## Tuesday, August 20, 2013

### Financial Mathematics: Geometric Series

For a refresher on sequences and series, see here.

A geometric sequence is a sequence in which the following term is a multiple of the previous term. For example:
$$(2, 6, 18, 54,162, . . .)$$
is an infinite geometric sequence. The following term is three times the previous. Obviously both this sequence (and the corresponding series) diverge. More generally, a geometric sequence is any sequence with variables $a$ and $r$ such that:

$$(a, ar, ar^2, ar^3, ar^4, . . .)$$
The series is simply the summation of the terms in the sequence.

And they don't have to be infinite, they may terminate after a finite number of terms. We'll start with the finite series since it's a little easier to work with.

Finite Geometric Series

Suppose we have a geometric series with $n$ terms:
$$S = a + ar + ar^2 + ar^3 + . . . + ar^{n-1} = \sum_{j=0}^{n-1} ar^j$$
I'm going to multiply both sides by $r$ to obtain:
$$Sr = ar + ar^2 + ar^3 + ar^4 + . . . + ar^{n-1}+ar^n$$
Next I'm going to subtract $S$ from $Sr$. What ends up happening is that all of the like terms will "cancel out" so what's left are the terms which differ between them. In other words, the two $ar$'s cancel out and the two $ar^2$'s cancel out and so on.What this ends up looking like is this:
$$Sr - S = ar^n - a$$
Factoring out an $S$ I get:
$$(r-1)S = ar^n - a$$
And then we can divide both sides by $(r-1)$ to . . . wait, this is only valid provided that $r-1\neq 0$ since we can't divide by 0. Therefore we'll assume that $r \neq 1$ and then divide both sides by $(r-1)$ to obtain:
$$S=\frac{ar^n - a}{r-1} =\frac{a-ar^n }{1-r}=a\frac{1-r^n}{1-r}$$
You can write it either way as I did above there. Sometimes it will be more convenient to use one over the other.

If you're really dying to know what happens if $r=1$ (I know you are!), the series looks like this:
$$S = a + a(1) + a(1)^2 + a(1)^3 + . . . + a(1)^{n-1} = an$$
(Exciting! Yes? Maybe? OK, probably not. But paying attention to assumptions matters.)

Infinite Geometric Series

We've actually already done most of the hard work (well, technically no... the hard work would require that I teach you some real analysis and proof techniques which is beyond the scope of this financial mathematics series. So I'm going to do some handwaving over the hard parts.)

Now the geometric series sometimes converges and sometimes diverges. It's important to know when. First we already saw what the series looks like when $r=1$ (for the finite case at least):
$$S = a + a + a + a + . . .$$
If you haven't guessed already, adding up an infinite number of $a$'s gets you to infinity. The same will apply if $r>1$.

Likewise, if we suppose that $r=-1$, you get something similar:
$$S = a - a - a -a -a -a - . . .$$
The result ends up being somewhere in the vicinity of negative infinity. So the only time the series converges is when $-1 < r < 1$ or $|r| < 1$.

So since we're interested in the series when it converges, we want to know what value it converges to. So we'll make the assumption that $|r| < 1$.

So let's start with the result we obtained in the finite case and take the limit as $n \to \infty$:

$$\lim_{n \to \infty} S = \lim_{n \to \infty} \frac{a-ar^n }{1-r} = \frac{a}{1-r}$$
Now there's some handwaving in the above result as it requires knowledge of limits (calculus). We didn't technically prove that it converges but it does. If you're really interested in that, feel free to pick up a calculus textbook.

Zeno proposed a number of paradoxes of motion. He proposed these paradoxes in order to prove that there is no motion. I frankly never found his arguments terribly moving (yes, lame joke... I know) but they are at least an interesting historical artifact. One of his arguments runs like this:

In order for me to move from point A to point B, I have to move half the distance to some point C. Then to move from points C to B, I have to move half the distance to some point D. Then to move from D to B I have to move half the distance to point E and so on (we could be here a while.)

The point here is that you'll never reach point B and therefore motion isn't really happening. . . or so Zeno claims. But it turns out that this is a geometric series:
$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . = \sum_{j=0}^\infty \frac{1}{2}\left(\frac{1}{2}\right)^j$$
In our expression, $a = \frac{1}{2}$ and $r=\frac{1}{2}$. Since $r < 1$ it converges to:
$$S = \frac{a}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$$

So that's a geometric series.