Financial Mathematics Text

Tuesday, October 8, 2013

Financial Mathematics: Perpetuities

So we're continuing our look at annuities. A perpetuity is an annuity that continues to make payments indefinitely or perpetually, hence the name perpetuity.

For simplicity, we shall make our usual assumptions:
  1. All payments are made in equal amounts.
  2. The payments are made at equal intervals (specified as a fraction of a year).
  3. There is no possibility to alter either the times or the amounts of the payments (e.g. no prepayments).
Scenario 1: Annual Payments

We'll begin by assuming that the payments are made once per year. We'll assume that each payment is of the amount $C$ and is discounted at the interest rate $i$.

The present value of a stream of payments that continue forever is given by:
PV &= \frac{C}{(1+i)} + \frac{C}{(1+i)^2} + \frac{C}{(1+i)^3} + . . .  \\
&= \sum_{j=0}^{\infty} \frac{C}{1+i} \frac{1}{(1+i)^j}
And we, once again, have our geometric series. Our $a=\frac{C}{1+i}$ and our $r=\frac{1}{1+i}$. But there's a catch here.

What we're actually doing is adding up an infinite number of items. What guarantees that this doesn't just add up to infinity?

Recall, that in order to guarantee that this series converges, $|r|<1$. That basically means that $i>0$. That's not going to usually be a problem here since we normally deal with positive interest rates.

Observing all of that we can now show that the series converges to:
PV &=  \frac{C}{1+i}\frac{1}{1-(1+i)^{-1}} \\
&= \frac{C}{i}
Now there was actually another way to derive this. Recall that for the finite series, the present value is given by:
$$PV = C\frac{1-(1+i)^{-t}}{i}$$
Now we take the limit as $t \to \infty$. What happens is that as $t$ gets very large, $(1+i)^{-t}$ gets very small, converging to 0. That ends up producing the same result we had above.

Scenario 2: Periodic Payments

So now we'll assume that there are $n$ payments per year in which each payment is $C/n$. The setup looks about like this:
PV &= \frac{C/n}{(1+\frac{i}{n})} + \frac{C/n}{(1+\frac{i}{n})^2} + \frac{C/n}{(1+\frac{i}{n})^3} + . . .  \\
&= \sum_{j=0}^{\infty} \frac{C}{n(1+\frac{i}{n})} \frac{1}{(1+\frac{i}{n})^j}
We have our friend the geometric series again where $a=\frac{C}{n(1+\frac{i}{n})}$ and $r=(1+\frac{i}{n})^{-j}$. And just as we did before, we need to assume that $i>0$ to make sure that this series converges. The result then is this:
PV &=  \frac{C}{n(1+\frac{i}{n})}\frac{1}{1-(1+\frac{i}{n})^{-1}} \\
&= \frac{C}{i}
So the periodic payments for the perpetuity is identical to the annual payments. The value is always the same.

And just like we did above, we can actually derive it using the result from the finite series:
PV &= \lim_{t \to \infty} C \frac{1-(1+\frac{i}{n})^{-t \times n}}{i} \\
&= \frac{C}{i}


Suppose that there is a perpetuity that pays $ \$10 $ per year and is currently selling for $ \$125 $. What's the effective interest rate on this perpetuity?

We can rearrange the equation as follows:
$$ i = \frac{C}{PV}$$
We then substitute $ C=\$10 $ and $ PV = \$125 $:
i &= \frac{ \$10}{ \$125} \\
&= 8 \%

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