13.1: The Value of an Annuity

Would you prefer a million dollars today or $50,000 a year for the rest of your life? On the one hand, instant gratification is nice. On the other hand, the total dollars received at $50K per year is much larger if you live long enough.

Formally, this is a question about the value of an annuity. An annuity is a financial instrument that pays out a fixed amount of money at the beginning of every year for some specified number of years. In particular, an \(n\)-year, \(m\)-payment annuity pays \(m\) dollars at the start of each year for \(n\) years. In some cases, \(n\) is finite, but not always. Examples include lottery payouts, student loans, and home mortgages. There are even firms on Wall Street that specialize in trading annuities.¹

A key question is, “What is an annuity worth?” For example, lotteries often pay out jackpots over many years. Intuitively, $50,000 a year for 20 years ought to be worth less than a million dollars right now. If you had all the cash right away, you could invest it and begin collecting interest. But what if the choice were between $50,000 a year for 20 years and a half million dollars today? Suddenly, it’s not clear which option is better.

The Future Value of Money

In order to answer such questions, we need to know what a dollar paid out in the future is worth today. To model this, let’s assume that money can be invested at a fixed annual interest rate \(p\). We’ll assume an 8% rate² for the rest of the discussion, so \(p = 0.08\).

Here is why the interest rate \(p\) matters. Ten dollars invested today at interest rate \(p\) will become \((1 + p) \cdot 10 = 10.80\) dollars in a year, \((1 + p)^2 \cdot 10 \approx 11.66\) dollars in two years, and so forth. Looked at another way, ten dollars paid out a year from now is only really worth \(1/(1 + p) \cdot 10 \approx 9.26\) dollars today, because if we had the $9.26 today, we could invest it and would have $10.00 in a year anyway. Therefore, \(p\) determines the value of money paid out in the future.

So for an \(n\)-year, \(m\)-payment annuity, the first payment of \(m\) dollars is truly worth \(m\) dollars. But the second payment a year later is worth only \(m/(1 + p)\) dollars. Similarly, the third payment is worth \(m/(1 + p)^2\), and the \(n\)-th payment is worth only \(m/(1 + p)^{n-1}\). The total value, \(V\), of the annuity is equal to the sum of the payment values. This gives:

\[\begin{align} \nonumber V &= \sum_{i = 1}^{n} \dfrac{m}{(1+p)^{i-1}} \\ \nonumber &= m \cdot \sum_{j = 0}^{n-1} (\dfrac{1}{1+p})^j & (\text{substitute } j = i - 1) \\ &= \label{13.1.1.} m \cdot \sum_{j = 0}^{n-1} x^j & (\text{substitute } x = 1/(1+p)). \end{align}\]

The goal of the preceding substitutions was to get the summation into the form of a simple geometric sum. This leads us to an explanation of a way you could have discovered the closed form (13.2) in the first place using the Perturbation Method.

The Perturbation Method

Given a sum that has a nice structure, it is often useful to “perturb” the sum so that we can somehow combine the sum with the perturbation to get something much simpler. For example, suppose

\[\nonumber S = 1 + x + x^2 + \cdots + x^n.\]

An example of a perturbation would be

\[\nonumber xS = x + x^2 + \cdots + x^{n+1}.\]

The difference between \(S\) and \(xS\) is not so great, and so if we were to subtract \(xS\) from \(S\), there would be massive cancellation:

\[\begin{aligned} S &= 1 + x + x^2 + x^3 + \cdots + x^n \\ -xS &= - x - x^2 - x^3 - \cdots - x^n - x^{n+1}. \end{aligned}\]

The result of the subtraction is

\[\nonumber S - xS = 1 - x^{n+1}.\]

Solving for \(S\) gives the desired closed-form expression in equation 13.2, namely,

\[\nonumber S = \dfrac{1 - x^{n+1}}{1 - x}.\]

We’ll see more examples of this method when we introduce generating functions in Chapter 15.

13.1.3 A Closed Form for the Annuity Value

Using equation 13.2, we can derive a simple formula for \(V\), the value of an annuity that pays \(m\) dollars at the start of each year for \(n\) years.

\[\begin{align} \label{13.1.2.} V &= m(\dfrac{1 - x^n}{1 - x}) & (\text{by equations } \ref{13.1.1.} \text{ and } 13.2) \\ \label{13.1.3.} &= m(\dfrac{1 + p - (1/(1+p))^{n-1}}{p}) & (\text{substituting } x = 1/(1+p)). \end{align}\]

Equation \(\ref{13.1.3.}\) is much easier to use than a summation with dozens of terms. For example, what is the real value of a winning lottery ticket that pays $50,000 per year for 20 years? Plugging in \(m = $50,000, n = 20,\) and \(p = 0.08\) gives \(V \approx $530,180\). So because payments are deferred, the million dollar lottery is really only worth about a half million dollars! This is a good trick for the lottery advertisers.

Infinite Geometric Series

We began this chapter by asking whether you would prefer a million dollars today or $50,000 a year for the rest of your life. Of course, this depends on how long you live, so optimistically assume that the second option is to receive $50,000 a year forever. This sounds like infinite money! But we can compute the value of an annuity with an infinite number of payments by taking the limit of our geometric sum in equation 13.2 as \(n\) tends to infinity

Theorem \(\PageIndex{1}\)

If \(|x| < 1\), then

\[\nonumber \sum_{i = 0}^{\infty} x^i = \dfrac{1}{1-x}.\]

Proof

\[\begin{aligned} \sum_{i = 0}^{\infty} x^i &::= \lim_{n \rightarrow \infty} x^i \\ &= \lim_{n \rightarrow \infty} \dfrac{1-x^{n+1}}{1-x} & (\text{by equation 13.2}) \\ &= \dfrac{1}{1-x}. \end{aligned}\]

The final line follows from the fact that \(\lim_{n \rightarrow \infty} x^{n+1} = 0\) when \(|x| < 1. \quad \blacksquare\)

In our annuity problem, \(x = 1/(1 + p) < 1\), so Theorem 13.1.1 applies, and we get

\[\begin{aligned} V &= m \cdot \sum_{j = 0}^{\infty} x^j & (\text{by equation 13.1.1.}) \\ &= m \cdot \dfrac{1}{1-x} & (\text{by Theorem 13.1.1.}) \\ &= m \cdot \dfrac{1+p}{p} & (x = 1/(1+p)). \end{aligned}\]

Plugging in \(m = $50,000\) and \(p = 0:08\), we see that the value \(V\) is only \($675,000\). It seems amazing that a million dollars today is worth much more than $50,000 paid every year for eternity! But on closer inspection, if we had a million dollars today in the bank earning 8% interest, we could take out and spend $80,000 a year, forever. So as it turns out, this answer really isn’t so amazing after all.

Examples

Equation 13.2 and Theorem 13.1.1 are incredibly useful in computer science.

Here are some other common sums that can be put into closed form using equation 13.2 and Theorem 13.1.1:

\[\begin{align} \label{13.1.4} 1 + 1/2 + 1/4 + \cdots = \sum_{i = 0}^{\infty}\left(\dfrac{1}{2}\right)^i = \dfrac{1}{1 - (1/2)} = 2 \\ \label{13.1.5} 0.99999 \cdots = 0.9\sum_{i = 0}^{\infty}\left(\dfrac{1}{10}\right)^i = 0.9\left(\dfrac{1}{1 - 1/10}\right) = 0.9\left(\dfrac{10}{9}\right) = 1 \\ \label{13.1.6} 1 - 1/2 + 1/4 - \cdots = \sum_{i = 0}^{\infty}\left(\dfrac{-1}{2}\right)^i = \dfrac{1}{1 - (-1/2)} = \dfrac{2}{3} \\ \label{13.1.7} 1 + 2 + 4 + \cdots + 2^{n-1} = \sum_{i = 0}^{n-1} 2^i = \dfrac{1-2^n}{1-2} = 2^n - 1 \\ \label{13.1.8} 1 + 3 + 9 + \cdots + 3^{n-1} = \sum_{i = 0}^{n-1} 3^i = \dfrac{1-3^n}{1-3} = \dfrac{3^n - 1}{2} \end{align}\]

If the terms in a geometric sum grow smaller, as in equation \(\ref{13.1.4}\), then the sum is said to be geometrically decreasing. If the terms in a geometric sum grow progressively larger, as in equations \(\ref{13.1.7}\) and \(\ref{13.1.8}\), then the sum is said to be geometrically increasing. In either case, the sum is usually approximately equal to the term in the sum with the greatest absolute value. For example, in equations \(\ref{13.1.4}\) and \(\ref{13.1.6}\), the largest term is equal to 1 and the sums are 2 and 2/3, both relatively close to 1. In equation \(\ref{13.1.7}\), the sum is about twice the largest term. In equation \(\ref{13.1.8}\), the largest term is \(3^{n-1}\) and the sum is \((3^{n} - 1) / 2\), which is only about a factor of 1:5 greater. You can see why this rule of thumb works by looking carefully at equation 13.2 and Theorem 13.1.1.

Variations of Geometric Sums

We now know all about geometric sums—if you have one, life is easy. But in practice one often encounters sums that cannot be transformed by simple variable substitutions to the form \(\sum x^i\).

A non-obvious but useful P way to obtain new summation formulas from old ones is by differentiating or integrating with respect to \(x\). As an example, consider the following sum:

\[\nonumber \sum_{i = 1}{n - 1} i x^i = x + 2x^2 + 3x^3 + \cdots + (n-1)x^{n-1}\]

This is not a geometric sum. The ratio between successive terms is not fixed, and so our formula for the sum of a geometric sum cannot be directly applied. But differentiating equation 13.2 leads to:

\[\label{13.1.9} \dfrac{d}{dx}\left(\sum_{i = 0}{n-1} x^i \right) = \dfrac{d}{dx}\left( \dfrac{1-x^n}{1-x} \right).\]

The left-hand side of equation \(\ref{13.1.9}\) is simply

\[\nonumber \sum_{i = 0}^{n-1} \dfrac{d}{dx}(x^i) = \sum_{i = 0}^{n-1} i(x^{i-1})\]

The right-hand side of equation \(\ref{13.1.9}\) is

\[\begin{aligned} \dfrac{-nx^{n-1}(1-x)-(-1)(1-x^n)}{(1-x)^2} &= \dfrac{-nx^{n-1} + nx^n + 1-x^n)}{(1-x)^2} \\ &= \dfrac{1-nx^{n-1} + (n-1)x^n}{(1-x)^2}. \end{aligned}\]

Hence, equation \(\ref{13.1.9}\) means that

\[\nonumber \sum_{i=0}^{n-1}ix^{i-1} = \dfrac{1-nx^{n-1} + (n-1)x^n}{(1-x)^2}.\]

Incidentally, Problem 13.2 shows how the perturbation method could also be applied to derive this formula.

Often, differentiating or integrating messes up the exponent of \(x\) in every term. In this case, we now have a formula for a sum of the form \(\sum ix^{i-1}\), but we want a formula for the series\(\sum ix^{i-1}\). The solution is simple: multiply by \(x\). This gives:

\[\label{13.1.10} \sum_{i=0}^{n-1}ix^{i} = \dfrac{x - nx^n + (n-1)x^{n+1}}{(1-x)^2}\]

and we have the desired closed-form expression for our sum. It seems a little complicated, but it’s easier to work with than the sum.

Notice that if \(|x| < 1\), then this series converges to a finite value even if there are infinitely many terms. Taking the limit of equation \(\ref{13.1.10}\) as \(n\) tends to infinity gives the following theorem:

Theorem \(\PageIndex{2}\)

If \(|x| < 1\), then

\[\sum_{i = 0}^{\infty} ix^i = \dfrac{x}{(1-x)^2}.\]

As a consequence, suppose that there is an annuity that pays \(im\) dollars at the end of each year \(i\), forever. For example, if \(m = $50,000\), then the payouts are $50,000 and then $100,000 and then $150,000 and so on. It is hard to believe that the value of this annuity is finite! But we can use Theorem 13.1.2 to compute the value:

\[\begin{aligned} V &= \sum_{i=1}^{\infty}\dfrac{im}{(1+p)^i} \\ &= m \cdot \dfrac{1/(1+p)}{(1-\frac{1}{1+p})^2} \\ &= m \cdot \dfrac{1+p}{p^2}.\end{aligned}\]

The second line follows by an application of Theorem 13.1.2. The third line is obtained by multiplying the numerator and denominator by \((1+p)^2\).

For example, if \(m = $50,000\), and \(p = 0.08\) as usual, then the value of the annuity is \(V = $8,437,500\). Even though the payments increase every year, the increase is only additive with time; by contrast, dollars paid out in the future decrease in value exponentially with time. The geometric decrease swamps out the additive increase. Payments in the distant future are almost worthless, so the value of the annuity is finite.

The important thing to remember is the trick of taking the derivative (or integral) of a summation formula. Of course, this technique requires one to compute nasty derivatives correctly, but this is at least theoretically possible!

¹Such trading ultimately led to the subprime mortgage disaster in 2008–2009. We’ll talk more about that in a later chapter.

²U.S. interest rates have dropped steadily for several years, and ordinary bank deposits now earn around 1.0%. But just a few years ago the rate was 8%; this rate makes some of our examples a little more dramatic. The rate has been as high as 17% in the past thirty years.