Essential Topic: Continuous cash flows



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

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

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.

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

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

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.

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 = 74.69 e.1+.25

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 = 29.39 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).

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

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

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) = 24.816 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) = 55.239

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