Decreasing and General Arithmetic Annuities

A decreasing annuity-immediate pays n at the end of period 1, n−1 at period 2, …, down to 1 at period n — the mirror image of the increasing annuity. Both are special cases of one general arithmetic pattern, P, P+Q, …, P+(n−1)Q, that lets a single toolkit price payments that shrink, grow, or hold level.

By the end you'll be able to value a decreasing annuity with \((Da)_n = (n-a_n)/i\), and split any arithmetic stream into a level piece worth \(P \cdot a_n\) plus a step piece that reuses the increasing annuity's own formula.

Predict: set Q negative so payments decrease — will the present value end up higher or lower than a level annuity of the first payment (\(P \cdot a_n\))? Drag Q below zero and check.

Each bar is one payment, \(P + (k-1)Q\), for periods 1 through n = 10. It splits into a level layer of P (the dashed line) and a step layer of size \((k-1)Q\) — solid on top when Q > 0, or a dashed outline showing the shortfall when Q < 0. Hover or focus a bar for its exact payment.

Set P = n and Q = −1 (or press the snap button) to see the pure decreasing case collapse to \((Da)_n = (n-a_n)/i\).

Level PV = P·a₁₀ = 58.8807   Step PV = 0.0000    Total PV = 58.8807

Payments 1–10 — level layer P plus step layer \((k-1)Q\)
level layer, P step layer, (k−1)Q

A decreasing annuity-immediate runs the increasing annuity in reverse — n, n−1, …, 1 instead of 1, 2, …, n — and both sit inside the same general arithmetic pattern P, P+Q, …, P+(n−1)Q, so one decomposition prices payments that shrink, grow, or hold level.

Formal

Pair the decreasing stream with the increasing one: payment k of the decreasing stream is n+1−k, and payment k of the increasing stream is k, so together every time point sums to the constant n+1 — giving \((Ia)_n + (Da)_n = (n+1) \cdot a_n\) and the closed form \((Da)_n = (n-a_n)/i\). More generally, a stream P, P+Q, …, P+(n−1)Q splits into a level layer of P (worth \(P \cdot a_n\)) and a step layer that amortizes by Q each period — worth Q times the excess of the increasing annuity over the level one, and built from the exact same \((\ddot{a}_n - n \cdot v^n)/i\) that prices \((Ia)_n\). Setting \(P=n, Q=-1\) collapses this general split back down to the compact \((Da)_n = (n-a_n)/i\) you see highlighted above.

Applied

A company retiring a bond issue with sinking-fund payments that shrink by a fixed amount each year, or a trust paying a stipend that steps down as a beneficiary's other income grows, both value the same way: split the stream into its starting level and its constant step, then price each piece with tools you already have — \(a_n\) for the level piece, \((Ia)_n\)'s machinery for the step.

Worked example

Payments 10, 9, …, 1 at the end of years 1–10, \(i = 6\%\). \(a_{10} = (1-1.06^{-10})/0.06 = 7.360087\). \((Da)_{10} = (10 - 7.360087)/0.06 = \) 43.9986.

Your turn

Same shape, different size: payments 6, 5, …, 1 for 6 years, \(i = 4\%\). \(a_6 = (1-1.04^{-6})/0.04 = 5.242137\). \((Da)_6 = (6 - 5.242137)/0.04 = \) ____

Reveal the answer

\((Da)_6 = 0.757863/0.04 = \) 18.9466. Set the snap button above with n adjusted mentally to 6 payments, or just check the shape: each bar should sit exactly one step below the last, bottoming out at 1.

More info — why the general formula isn't just "P times level, Q times increasing"

It's tempting to guess the step layer is worth \(Q \cdot (Ia)_n\) outright, but that overcounts: \((Ia)_n\) already prices payments 1, 2, …, n, while the step layer here pays 0, Q, 2Q, …, (n−1)Q — one period behind. The correct step value is Q times the increasing annuity's excess over the level one, \(Q \cdot [(Ia)_n - a_n]\). Check it against the n=1 case: with a single payment, there's no room for a step at all, and only this corrected version gives back the trivial answer \(P \cdot a_1\). See the Marcel Finan guide in Dive deeper for the full derivation.

Away from the pure P=n, Q=−1 case, the split still works for any level-plus-step mix: \(P=5000, Q=-500, n=10, i=5\%\) gives level PV \(= 5000 \cdot a_{10} = 38{,}608.67\), step PV \(= -500 \cdot [(Ia)_{10}-a_{10}] = -15{,}826.02\), and total PV \(= 22{,}782.65\).

Check your understanding

Question 1 of 4

PV of payments 12, 11, …, 1 at the end of years 1–12, at \(i = 7\%\). Using \((Da)_{12} = (12 - a_{12})/i\), what is the present value?

Question 2 of 4

In the level-plus-step decomposition of a general arithmetic annuity, what happens to the present value if you set the step size \(Q = 0\)?

Question 3 of 4

Which other named annuity's whole present-value formula, \((\ddot{a}_n - n \cdot v^n)/i\), gets reused inside the step layer when valuing a general arithmetic annuity?

Question 4 of 4

A stream paying 8, 7, 6, …, 1 at the end of 8 years is the general arithmetic annuity \(P, P+Q, \ldots, P+(n-1)Q\) (with \(n=8\)) for which P and Q?

Recap

  • A decreasing annuity-immediate pays n, n−1, …, 1; its present value is \((Da)_n = (n-a_n)/i\).
  • \((Ia)_n + (Da)_n = (n+1) \cdot a_n\) — the increasing and decreasing streams sum to the constant n+1 at every payment date.
  • A general arithmetic annuity P, P+Q, …, P+(n−1)Q splits into a level layer worth \(P \cdot a_n\) plus a step layer worth \(Q \cdot [(Ia)_n - a_n]\).
  • Q < 0 shrinks payments (decreasing), Q > 0 grows them (increasing), and Q = 0 collapses to the plain level annuity.

Dive deeper

Sources

  • Decreasing and General Arithmetic Annuities