Notes on the text of Math 4065 Text: The Theory of Interest, Stephen G.~Kellison, 2nd Edition *********************************************************************** Look at the "Principal repaid" column of Table 6.2 on p. 171. Here the effective rate of interest annually is 8%. Note that each entry in that column is 1.08 times the preceding entry. Look at the "Amount for amortization of premium" column of Table 7.2 on p. 217. Here the effective rate of interest semiannually is 3%. (Same as 6% convertible semiannually.) Note that each entry in that column is 1.03 times the preceding entry. Look at the "Amount for accumulation of discount" column of Table 7.3 on p. 218. Here the effective rate of interest semiannually is 5%. (Same as 10% convertible semiannually.) Note that each entry in that column is 1.05 times the preceding entry. Look at column labeled "D_t" in the table appearing in Example 8.9 on p. 275. Here the effective rate of interest on the sinking fund is 5%. Note that each entry in the column is 1.05 times the preceding entry. Here is an explanation for the phenomenon in Table 7.2, and there are similar explanations for the other tables. By the premium/discount formula (formula 7.2, p. 212), we have that the book value of the bond after the t'th coupon is C+(Fr-Ci)a_{n-t}. Thus the change in the book value over the t'th coupon period is [ C + (Fr-Ci)(a_{n-t+1}) ] - [ C + (Fr-Ci)(a_{n-t)})]. This is (Fr-Ci)(\nu^{n-t+1}). This is what appears in the t'th entry of the column labeled "Amount for amortization of premium". Thus each entry is (1/\nu) times the preceding entry. Since i=.03, we get \nu=1/1.03, so 1/\nu=1.03. As a result of these observations, whenever you know one of the entries in a column that gives principal adjustment of a loan during each payment period, you can find out all the other entries, as long as you know the effective rate of interest per payment period. *********************************************************************** Explanation of Equation 8.19 on p. 271. Assume that you borrow A at time 0 at an interest rate of i from someone who demands payment of the interest iA at times 1,2,...,n and who demands that you pay back the loan amount A at time n. You go out and use the borrowed A to invest in some equipment which produces returns of R at times 1,2,...,n. Each time you receive a return of R, after paying off iA, you can invest the remainder in a sinking fund earning a rate j. Then, at time n, you sell the equipment at a price of S. You must also have accumulated and amount of A-S in the sinking fund. That way, you can pay off the loan amount of A at time n, by combining S with A-S. Then R-Ai is the amount going into the sinking fund each year, and it must accumulate to A-S. Thus we have (R-iA)(s_{n,j})=A-S. Solving for R gives Equation 8.19. *********************************************************************** p. 134, from line 13 from the bottom to line 10 from the bottom "Actually, yield rates are unique under a broader set of conditions than given above. It is possible to show that if the outstanding investment balance is positive at all points througout the period of investment, then the yield rate will be unique." What is being said is that if there is an interest rate i_0 for which B_0,...,B_{n-1} are all positive and for which B_n=0, then it is a yield rate (because B_n=0) and there is no other yield rate that is >= -1. Proof: We follow the notation of Section 5.3, where the costs at times 0,...,n are C_0,...,C_n, respectively. Define B_0(x)=C_0, so B_0 is a constant function of x. Define B_t(x) = x [B_{t-1}(x)] + C_t, for t=1,2,...,n. Then B_t(1+i) is the outstanding investment balance at time t, assuming interest rate i. We are assuming that B_0(1+i_0) > 0, ... , B_{n-1}(1+i_0) > 0 and that B_n(1+i_0) = 0. We wish to show, for any i >= -1, that: if B_n(1+i) = 0, then i=i_0. Let x_0 := 1 + i_0. We are assuming that B_0(x_0) > 0, ... , B_{n-1}(x_0) > 0 and that B_n(x_0) = 0. We wish to show, for any x >= 0, that: if B_n(x) = 0, then x = x_0. Let a_0 := B_0(x_0), ... , a_{n-1}:= B_{n-1}(x_0). Then a_0 > 0, ..., a_{n-1} > 0. Let f(x) := a_0 x^{n-1} + a_1 x^{n-2} + ... + a_{n-2} x + a_{n-1}. Then f(x) > 0, for all x >= 0. We calculate that B_n(x) = C_0 x^n + ... + C_n. We have B_n(x_0)=0, and, after factoring out x-x_0, we calculate that B_n(x) = [x-x_0][ f(x) ]. So fix x >= 0 and assume that B_n(x) = 0. We wish to show that x = x_0. We have 0 = B_n(x) = [x-x_0][ f(x) ]. We also have f(x) > 0. Then 0 = [B_n(x)] / [f(x)] = x-x_0. Then x = x_0, as desired. QED. *********************************************************************** Explanation of equation 3.6 on p. 61: The total of 1 at time 0, i at times 1,2,...,n and -1 at time n is worth zero, since one can imagine depositing 1 at time 0, withdrawing the interest at the end of each year and withdrawing 1 at time n. The 1 at time 0 is worth 1/a_n at times 1,...,n. The -1 at time n is worht -1/s_n at times 1,...,n. Therefore a payment of i + (1/a_n) - (1/s_n) at times 1,...,n is worth zero. *********************************************************************** Even though Problem 15 in Chapter 5 is technically associated to Section 5.5, there's no reason that the students can't handle it using information from sections 5.1-5.4. In 2000, I canceled this problem, but in future years, I think I'll let it go. *********************************************************************** There is a much better proof of the uniqueness result for yield rate, exposed (poorly) in the text in Section 5.3, pp. 133-136. Assume, as in the text, that R_0,...,R_k <= 0 and that R_{k+1},...,R_n >= 0. Then C_0,...,C_k >=0, since we have C_i=-R_i. Let f(x) = C_0 x^{-k} + C_1 x^{-k+1} + ... + C_k and let g(x) = R_{k+1} x + R_{k+2} x^2 + ... + R_n x^{n-k}. Then f is decreasing, g is increasing. We also have f(0)=+\infty, f(+\infty)=C_k, g(0)=0 and g(+\infty)=+\infty, where three of these four statements are actually limit statements. As a consequence, there is a unique \nu>0 such that f(\nu)=g(\nu). The equation f(\nu)=g(\nu) is equivalent to R_0 + R_1 \nu + R_2 \nu^2 + ... + R_n \nu^n = 0, so there is a unique \nu that satisfies the equation for the yield rate. That is, there is a unique yield rate. *********************************************************************** In Chapter 5, Problem 2 on p. 160, it should state that C_0=$3000, that R_1=$1000 and that R_2=$4000. As written, it states that C_1=$1000 and that R_1=$2000, which (in the notation introduced earlier in the text) is contradictory, since we should have C_i=-R_i, for all i. *********************************************************************** Suppose you have a series of payments, paying a_t at time t, with t ranging over s,s+1,s+2,...,u. Define a_t=0 for tu. Let V be the value of this series of payments at time r. Let i be the effective interest rate over 1 unit of time and let d be the effective discount rate over 1 unit of time. (We are assuming compound interest.) For all t, let b_t=(1/i)(-a_t+a_{t+1}) and let c_t:=(1/d)(a_t-a_{t-1}). Then V is the value at time r of the series of payments given by {b_t} and V is also the value at time r of the series of payments given by {c_t}. The result about b_t follows from the fact that 1/i at time 0 is worth a perpetuity immediate of 1 at times 1,2,3,.... The result about c_t follows from the fact that 1/d at time 0 is worth a perpetuity due of 1 at times 0,1,2,3,.... These results can be used to prove easily that a_n=[1-\nu^n]/i, that s_n=[(1+i)^n-1]/i, that \"a_n=[1-\nu^n]/d and that \"s_n=[(1+i)^n-1]/d. They also give easy proofs of the formulas for (Ia)_n, (Is)_n, (Da)_n, (Ds)_n, etc. They can also be used to work out formulas for annuites where payments are more or less frequent than interest conversions. Just keep in mind the basic rule: If i is the effective interest rate over one conversion period, then i^{(m)}/m is the effective interest rate over 1/m conversion periods and i s_k is the effective interest rate over k conversion periods. *********************************************************************** If you have a annuity which pays 1 at time t, 1 at time t+1, and so on, until 1 at time u, be aware of the following: There are n:=u-t+1 payments. The value at time t-1 is a_n. The value at time t is \"a_n. The value at time u is s_n. The value at time u+1 is \"s_n. Present values are computed either at t-1 or at time t. The earlier one is t-1, and that one is an annuity immediate symbol, a_n. The later one is t, and that one one is an annuity due symbol, \"a_n. Accumulated values are computed either at u or at u+1. The earlier one is u, and that one is an annuity immediate symbol, s_n. The later one is u+1, and that one one is an annuity due symbol, \"s_n. Thus it is best to think of "immediate" and "due" as refering to the time at which the value of the annuity is computed. In each case (present value or accumulated value), there are two choices and the earlier one is called "immediate", while the later one is called "due". Present values are denoted by a, accumulated values are denoted by s. Due is indicated by a double dot (denoted here by \"). Stay away from using several different "clocks", since that causes confusion. By using the system described above, one does not need to think of "present value" as referring to time 0, and this alleviates the need for having many different systems of time, i.e., many different clocks. *********************************************************************** Present values have 1 - \nu^n in the numerator. Accumulated values have (1+i)^n - 1 in the numerator. Annuities immediate have an i in the denominator. Annuities due have an d in the denominator. Thus the formulas are a_n = [1 - \nu^n] / i \"a_n = [1 - \nu^n] / d s_n = [(1+i)^n - 1] / i \"s_n = [(1+i)^n - 1] / d. Thus, keep in mind that "due" corresponds to "double dot" (which means \"), and also corresponds to a "d" in the denominator. Also, "immediate" corresponds to an "i" in the denominator. *********************************************************************** The last word of Example 4.1 on p. 96, which appears on the 16th line from the bottom of that page, should be changed from "quarterly" to "monthly". *********************************************************************** On p. 15, line 17 to line 18: "Consider a situation in which the amount of discount earned during each period is constant." This is misleading at best. On p. 13, the "amount of discount" is defined to be I_n = A(n)-A(n-1). In simple discount, it is not true that I_n is constant (i.e., independent of n). What is true instead is that 1/(A(n)) is linear. That is, the rate of change in the discount function is constant. That is, if we define J_n := [1/(A(n))]-[1/(A(n-1))], then J_n is constant (i.e., independent of n). So we need to redefine the "amount of discount" to be J_n rather than A_n in order for the statement to be true. *********************************************************************** In the problems for Chapter 2, #17 on p. 55 has a typo: Instead of T_1-T_2, they should say T_2-T_1. In the problems for Chapter 3, #35 on p. 91, one needs the omitted fact that the withdrawals are $1000 per year to get the answer in the book. *********************************************************************** Example 3.7 on p. 74 of the text shows that there is sometimes no good answer to an unknown time question. In this problem we are to accumulate $25,000 in a bank account earning 8% effectively. We pay $1000 at the end of every year with a final payment that is less than $1000 and is made one year after the last regular payment. The difficulty is that, the amount accumulated in the account immediately after the 14th payment is $1000 s_{14} = $24,215, and the amount in the account immediately before the 15th payment would be ($24,215)(1.08)=$26,152. This means we can either pay $785 (which is $25,000-24,215) at the moment of the 14th regular payment or make the 14th regular payment, wait until the end of the 15th year, and take out $1,152 (which is the $26,152-25,000) at that time. Under the first choice, our irregular payment would be $1000+785=$1,785, which is more than $1000, and it is made at the end of the 14th year. Under the second choice, our irregular payment would be -$1,152, which is negative, and is made at the end of the 15th year.