7. Continuously Varying Interest Rates 7.1 The Continuous Varying Interest Rate Formula. Suppose that interest is continuously compounded with a rate which is changing in time. Let the present time be time, and let r(s denote the interest rate per unit at time s units, s. The quantity r(s is called the spot or the instantaneous interest rate at time s. Let D(t be the amount that you will have in your account at time t if you deposit P at time. In order to determine D(t in terms of the interest rates r(s, s t, note that by the Simple Interest Formula 1.8, for small h, we have D(s + h D(s(1 + r(s h,
( means is approximately equal to D(s + h D(s + D(s r(s h, D(s + h D(s D(s r(s h, D(s + h D(s h D(s r(s. By the definition of the derivative 4.7, taking the limit as h, we have or D (s = D(s r(s D (s D(s = r(s. To solve this differential equation, integrate both sides: Thus D (s D(s ds = [ln D(s] t =.
or ln D(t ln D( = r(sds. Since D( = P, we obtain from the preceding equation that ln D(t ln P = exp (ln D(t ln P = exp ( exp (ln D(t exp ( ln P 1 = exp D(t P 1 = exp and D(t = P exp ( ( (.
7.2 Example. Suppose that interest is continuously compounded with a rate which is changing in time. Let the present time be time, and let r(s =.1.5 1 1 + s be the interest rate per unit at time s units, s. (i Draw a graph of the function r(s. (ii Find the amount D(t that you will have in your account at time t, t, if you deposit 1, at time. (iii Suppose the unit is equal to one year. How much money do you have in your account after 1 years?
Solution. (i The graph of the function y 1 (s =.5 1 1+s is The graph of the function y 2 (s =.5 1 1+s is The graph of the function r(s =.1.5 1 1+s is
(ii By the Continuous Varying Interest Rate Formula 7.1, Note that D(t = 1, exp r(sds = (.1.5 1 1 + s ds (by the linearity of the integrals Hence =.1ds.5 1 1 + s ds =.1 t.5 [ln 1 + s ] t =.1 t.5 (ln 1 + t ln 1 =.1 t.5 ln 1 + t. D(t = 1, exp (.1t.5 ln 1 + t ( D(t = 1, e.1t+ln 1 +t.5 = 1, e.1t eln 1 +t.5.
= 1, e.1t 1 + t.5. (iii For t = 1, D(1 = 1, e.1 1 (1 + 1.5 = 1, e 11 (Recall that e 2.71828182..5 = 1, 2.71828182.8871378 = 2, 411.15. 7.3 The Continuous Compounded Interest Formula. The Continuous Compounded Interest Formula 4.1 is a special case of the Continuous Varying Interest Rate Formula 7.1, arising when the interest rate r(s, s, is always equal to r 1. Let a principal P be borrowed at an annual rate of interest r% compounded continuously for t years. Then, according to 7.1, for D( = P and r(s = r 1, s, the amount A owed on the principal P at time t is A = D(t = P exp (
( = P exp = P exp r 1 ds ( r 1 t = P e r 1 t. 7.4 The Present Value. The Continuous Varying Interest Rate Formula 7.1 states that a principal P earning a continuously compounded interest r(s per unit at time s units will be worth D(t = P exp at time t. Therefore [ P = D(t exp = D(t exp ( ( ] 1 [ ( ] r(sds.
Definition. The present value P (t corresponding to an amount A and to a variable interest rate r(s, s, is given by [ ( P (t = A exp r(sds, t, and is equal to the amount to be invested at time to accumulate A at time t by continuous compounding at the interest rate r(s, s. ] 7.5 Definition. The function [ ( P (t = exp r(sds, t, which is at each point t equal to the present value P (t corresponding to the amount 1 and to a variable interest rate r(s, s, is called the present value function. ] In other words, the present value function at t is equal to the amount to be invested at time to accumulate 1 at time t by continuous compounding at the interest rate r(s, s.
7.6 Example. How much money should be invested at time at a continuously compounded interest rate r(s =.1.5 1 1 + s per unit at time s units, s, in order to accumulate 5,, at time t? Solution. By 7.4, the present value P (t of 5,, is P (t = 5,, exp [ ( ]. In Example 7.2 we showed that r(sds = 1.1.5 1 + s ds =.1 t.5 ln 1 + t. Hence the present value P (t is 5,, exp [ (.1 t.5 ln 1 + t ]
= 5,, exp [.1 t +.5 ln 1 + t ] = 5,, exp (.1 t exp ( ln 1 + t.5 = 5,, exp (.1 t 1 + t.5. Suppose t = 1: then P (1 = 5,, exp (.1 1 1 +1.5 = 5,, 1 e 11.5 = 2, 73, 696.33. 7.7 Definition. The average of the spot interest rates up to time t r(t = 1 t is called the yield curve. r(sds, t,
7.8 Example. Find the yield curve and the present value function if r(s = r 1 1 1 + s + r 2 s 1 + s. Solution. Note that we can rewrite r(s as r(s = r 1 + r 2 s 1 + s = r 1 r 2 + r 2 (1 + s 1 + s = r 1 r 2 1 + s + r 2. Hence, for t, r(sds = r 1 r 2 1 + s + r 2ds = r 1 r 2 1 + s ds + r 2ds = (r 1 r 2 1 1 + s ds + r 2 ds
= (r 1 r 2 [ln 1 + s ] t + r 2[s] t = (r 1 r 2 ln(1 + t + r 2 t. Therefore, for t, the yield curve is r(t = 1 t = 1 t [(r 1 r 2 ln(1 + t + r 2 t] = r 1 r 2 t ln(1 + t + r 2. Consequently, for t, the present value function is [ ( ] P (t = exp r(sds = exp [ ((r 1 r 2 ln(1 + t + r 2 t] = exp ( ln[(1 + t (r 2 r 1 ] exp ( r 2 t = (1 + t (r 2 r 1 exp ( r 2 t.