru) du, log P t) = log P 0) + 0 t 0 ru) du, t ) P t) = P 0) exp ru) du, 3) 0 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 0)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. 2.2.2 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 0 ru) du, turns out to be useful. Note that rt) is just the average interest rate on the interval 0, t). We have t ) P t) = P 0) exp ru) du = P 0) exprt) t). Example 2.11. Let the time-varying instantaneous interest rate be given by 0 rt) = 1 1 + t r 1 + t 1 + t r 2, where 0 < r 1 < r 2 < 1. In order to get a rough idea what the graph of this function looks like note that r0) = r 1 and lim t rt) = r 2. Moreover, rewriting we see that, for t > 0, rt) = r 2 r 2 r 1 1 + t, r t) = r 2 r 1 1 + t) 2 > 0.

22 and 0 r t) = 2 r 2 r 1 1 + t) 3 < 0. Thus rt) is a concave function for positive arguments, starting at r 1 for t = 0, 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 2.12. Suppose rt) = r + A cos ωt, where 0 < A < r and ω > 0, which can be thought of as a simple model of fluctuating interest rates. Now, if at time 0 you have P 0) pounds in the bank, then at time t you have t ) P t) = P 0) exp ru) du 0 t ) = P 0) exp r + A cos ωu) du 0 [ ] ) u=t A sin ωu = P 0) exp r u + ω u=0 ) A sin ωt = P 0) exp r t +. ω The yield curve is Note that rt) = 1 t t 0 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 10%. If somebody gives you 100 pounds today, you can put it in the bank and after a year you will have 110 pounds. Thus 100 pounds received today have a value of 110 pounds in a year s time, or conversely, 110 pounds received in a year s time have a present value of 100 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.

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 0 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 a)1 + r) a 1 = a i 1 + r) i ) 1 + r) a 1 = a 1 + a 2 1 + r) + a 3 1 + r) 2 + + a n 1 + r) n 1 a 1 = a 2 1 + r) + a 3 1 + r) 2 + + a n 1 + r) n 1. At the end of year 2 the amount in the bank is ) a2 1 + r) + a 3 1 + r) + + a n 1 + r) a 2 1 + r) n 1 2 = a 3 1 + r) + a 4 1 + r) 2 + + 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 2.13. You are offered three different jobs. The salary paid at the end of each year in thousands of pounds) is Job 1 2 3 4 5 A 32 34 36 38 40 B 36 36 35 35 35 C 40 36 34 32 30 Which job pays best if the nominal rate is r = 0.1, r = 0.2, and r = 0.3?

24 Solution. We shall compare the present values of the cash flow streams. Now, the present value for job A is P V A) = 32 1 + r + 34 1 + r) 2 + 36 1 + r) 3 + 38 1 + r) 4 + 40 1 + r) 5, while the present value for job B is P V B) = 36 1 + r + 36 1 + r) 2 + 35 1 + r) 3 + 35 1 + r) 4 + 35 1 + r) 5. Finally, the present value for job C is P V C) = 40 1 + r + 36 1 + r) 2 + 34 1 + r) 3 + 32 1 + r) 4 + 30 1 + r) 5. Taking the different interest rates into account we get the following present values. Thus, r PV A PV B PV C 0.1 135.03 134.41 132.14 0.2 105.51 106.20 105.50 0.3 85.20 86.61 86.83 if r = 10%, then A pays best, then B, then C; if r = 20%, then B pays best, then A, then C; if r = 30%, 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 2.14. Let b 1 and let n be a positive integer. Then i=0 b i = 1 + b + b 2 + + b n = 1 bn+1 1 b = bn+1 1 b 1. Proof. Let Then so Thus as claimed. x = 1 + b + b 2 + + b n. bx = b + b 2 + + 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.

25 Corollary 2.15. If b < 1, then Proof. Note that if b < 1, then Thus i=0 b i = lim n i=0 i=0 b i = 1 1 b. lim n bn = 0. b i = lim n 1 b n+1 1 b = 1 0 1 b = 1 1 b. Another corollary follows. Corollary 2.16. 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=0 = ra r b+1. 1 r We now return to the present value of cash flow streams that go forever. Example 2.17. 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 1 1 + r) i 1 1 + r) i i=0 c 1 = 1 + r) 1 1 + r) 1 ) c = 1 + r) 1 = c r.

26 Remark 2.18. 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. 2.3.1 An Example on transferring cash flows on the time line. We close this section with the following example. Example 2.19. Suppose that John is self-employed and wants to save for his retirement in 20 years. From the time of his retirement onwards he wants to withdraw 1,000 every month at the beginning of each month for 30 years. What amount of money does he have to save every month at the beginning of each month for the next 20 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 20 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 20. Set the monthly compounding factor as α = 1 + r/12 = 1.005. If A is deposited at the beginning of each month i, then the forward value of each such deposit by the end of year 20 will be Aα 240 i, for i = 0, 1, 2,..., 239. Hence by the end of the 20 year period there will be 240 monthly deposits 20 years) with the total value Aα 240 + Aα 239 +... + Aα 2 + Aα = Aα α240 1 α 1. Next, set the monthly discounting factor β = α 1, since it is the inverse of the monthly compounding. If 1,000 is withdrawn at the beginning of each month 240 + j, then the value of each such withdrawal at the beginning of month 241 beginning of year 21) is 1000β j, for j = 0, 1,..., 359. Hence, the total value of the 360 withdrawals 30 years) at the beginning of the 30 year period is 1000 + 1000β + 1000β 2 +... + 1000β 359 = 1000 1 β360 1 β. So, we equate the forward value of the deposits with the discounted value of all the withdrawal, to get A α α240 1 α 1 1 β360 = 1000 1 β and, taking into account that α/α 1) = 1/1 β), we obtain A = 1000 1 β360 α 240 1 = 360.9.

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 = 1 1 + 0.06 12 Then, since there are 240 months in 20 years, = 1 1.005. P V A = A + Aβ + Aβ 2 + + Aβ 239 = A1 + β + β 2 + + β 239 ) = A 1 β240 1 β. Let W = 1000 denote his monthly withdrawals in pounds. Since his first withdrawal will take place in 240 months time, and since there are 360 months in 30 years, the present value P V W of all his withdrawals is P V W = W β 240 + W β 241 + + W β 240+359 = W β 240 1 + β + + β 359 240 1 β360 ) = W β 1 β. Clearly, the present value of his deposits must equal the present value of his withdrawals. Thus so hence A 1 β240 1 β A = W β P V A = P V W, = W β240 1 β360 1 β, 240 1 β360 = 360.9, 1 β240 that is, he has to deposit 361 every month to fund his retirement. 2.3.2 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 100 pounds now, will cost 103 pounds next year. From a different point of view, the purchasing power of 103 pounds in one year is 100 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.

28 - Suppose you put P in the bank now. - Then after one year you will have 1 + r)p in the bank. - But the purchasing power of the amount 1 + r)p in one year from now, taking into account the inflation, is the same as the amount 1 + r)p 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 2.20. 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%. 2.3.3 Discount Factors We can write present value for all types of interest by the use of discount factors. Definition 2.21. The discount factor to T years, DT ), is the inverse of what an account holding P = 1 will grow to in time T. Remarks 2.22. 1. If r is the yearly compounded interest rate i.e. the one year rate), then the discount factor to T = n years is Dn) = 1 1 + 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 = ) 12) 1 + r 1 n. 12 3. 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

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 2.23. 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 2.24. 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 2.4.1 Definition Suppose you invest amount a and after one year you get back amount b. Recall from Section 3.1.1 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) = 0, for the function fr) = a + b i 1 + r) i. 2.4.2 Existence and uniqueness of the internal rate of return The definition above is meaningful, if the equation fr) = 0 has a unique solution 1 < r <. The following result proves this precise property.

30 Theorem 2.25. If a > 0 and b i > 0 for i = 1,..., n, then the equation fr) = 0 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 < 0. 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 < 0 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 < 0. 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. 2.4.3 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 2.26. Suppose that at the beginning of 2012 an investor bought shares of a certain company worth 1,000, which managed to sell again at the beginning of 2014 for 1,050. At the end of 2012 and 2013 the investor also received dividends of 10 and 20, 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 10% income tax on the dividends and 18% capital gains tax? Solution. a) The internal rate of return is a solution of the equation 1000 = 10 1 + r) + 20 1 + r) 2 + 1050 1 + r) 2 = 10 1 + r) + 1070 1 + r) 2.

31 By rewriting this, we get 10001 + r) 2 101 + r) 1070 = 0, which is a quadratic equation in 1 + r) with solutions This implies that the solutions r are 1.0394 and 1.0294. 0.0394 and 2.0294 reject, since r > 0). Thus the internal rate of return of this investment is r = 3.94%. b) Since the two dividend payments were taxed at 10%, the investor received only 10 1 = 9 at the end of 2012 and 20 2 = 18 at the end of 2013. As the investor made a profit of 1,050 1,000 = 50 after selling the shares, he had to pay 0.18 50 = 9 capital gains tax. This reduces the investor s final pay-off to 1050 9 = 1,041. Thus the internal rate of return in this case is a solution of the equation 1000 = 9 18 + 1041 +. 1 + r) 1 + r) 2 Again, this is a quadratic equation in r with solutions 0.0335 and 2.0245 reject, since r > 0). 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. 2.5 Immunisation of cash-flows A company with several cash-flows in the future is exposed to interest rate fluctuation. This chapter will explore how cash-flows can be structured in a way that minimises the exposure to interest rate variability. For the sake of simplicity we initially presume that the interest rate is continuously compounded. Therefore the discount factor to a date T is just DT ) = exp rt ). We will assume that the company has a sequence of positive future cash flows, called assets, whose net present value is V A r) and a sequence of negative future cash flows, called liabilities, with net present value V L. The net present value of these cash-flows is it will have the form V r) = V A r) V L r) V r) = C i e rt i where C i is a cash-flows transacted at time t i. The positive C i s will appear in V A whereas the negative ones come from V L r): V A r) = C i e rt i 0 and V L r) = C i )e rt i 0. C i >0 C i <0

32 2.5.1 The effective duration A treasurer of a corporation will be worried that V r) might vary wildly if interest rates happen to change. Question: How can we measure how much our PV, V r), can vary if the rates r vary? Answer: By looking at V r). For technical reasons explained below investors prefer to use the following slightly different quantity: Definition 2.27. The effective duration is defined as: note this is only defined if V r) 0). ν = V r) V r) Remarks 2.28. 1. If the PV, V r), is of the order of millions then we expect the derivative to be the same kind of units. This is the reason why in the previous definition we divide by V r); to scale things to a more sensible size. 2. Note that for positive cash-flows we expect V r) to be negative as when the rates increase the value of future cash diminishes. This is the reason why the definition above has a negative sign. This way the effective duration will be a positive number for positive cash flows, and negative for negative cash-flows. 3. The reason why effective duration is a duration is that for a single cash-flow it gives back the time to maturity. A different quantity called duration will be defined later on.) To see this consider a single cash flow paying C at time T, then its present value is V r) = C exp rt ) and the duration is: V r) V r) = T Ce rt Ce rt = T. 4. In the case of more than one cash-flow the effective duration as defined above can be interpreted as some form of average time to the several cash-flows. Take for example the case n = 2 then V r) = C 1 e rt 1 + C 2 e rt 2 and V r) = C 1 t 1 )e rt 1 + C 2 t 1 )e rt 2. The duration is therefore ν = C 1e rt 1 t 1 + C 2 e rt 2 t 2 C 1 e rt 1 + C2 e rt 2 which can be interpreted as a weighted average of t 1 and t 2 ν = ω 1t 1 + ω 2 t 2 ω 1 + ω 2 with weights ω 1 = C 1 exp rt 1 ) and ω 2 = C 2 exp rt 2 ). This average will be closer to t 1 if C t exp rt 1 ) is much larger than C 2 exp rt 2 ), and closer to t 2 if C 2 exp rt 2 ) is much larger. If C 1 exp rt 1 ) = C 2 exp rt 2 ) then is the usual average ν = t 1 +t 2 )/2. 5. The interpretation of the effective duration as some form of average time to cash flow is only valid if all cash-flows are positive. Otherwise ν can be negative or zero and cannot be interpreted as above.

33 Proposition 2.29. For a sequence of cash flows C 1,..., C n at times t 1,..., t n the duration is ν = C 1e rt 1 t 1 + + C n e rtn t n C 1 e rt 1 + + Cn e rtn In the case all cash flows are positive this can be interpreted as a weighted average of the times t 1,..., t n : ν = ω 1t 1 + + ω n t n ω 1 + + ω n weighted by the PV of each cash flow ω 1 = C 1 e rt 1,..., ω n = C n e rtn. Proof. The proof is very similar as the one for n = 2 discussed in Remarks 2.204). 2.5.2 Reddington Immunisation As discussed in the previous lesson the objective in this section is to explore how a treasurer can minimise the risk of interest rates moves unpredictably affecting the present value of a sequence of positive assets) and negative liability) future cash-flows. As in the previous subsections we write V r) for the PV of our cash flows: V r) = C i e rt i This can be split into the sum of terms with positive and negative terms, V r) = V A r) V L r): V A r) = C i e rt i 0 and V L r) = C i >0 C i )e rt i 0. We assume that the current interest rate is r = r 0 and wish to explore what happens when r moves to a different value r = r 0 + ε where ε is assumed to be a small positive or negative number. C i <0 Matching PV of assets and liabilities. Order zero immunization. A treasurer will attempt to match the PV of assets and liabilities, i.e. they will try to set up the payments and receipts in a way that their present value is the same. This is encapsulated in the following definition: Definition 2.30. A sequence of cash-flows is said to be zero order immune if the present value of assets and liabilities is the same: V A r 0 ) = V L r 0 ). Note that this can only be achieved for the current level of interest rates, r = r 0, because as soon as rates change, the present values of assets and liabilities will change to V A r 0 + ε) and V L r 0 + ε) which might no longer be equal.

34 Matching duration of assets and liabilities. First Order immunization. As stated above even if we have V A r 0 ) = V L r 0 ) as soon as rates change to r = r 0 + ε it will no longer be necessarily the case that V A r 0 + ε) = V L r + ε). We can use the Taylor expansion to approximate these quantities: V A r 0 + ε) V A r 0 ) + V Ar 0 )ε = V A r 0 ) V A r 0 )ν A ε = V A r 0 ) 1 ν A ε) 5) V L r 0 + ε) V L r 0 ) + V Lr 0 )ε = V L r 0 ) V L r 0 )ν L ε = V L r 0 ) 1 ν L ε) 6) where ν A = V A r 0)/V A r 0 ) and ν L = V L r 0)/V L r 0 ) indicate the effective durations of the assets and liabilities. Definition 2.31. A sequence of cash-flows is said to be first order immune if it is immune to zero order i.e. V A r 0 ) = V L r 0 )) and the duration of assets and liabilities coincide: ν A = ν L. In this case by the equations above the net present values of assets and liabilities continue to be similar after a small change in the interest rates. Remarks 2.32. 1. For a sequence of cash-flows that are zero order immune, we have that V r 0 ) = V A r 0 ) V L r 0 ) = 0. If it is first order immune then as rates move by a small amount the two quantities in 5) and 6) are the same so that V r 0 + ε) = V A r 0 + ε) V L r 0 + ε) 0, i.e. the PV does not vary. 2. The approximation above consists of disregarding the Taylor terms 1 2 V A r 0)ε 2, 1 2 V L r 0)ε 2 and terms with higher powers of ε. Typically these will be small. For example for ε = 0.1% = 0.001 we have ε 2 = 0.000001. Convexity. Reddington immunisation One can refine the considerations above by introducing another measure of risk called the convexity. Definition 2.33. If the PV of a sequence of cash-flows, V r), is different from zero, V r) 0, we define its convexity as: c = V r) V r). Remark 2.34. If our cash-flows are C 1,..., C n to be transacted at time t 1,..., t n then the convexity is c = C 1e rt 1 t 2 1 + C n e rtn t 2 n C 1 e rt 1 + Cn e rtn and so, if the cash-flows are positive can be interpreted as some form of weighted average of t 2 1,..., t 2 n. This does not have an immediate financial meaning. The motivation for its definition is that it appears in by equations 7) and 8) below.

35 We can use the convexity to refine our previous approximations 5) and 6) as follows: V A r 0 + ε) V A r 0 ) + V Ar 0 )ε + 1 2 V Ar 0 )ε 2 = V A r 0 ) V A r 0 )ν A ε + 1 2 V Ar 0 )c A ε 2 = = V A r 0 ) 1 ν A ε + 12 ) c Aε 2 7) V L r 0 + ε) V L r 0 ) + V Lr 0 )ε + 1 2 V L r 0 )ε 2 = V L r 0 ) V L r 0 )ν L ε + 1 2 V Lr 0 )c L ε 2 = = V L r 0 ) 1 ν L ε + 12 ) c Lε 2 8) where c A = V A r 0)/V A r 0 ) and c L = V L r 0)/V L r 0 ) indicate respectively the convexity of the assets and the liabilities. We could demand that our cash-flows on top of being first order immune further verify that c A = c L so that the approximation V A r 0 + ε) = V L r 0 + ε) is even more accurate. In practice it is often sufficient to demand that c A c L which means that V r 0 + ε) = V A r 0 + ε) V L r 0 + ε) V A r 0 ) 1 ν A ε + 12 ) c Aε 2 = V A r 0 ) 1 2 c A c L )ε 2 V L r 0 ) 1 ν L ε + 12 ) c Lε 2 = which is greater than V r 0 ) = 0. This means that any changes in interest rates result in a gain. Definition 2.35. Reddington immunisation consists on arranging the cash-flows in a way that: 1. The present value of the asset cash flow equals the present value of the liabilities cash flow at the present interest rate r 0 : V A r 0 ) = V L r 0 ). 2. The duration of the asset cash-flow is the same as that of the liabilities: ν A r 0 ) = ν L r 0 ). 3. The convexity of the asset cash-flow is larger than the convexity of the liabilities cash-flow: c A c L. Remark 2.36. Engineering the financial cash-flow and commitments so as to minimise exposure to adverse moves is generally called risk management. The action of entering into transactions of the purpose of managing the risk is sometimes referred to as hedging.

36 2.5.3 Immunisation with compound interest Suppose now that interest is copounded with effective rate r. Consider a cash-flow with payments C i made at times t i, for i = 1,..., n. The present value of the cash-flow is V r) = C i β t i where β = 1 1 + r. Note that and therefore and ) ti 1 [ d dr βt i 1 = t i 1 ] = t 1 + r 1 + r) 2 i β t i+1 V r) = V r) = t i C i β t i+1 t i t i + 1)C i β t i+2 A t-year zero-coupon bond is an asset paying a fixed amount in t years. Bonds will be discussed in more detail later.) Example 2.37. A fund must make paments of 50, 000 at the end of the sixth and eigth year from now. Show that immunisation can be acheived for 7% interest with a comination of 5-year zero-coupon bonds and 10-year zero-coupond bonds. Solution. Suppose we purchase 5-year bonds paying P and 10-year bonds paying Q. Then, with β = 1.07) 1, V A 0.07) = P β 5 + Qβ 10. We also have and so first order immunisation implies that The effective duration of the assets is while the effective duration fo the liabilities is Setting ν A = ν L gives Solving 9) and 10) for P and Q gives V L 0.07) = 50, 000β 6 + β 8 ) = 62, 418 P β 5 + Qβ 10 = 62, 418. 9) ν A = 5P β6 + 10Qβ 11 62418 ν L = 50, 0006β7 + 8β 9 ) 62418 = 404398 62418. 5P β 6 + 10Qβ 11 = 404398. 10) P = 53, 710, Q = 47, 454. 11)

37 We must still show that the third immunisation condition is satisfied. The convexity of the assets is c A = V A 0.07) V A 0.07) = 30P β7 + 110Qβ 12 = 53.21 62418 while the convexity of the liabilities is c L = V L 0.07) V L 0.07) = 5000042β8 + 72β 10 ) 62418 = 48.90. Since c A C L, the third condition is satisfied and 11) gives immunisation. Instead of the effective duration, we can also work with a related quantity. Definition 2.38. The duration is defined as: τ = n t ic i β t i n C. iβ t i We see that τ is a weighted average of the t i and that τ = ν/β, so that setting τ A = τ L is equivalent to setting ν A = ν L. Example 2.39. An investor has a liability of 20, 000 to be paid in 3.5 years and another of 18, 000 to be paid in 6 years. Immunize these liabilities at 10% per annum using 4-year zero-coupon bonds and 7-year zero-coupon bonds. Solution. Let X be the present value of the 4-year bonds and Y be the present value of the 7-year bonds. The first order immunisation gives Setting τ A = τ B gives X + Y = 20000β 3.5 + 18000β 6 = 24488. 4X + 7Y = 3.5 20000β 3.5 + 6 18000β 6 = 111, 108. Solving the two previous equations for X and Y produces X = 20, 103 Y = 4385. We next calculate the convexities. We find that and c A = 4)5)Xβ2 + 7)8)Y β 2 24488 = 21.86 c L = 3.5)4.5)20000)β5.5 + 6)7)18000)β 8 = 22.02 24488 We see that c L > c A and so immunisation is not possible.