Essential Topic: Continuous cash flows

Transcription

1 Essential Topic: Continuous cash flows Chapters 2 and 3 The Mathematics of Finance: A Deterministic Approach by S. J. Garrett

2 CONTENTS PAGE MATERIAL Continuous payment streams Example Continuously paid annuities Example SUMMARY

3 CONTINUOUS PAYMENT STREAMS Continuously paid cash flows are an important theoretical construction. Although they do not appear in practice, a frequently paid discrete cash flow can be approximated by a continuous payment stream over the long term. For example, consider a pension paid weekly for an extended time period, say, 2 years. The natural time unit here is the week and we should use nominal rates i (52) (t) per annum in the present-value calculation involving 2 52 = 14 payments. Recalling that (t) = lim i (p) (t) we see that i (52) and the importance of the force of interest in the approximation of very frequently paid cash flows is clear.

4 CONTINUOUS PAYMENT STREAMS We define the rate of payment of a continuously paid cash flow stream, ρ(t), such that ρ(t) = dm(t) dt for all t where M(t) is the total payment between time and t. Between times t and t + dt with dt, the total payment is therefore such that lim {M(t + dt) M(t)} = ρ(t)dt. dt

5 PRESENT VALUE AND ACCUMULATION We consider the payment stream as a collection of payment elements, ρ(t)dt, at each time t within the cash flow. In its general form, each element has present value [ ν t ρ(t)dt = ρ(t) exp t ] (s)ds dt We then integrate (i.e. sum) these to give the total present value of the stream [ t ] ν t ρ(t)dt = ρ(t) exp (s)ds dt The equivalent expression for the accumulation of the stream to time t = n is n [ n ] ρ(t) exp (s)ds dt t

6 EXAMPLE If the force of interest at time t is given by {.2 for t < 5 (t) =.2 (t 5) for t 5 calculate a.) the value at time of 1 due at time t = 1, b.) the accumulated value at time t = 2 of a payment stream of rate ρ(t) = 1.5t paid continuously between t = 8 and t = 1.

7 EXAMPLE a.) In this case we have a single payment and should evaluate the present value by breaking the period into two subintervals We have 1 A(, 1) = 1 A(, 5) A(5, 1) A(, 5) = exp (.2 5) = e.1 ( 1 ) A(5, 1) = exp.2 (t 5)dt = e.25 5 Therefore 1 A(, 1) = 1 = e.1+.25

8 EXAMPLE b.) Here we need to consider the payment stream, ρ(t) = 1.5t. Similar arguments to those above can be used to construct the accumulation as 1 ( 1 ) ρ(t) exp (s)ds dt 8 1 = 8 1.5t exp t ( 1 t ).2(s 5)ds dt = Note that here the piecewise nature of (t) does not matter as the integrals are constrained within 8 t 1 where (t) =.2 (t 5).

9 CONTINUOUSLY PAID ANNUITIES Consider the particular case that (t) = and the payment stream is ρ(t) = 1 for t n. This payment stream represents a continuously paid, unit n-year annuity. In order to construct an expression for the present value of the annuity we follow the procedure above n ( 1 exp t ) ds dt = 1 e n It is usual to denote the present value of this annuity as ā n = 1 νn

10 ACCUMULATED VALUE The accumulated value at time n of the continuously paid annuity, s n, can be calculated in a number of ways. One approach is to construct the accumulation of the payment stream as above n ( n ) s n = 1 exp ds dt = (1 + i)n 1 t Alternatively, one can simply accumulate the present value at time t = to time t = n s n = (1 + i) n ā n = (1 + i)n 1

11 EXAMPLE If the effective rate of interest is i = 4% per annum, calculate the following quantities. a.) 3ā 1 b.) 1 s 5 Answers a.) This is the present value of a continuous annuity paying 3 per annum for 1 years, 3ā 1 = 3 1 ν1 = 3 1 (1.4) 1 ln(1.4) = b.) This is the accumulated value at t = 5 of a continuous annuity paying 1 per annum for 5 years, 1 s 5 = 1 (1 + i)5 1 = 1 (1.4)5 1 ln(1.4) =

12 SUMMARY The present value and accumulation of a continuously paid cash flow can be calculated by constructing integral expressions in terms of ρ(t) and (t). A continuously paid unit n-year annuity is the particular case that ρ(t) = 1 for t n years. If (t) =, we denote the present value of this stream as ā n = 1 νn and the accumulated value at time t = n as s n = (1 + i)n 1