Sections 4.5, 4.6 An annuity where the frequency of payment becomes infinite (i.e., continuous) is of considerable theoretical and analytical significance. Formulas corresponding to such annuities are useful as approximations corresponding to annuities payable with great frequency (e.g., daily). The present value of an annuity payable continuously for n interest conversion periods so that 1 is the total amount paid during each interest conversion period is n n _ a – n| vt —— = ln(v) vn – 1 —— = ln(v) vn – 1 —————— = ln(1) – ln(1+i) 1 – vn ——  = vt dt = (The formula could could also be derived by taking the limit as m goes to  of either or .) (m) a – n| ..(m) a – n| Note that certain values of this first factor appear in the interest tables. _ a – n| i —  Observe that = a – n|

The accumulated value of a continuous annuity at the end of n interest conversion periods so that 1 is the total amount paid during each interest conversion period is n n _ s – n| (1 + i)t ——— = ln(1 + i) (1 + i)n – 1 ———— = ln(1 + i) (1 + i)n – 1 ————  = (1 + i)t dt = (The formula could could also be derived by taking the limit as m goes to  of either or .) (m) s – n| ..(m) s – n| _ s – 1| Note that this first factor is . _ s – n| i —  Observe that = s – n| _ a – n| 1 – e–n ———  _ s – n| en – 1 ———  Observe that = and =

Find the force of interest at which the accumulated value of a continuous payment of 1 every year for 8 years will be equal to four times the accumulated value of a continuous payment of 1 every year for four years. _ s – 8| _ s – 4| = 4 e8 – 1 ——— =  e4 – 1 4 ———  e8 – 1 = 4e4 – 4 e8 – 4e4 + 3 = 0 (e4 – 3)(e4 – 1) = 0 e4 = 3   = ln(3) / 4  0.275 e4 = 1   = 0 which is not a possibility.

Consider an annuity-immediate with a term of n periods where the interest rate is i per period, and where the first payment is P (>0) with successive payments each having Q (possibly negative in certain situations) added to the previous payment, i.e., the payments form an arithmetic progression. Payments P P+Q P+2Q P+(n–2)Q P+(n–1)Q Periods 1 2 3 … n – 1 n If A is the present value for this annuity, then A = Pv + (P + Q)v2 + (P + 2Q)v3 + … + [P + (n – 2)Q]vn–1 + [P + (n – 1)Q]vn (1 + i)A = P + (P + Q)v + (P + 2Q)v2 + … + [P + (n – 2)Q]vn–2 + [P + (n – 1)Q]vn –1

A = Pv + (P + Q)v2 + (P + 2Q)v3 + … + [P + (n – 2)Q]vn–1 + [P + (n – 1)Q]vn (1 + i)A = P + (P + Q)v + (P + 2Q)v2 + … + [P + (n – 2)Q]vn–2 + [P + (n – 1)Q]vn –1 Subtracting the first equation from the second, we have iA = P – Pvn + Q(v + v2 + v3 + … + vn–1) – (n – 1)Qvn iA = P(1 – vn)+ Q(v + v2 + v3 + … + vn–1 + vn) – Qnvn 1 – vn Qnvn – nvn A = P ——– + Q ——— – —— = P + Q ————— i i i i a – n| a – n| a – n| The accumulated value of this annuity over the n periods is – n P + Q ———— i s – n| A(1 + i)n = s – n|

If P = Q = 1, then the annuity is called an increasing annuity, and the present value of such an annuity is .. a – n| – nvn ————— = i 1 – vn + – nvn ———————— = i (Ia) – n| – nvn ————— i = a – n| + a – n| a – n| The accumulated value of this annuity over the n periods is .. s – n| s––– n+1| i – (n + 1) ——————— i – n ———— i (Is) – n| (Ia) – n| = (1 + i)n = = (Ia) – n| Observe that can be interpreted as a sum of level deferred annuities after realizing that .. a – n| n–1  t=0 n–1  t=0 1 – vn–t ——– = i – nvn ————— = i (Ia) – n| vt = vt a ––– n–t| .. a – n| (Ia) – n| Writing the last equation as suggests the following verbal interpretation: The present value of an investment of 1 at the beginning of each period for n periods is equal to the sum of the interest to be earned and the present value of the return of the principal. i + nvn =

If P = n and Q = –1, then the annuity is called a decreasing annuity, and the present value of such an annuity is – nvn ————— = i n – nvn – + nvn ————————— = i a – n| n – ——— i (Da) – n| = n a – n| a – n| – a – n| The accumulated value of this annuity over the n periods is s – n| n(1 + i)n – —————— i (Ds) – n| (Da) – n| = (1 + i)n = (Da) – n| Observe that can be interpreted as a sum of level annuities after realizing observing that n  t=1 n  t=1 a – n| 1 – vt ——– = i n – ——— = i = (Da) – n| a – t| Note: changing i to d in the denominator of any of the formulas derived for an annuity-immediate with payments in an arithmetic progression will give a corresponding formula for an annuity-due.

Consider a perpetuity with a payments that form an arithmetic progression (and of course P > 0 and Q > 0). The present value for such a perpetuity with the first payment at the end of the first period is – nvn P + Q ————— = i (Ia) ––  | a – n| = lim n a – n| – nvn P + Q ————————— = i lim n lim n a – n| lim n a – n| 1 — i – 1 — i ———— = i P Q — + — i i2 P + Q

When working with varying annuities, it can be useful to make use of the following quantities: Fn = Gn = Hn = the present value of a payment of 1 at the end of n periods = vn the present value of a level perpetuity of 1 per period with the first payment at the end of n periods = 1 vn —— = 1 – v vn — d vn + vn+1 + vn+2 + … = vn(1 + v + v2 + v3 + …) = the present value of an increasing perpetuity of 1, 2, 3, …, with the first payment at the end of n periods = vn + 2vn+1 + 3vn+2 + … = vn(1 + 2v + 3v2 + 4v3 + …) = d vn — (1 + v + v2 + v3 + …) = dv d 1 vn — —— = dv 1 – v 1 vn ——— = (1 – v)2 vn — d2 Example 4.11 in the textbook uses these quantities to re-derive the formulas for and . (Ia) – n| (Da) – n|

Find the present value of a perpetuity-immediate whose successive payments are 1, 2, 3, 4, … at an effective rate of 6%. 1 1 —— + ——— = 0.06 (0.06)2 $294.44 We can find this either from 1 / 1.06 ————— = (0.06 / 1.06)2 $294.44 or from H1 =

Find the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of n, and then decrease by 1 each period up a final payment of 1. .. a – n| a ––– n–1| – nvn ————— i (n – 1) – ——————— i (Ia) – n| + vn (Da) ––– n–1| = + vn = .. a – n| .. a – n| – nvn + nvn – vn – vn ———————————————— = i a ––– n–1| – vn (1 + ) ——————————— = i a ––– n–1| .. a – n| .. a – n| – vn ——————— = i .. a – n| 1 – vn ——— = i .. a – n| a – n| In one of the homework exercises, you are asked to give a verbal interpretation of this formula.

Find the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of n, and then decrease by 1 each period up a final payment of 1. (an alternative derivation) The present value of the annuity can be written as H1 – 2Hn+1 + H2n+1 . H1 1 2 … n – 1 n n + 1 … 2n – 1 2n 2n + 1 2n + 2 Hn+1 1 … n – 1 n n + 1 n + 2 H2n+1 1 2 1 2 … n – 1 n n – 1 … 1 1 2 … n – 1 n n + 1 … 2n – 1 2n 2n + 1 2n + 2 .. a – n| An algebraic proof that H1 – 2Hn+1 + H2n+1 = is given in Example 4.13 in the textbook. a – n| Look at Example 4.14 (page 116) in the textbook.

Consider an annuity-immediate with a term of n periods where the interest rate is i per period, and where the first payment is 1 with successive payments each having 1 + k multiplied by the previous payment, i.e., the payments form a geometric progression. The present value for this annuity is v + (1 + k)v2 + (1 + k)2v3 + … + (1 + k)n–1vn = n n 1 + k —— 1 + i 1 + k —— 1 + i 1 – ————— = 1 – ————— i – k 1 —— 1 + i if k  i 1 + k —— 1 + i if k = i nv

The first of 30 payments of an annuity occurs in exactly one year and is equal to $500. The payments increase so that each payment is 5% greater than the preceding payment. Find the present value of this annuity with an annual effective rate of interest of 8%. 30 1.05 —— 1.08 1 – ————— = 0.08 – 0.05 $9508.28 500

Consider a perpetuity with a payments that form a geometric progression where 0 < 1 + k < 1 + i. The present value for such a perpetuity with the first payment at the end of the first period is The present value for such a perpetuity with the first payment at the beginning of the first period is 1 —— i – k 1 + i —— i – k Observe that the value for these perpetuities cannot exist if 1 + k  1 + i.