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Deriving The Perpetuity Formula

By fizzbuzzer on August 26, 2018

Perpetuity refers to an infinite amount of time (\lim_{n\to\infty}). In finance, perpetuity is a constant stream of identical cash flows, (C), with no end.

The present value (PV) of a security with perpetual cash flows can be determined as:

PV = \frac{C}{(1 + d)} + \frac{C}{(1 + d)^2} + \frac{C}{(1 + d)^3} + \dots + \frac{C}{(1 + d)^n} = \frac{C}{d}

with d being the discount rate or cost of capital. Present value just states:

How much money would you need to deposit into an interest earning account (with rate d) or investment today, in order to get C amount of money in n years.

In other words, present value is the result of interest being deducted or discounted from a future amount (compounding in reverse).

So back to our original formula. Why can we rewrite it as follows?

PV = \frac{C}{d}

If we look at the original formula we can see that it is a geometric series:

s = a + ar + ar^2 + ar^3 + \dots

with a=\frac{C}{(1 + d)} and r=\frac{1}{(1 + d)}.

Since for n\to\infty:

a + ar + ar^2 + ar^3 + \dots + ar^n = \frac{a}{(1 - r)}

we can easily see that:

PV = \frac{C}{(1 + d)} + \frac{C}{(1 + d)^2} + \frac{C}{(1 + d)^3} + \dots + \frac{C}{(1 + d)^n}=\frac{\frac{C}{(1 + d)}}{(1 - \frac{1}{(1 + d)})} = \frac{C}{(1+d)(1 - \frac{1}{(1+d)})} = \frac{D}{d}

Proof that a + ar + ar^2 + ar^3 + \dots + ar^n = \frac{a}{(1 - r)} for n\to\infty

Let

s = a + ar + ar^2 + ar^3 + \dots + ar^{n-1} = \sum\limits_{i=0}^{n-1}ar^i

Multiplying s with r we get:

rs = ar + ar^2 + ar^3 + \dots + r^n

Then:

s - rs = a - ar^n

Solving this for s we get:

s= a\frac{(1 - r^n)}{(1 - r)}

Using this we can \lim_{n\to\infty} (a + ar + ar^2 + ar^3 + \dots + ar^n):

\lim_{n\to\infty} (a + ar + ar^2 + ar^3 + \dots + ar^n) = \lim_{n\to\infty}a\frac{1 - r^{n+1}}{1 - r}

Above we used r^{n+1} simply because our formula \sum\limits_{i=0}^{n-1}ar^i=a\frac{(1 - r^n)}{(1 - r)} is for i=0\dots (n-1).

For r \textless 1, which is our case because r=\frac{1}{(1 + d)} we get:

\lim_{n\to\infty}a\frac{(1 - r^{n+1})}{(1 - r)} = \frac{a}{(1-r)}

Similarly we can derive the Present Value of Growing Perpetuity where periodic payments grow at a proportionate rate g:

PV = \frac{C}{(1 + d)} + \frac{C(1 + g)}{(1 + d)^2} + \frac{C(1 + g)^2}{(1 + d)^3} + \frac{C(1 + g)^3}{(1 + d)^4} + \dots = \frac{C}{(d-g)}

which can be rewritten as:

PV = \frac{C}{(1 + d)} + \frac{C}{(1 + d)}(\frac{(1 + g)}{(1 + d)})+\frac{C}{(1 + d)}(\frac{(1 + g)}{(1 + d)})^2 + \dots

It is simply a geometric series:

s = a + ar + ar^2 + ar^3 + \dots

with a=\frac{C}{(1 + d)} and r=\frac{(1+g)}{(1 + d)}.

Published in How-To Guides

fizzbuzzer
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