The Level Annuity-Immediate

A annuity-immediate pays a level amount — normalized to 1 — at the end of each of n periods. "Immediate" is a historical label: it does not mean payments start now, only that each one lands at the close of its period.

By the end you'll be able to value that stream two ways — as a present value \(a_n=(1-v^n)/i\) and as an accumulated value \(s_n=((1+i)^n-1)/i\) — and predict how each one moves when the interest rate changes.

Predict: if you raise i, does the present value \(a_n\) go up or down? Drag i and check — and note what happens to \(s_n\) at the same time.

The timeline below shows n payments of 1, one at the end of every period. Each bar is one payment's value seen from a single vantage point: in the present value view, bar k is the discount factor \(v^k\) — how much that payment is worth today; in the accumulated value view, bar k is \((1+i)^{n-k}\) — how much it grows to by time n. Stacking (summing) every bar gives the dashed reference line: \(a_n\) or \(s_n\). Hover or focus a bar or the payment dots for exact readouts.

aₙ = 7.7217    sₙ = 12.5779

Payment timeline — n payments of 1, discounted (aₙ) or accumulated (sₙ)
payment of 1, at the end of each period
Intuitive

Line up n payments of 1, one at the end of each period. Each one is worth less today the farther out it sits, because money now can earn interest that money later can't. Add up all n of those "worth today" amounts and you get aₙ, the annuity's PV. Roll the same n payments forward to time n instead, letting each one earn interest until then, and you get sₙ, the annuity's AV.

Formal

Payments land at times \(1, 2, \ldots, n\), valued with the per-period effective rate i and its discount factor \(v = 1/(1+i)\). \(a_n\) is a finite geometric series of these discount factors: \(a_n = v + v^2 + \cdots + v^n\), first term v, ratio v, which sums to the standard result \(a_n = (1-v^n)/i\). Rolling that same stream forward n periods — multiplying by \((1+i)^n\) — gives \(s_n = (1+i)^n \cdot a_n = ((1+i)^n-1)/i\). They differ by exactly \((1+i)^n\): the same n payments, valued at two different points on the timeline.

Applied

An insurer pricing a structured settlement that pays a level amount at the end of every year for n years needs to know its cost today: that's \(a_n\) times the payment size. A pension fund tracking what those same payments will have grown to by the year the last one is made — say, to check it against a target balance — uses \(s_n\) instead. Same cash flows, different question, different formula.

Worked example

Present value of an annuity-immediate paying 500 at the end of each year for 8 years, \(i = 5\%\). \(v = 1/1.05\), \(v^8 = 0.676839\). \(a_8 = (1-0.676839)/0.05 = 6.46321\). \(\text{PV} = 500 \times 6.46321 = \) 3231.61.

Your turn

A different stream: 300 at the end of each year for 6 years, \(i = 7\%\). \(v = 1/1.07\), \(v^6 = 0.666342\). \(a_6 = (1-0.666342)/0.07 = 4.766543\). \(\text{PV} = 300 \times 4.766543 = \) ____

Reveal the answer

\(\text{PV} = 300 \times 4.766543 = \) 1429.96. Set n = 6 and i = 7% on the sliders above (present-value view) and check the dashed line against this number.

More info — what happens right at i = 0?

The slider above stops at i = 1% to keep the chart well-behaved, but it's worth knowing what \(a_n = (1-v^n)/i\) does as \(i \to 0\): both the top and the bottom of the fraction go to 0 at the same time, so it's a 0/0 form, not an automatic blow-up or collapse. Reasoning about it directly is easier than taking the limit: with no discounting at all, every one of the n payments keeps its full face value of 1, so \(a_n \to n\). The same logic applies to \(s_n\). See the Corporate Finance Institute link in Dive deeper below for more on the ordinary-annuity convention this builds on.

Check your understanding

Question 1 of 4

A 10-payment annuity-immediate pays 1 at the end of each year, \(i = 5\%\). What is \(a_{10}\)?

Question 2 of 4

Holding \(n\) fixed, what does \(a_n = (1-v^n)/i\) approach as \(i \to 0\), and why?

Question 3 of 4

\(a_n = v + v^2 + \cdots + v^n\), where \(v = 1/(1+i)\) is the one-period discount factor. If \(i\) rises, what happens to \(v\), and therefore to each term \(v^k\) and the sum \(a_n\)?

Question 4 of 4

Contrast: while \(a_n\) falls as \(i\) rises (n fixed), what happens to \(s_n = ((1+i)^n-1)/i\), and why?

Recap

  • A level annuity-immediate pays 1 at the end of each of n periods, at times \(1, 2, \ldots, n\).
  • Present value: \(a_n = v + v^2 + \cdots + v^n = (1-v^n)/i\), where \(v = 1/(1+i)\).
  • Accumulated value: \(s_n = (1+i)^n \cdot a_n = ((1+i)^n-1)/i\) — the same stream valued at time n instead of time 0.
  • As i rises (n fixed), \(a_n\) falls (every discount factor shrinks) while \(s_n\) rises (the accumulation factor outgrows the extra i in the denominator).
  • As \(i \to 0\), both \(a_n\) and \(s_n\) approach n — no discounting, no growth.

Dive deeper

Sources

  • The Level Annuity-Immediate